Selina Concise Solutions for ICSE Class 9 Mathematics Chapter 6 Simultaneous Linear Equations Including Problems

Exercise 6(A)

Question 1. Solve:
\( 8x + 5y = 9 \)
\( 3x + 2y = 4 \)
Answer:
We are given the following equations:
\( 8x + 5y = 9 \) ...(1)
\( 3x + 2y = 4 \) ...(2)

From equation (1), we can express \(y\) in terms of \(x\):
\( \implies y = \frac{9 - 8x}{5} \)

Let's plug this value of \(y\) into equation (2):
\( \implies 3x + 2\left(\frac{9 - 8x}{5}\right) = 4 \)
\( \implies 15x + 18 - 16x = 20 \)
\( \implies -x = 2 \)
\( \implies x = -2 \)

Now, substituting the value of \(x\) back into the expression for \(y\):
\( \implies y = \frac{9 - 8(-2)}{5} \)
\( \implies y = \frac{9 + 16}{5} \)
\( \implies y = \frac{25}{5} = 5 \)

Thus, the solution is \(x = -2\) and \(y = 5\).
In simple words: Express one variable in terms of the other from one equation, then substitute it into the second equation to solve for both variables step-by-step.

Exam Tip: Be careful with signs when substituting negative values. Always plug your final answers back into both equations to verify they are correct.

Question 2. Solve:
\( 2x - 3y = 7 \)
\( 5x + y = 9 \)
Answer:
Consider the given system of equations:
\( 2x - 3y = 7 \) ...(1)
\( 5x + y = 9 \) ...(2)

From equation (2), we can express \(y\) in terms of \(x\):
\( \implies y = 9 - 5x \)

Substituting this value of \(y\) into equation (1):
\( \implies 2x - 3(9 - 5x) = 7 \)
\( \implies 2x - 27 + 15x = 7 \)
\( \implies 17x = 34 \)
\( \implies x = 2 \)

Now, substitute \(x = 2\) into the equation for \(y\):
\( \implies y = 9 - 5(2) \)
\( \implies y = 9 - 10 \)
\( \implies y = -1 \)

Hence, the required values are \(x = 2\) and \(y = -1\).
In simple words: Find \(y\) from the second equation since it has a coefficient of 1, then plug it into the first equation to solve for \(x\).

Exam Tip: Substituting a variable with a coefficient of 1 or -1 is always the easiest route because it avoids working with fractions early in the solution.

Question 3. Solve:
\( 2x + 3y = 8 \)
\( 2x = 2 + 3y \)
Answer:
The equations are:
\( 2x + 3y = 8 \) ...(1)
\( 2x = 2 + 3y \) ...(2)

Since equation (2) gives a direct expression for \(2x\), we substitute this into equation (1):
\( \implies (2 + 3y) + 3y = 8 \)
\( \implies 2 + 6y = 8 \)
\( \implies 6y = 6 \)
\( \implies y = 1 \)

Substituting this value of \(y\) back into equation (2):
\( \implies 2x = 2 + 3(1) \)
\( \implies 2x = 5 \)
\( \implies x = 2.5 \)

So, the solution is \(x = 2.5\) and \(y = 1\).
In simple words: Since \(2x\) is already written in terms of \(y\) in the second equation, we can substitute the whole term \(2x\) directly into the first equation.

Exam Tip: Look for identical terms like \(2x\) in both equations. Substituting the entire term directly saves time and reduces calculation errors.

Question 4. Solve:
\( 0.2x + 0.1y = 25 \)
\( 2(x - 2) - 1.6y = 116 \)
Answer:
The given equations are:
\( 0.2x + 0.1y = 25 \) ...(i)
\( 2(x - 2) - 1.6y = 116 \) ...(ii)

From equation (i), we express \(x\) in terms of \(y\):
\( \implies 0.2x = 25 - 0.1y \)
\( \implies x = \frac{25 - 0.1y}{0.2} \) ...(iii)

Substituting this value of \(x\) into equation (ii):
\( \implies 2\left(\frac{25 - 0.1y}{0.2} - 2\right) - 1.6y = 116 \)
\( \implies 10(25 - 0.1y) - 4 - 1.6y = 116 \)
\( \implies 250 - y - 4 - 1.6y = 116 \)
\( \implies 246 - 2.6y = 116 \)
\( \implies -2.6y = 116 - 246 \)
\( \implies -2.6y = -130 \)
\( \implies y = 50 \)

Substitute this value of \(y\) back into equation (iii):
\( \implies x = \frac{25 - 0.1(50)}{0.2} \)
\( \implies x = \frac{25 - 5}{0.2} \)
\( \implies x = \frac{20}{0.2} = 100 \)

Hence, the solution is \(x = 100\) and \(y = 50\).
In simple words: Rearrange the decimal equation to find \(x\), plug it into the other equation, simplify the fractions, and then solve for both variables.

Exam Tip: Simplify terms like \( \frac{2}{0.2} = 10 \) first to get rid of fractions and make the algebra much cleaner.

Question 5. Solve:
\( 6x = 7y + 7 \)
\( 7y - x = 8 \)
Answer:
The system of equations is:
\( 6x = 7y + 7 \) ...(1)
\( 7y - x = 8 \) ...(2)

From equation (2), express \(x\) in terms of \(y\):
\( \implies x = 7y - 8 \)

Substituting this value of \(x\) into equation (1):
\( \implies 6(7y - 8) = 7y + 7 \)
\( \implies 42y - 48 = 7y + 7 \)
\( \implies 42y - 7y = 7 + 48 \)
\( \implies 35y = 55 \)
\( \implies y = \frac{11}{7} \)

Substituting this value of \(y\) back into our expression for \(x\):
\( \implies x = 7\left(\frac{11}{7}\right) - 8 \)
\( \implies x = 11 - 8 = 3 \)

Thus, the solution is \(x = 3\) and \(y = \frac{11}{7}\).
In simple words: Express \(x\) in terms of \(y\) from the second equation since it is simple to isolate, then substitute it into the first equation.

Exam Tip: Fractions in solutions are common. Keep them as improper fractions rather than decimals unless requested, as they often cancel out nicely in other steps.

Question 6. Solve:
\( y = 4x - 7 \)
\( 16x - 5y = 25 \)
Answer:
Given equations:
\( y = 4x - 7 \) ...(1)
\( 16x - 5y = 25 \) ...(2)

Since \(y\) is already expressed in terms of \(x\) in equation (1), substitute it into (2):
\( \implies 16x - 5(4x - 7) = 25 \)
\( \implies 16x - 20x + 35 = 25 \)
\( \implies -4x = 25 - 35 \)
\( \implies -4x = -10 \)
\( \implies x = \frac{5}{2} \)

Substituting this value of \(x\) back into equation (1):
\( \implies y = 4\left(\frac{5}{2}\right) - 7 \)
\( \implies y = 10 - 7 = 3 \)

Therefore, the solution is \(x = \frac{5}{2}\) and \(y = 3\).
In simple words: Substitute the expression for \(y\) from the first equation directly into the second equation, solve for \(x\), and then use it to find \(y\).

Exam Tip: When substituting a fraction like \(x = \frac{5}{2}\), cancel the common factors (e.g., \(4 \times \frac{5}{2} = 2 \times 5 = 10\)) to simplify your arithmetic.

Question 7. Solve:
\( 2x + 7y = 39 \)
\( 3x + 5y = 31 \)
Answer:
Consider the equations:
\( 2x + 7y = 39 \) ...(1)
\( 3x + 5y = 31 \) ...(2)

From equation (1), express \(x\) in terms of \(y\):
\( \implies x = \frac{39 - 7y}{2} \)

Substituting this value of \(x\) into equation (2):
\( \implies 3\left(\frac{39 - 7y}{2}\right) + 5y = 31 \)
\( \implies \frac{117 - 21y}{2} + 5y = 31 \)
\( \implies 117 - 21y + 10y = 62 \)
\( \implies 117 - 11y = 62 \)
\( \implies -11y = 62 - 117 \)
\( \implies -11y = -55 \)
\( \implies y = 5 \)

Substituting \(y = 5\) back into the expression for \(x\):
\( \implies x = \frac{39 - 7(5)}{2} \)
\( \implies x = \frac{39 - 35}{2} \)
\( \implies x = \frac{4}{2} = 2 \)

Hence, the solution is \(x = 2\) and \(y = 5\).
In simple words: Express \(x\) using the first equation, then substitute it into the second. Solve for \(y\), and finally calculate \(x\).

Exam Tip: Be sure to multiply the term not containing the fraction (in this case, \(5y\)) by the denominator when clearing fractions to maintain equality.

Question 8. Solve:
\( 1.5x + 0.1y = 6.2 \)
\( 3x - 0.4y = 11.2 \)
Answer:
The system of equations is:
\( 1.5x + 0.1y = 6.2 \) ...(i)
\( 3x - 0.4y = 11.2 \) ...(ii)

From equation (i), we can isolate \(x\):
\( \implies 1.5x = 6.2 - 0.1y \)
\( \implies x = \frac{6.2 - 0.1y}{1.5} \) ...(iii)

Substituting this expression for \(x\) into equation (ii):
\( \implies 3\left(\frac{6.2 - 0.1y}{1.5}\right) - 0.4y = 11.2 \)
\( \implies 2(6.2 - 0.1y) - 0.4y = 11.2 \)
\( \implies 12.4 - 0.2y - 0.4y = 11.2 \)
\( \implies 12.4 - 0.6y = 11.2 \)
\( \implies -0.6y = 11.2 - 12.4 \)
\( \implies -0.6y = -1.2 \)
\( \implies y = 2 \) ...(iv)

Substituting \(y = 2\) back into our equation (iii):
\( \implies x = \frac{6.2 - 0.1(2)}{1.5} \)
\( \implies x = \frac{6.2 - 0.2}{1.5} \)
\( \implies x = \frac{6}{1.5} = 4 \)

So, the solution is \(x = 4\) and \(y = 2\).
In simple words: Express \(x\) in terms of \(y\) from the first decimal equation. Substitute this into the second equation and solve.

Exam Tip: Notice that \( \frac{3}{1.5} = 2 \). Simplifying fractions with decimals early prevents complex calculations later.

Question 9. Solve:
\( 2(x - 3) + 3(y - 5) = 0 \)
\( 5(x - 1) + 4(y - 4) = 0 \)
Answer:
The equations are given as:
\( 2(x - 3) + 3(y - 5) = 0 \) ...(1)
\( 5(x - 1) + 4(y - 4) = 0 \) ...(2)

First, let's simplify equation (1):
\( \implies 2x - 6 + 3y - 15 = 0 \)
\( \implies 2x + 3y - 21 = 0 \)
\( \implies 2x = 21 - 3y \)
\( \implies x = \frac{21 - 3y}{2} \)

Next, simplify equation (2):
\( \implies 5x - 5 + 4y - 16 = 0 \)
\( \implies 5x + 4y - 21 = 0 \) ...(3)

Substituting our simplified expression for \(x\) into equation (3):
\( \implies 5\left(\frac{21 - 3y}{2}\right) + 4y - 21 = 0 \)
\( \implies \frac{105 - 15y}{2} + 4y - 21 = 0 \)
\( \implies 105 - 15y + 8y - 42 = 0 \)
\( \implies -7y + 63 = 0 \)
\( \implies 7y = 63 \)
\( \implies y = 9 \)

Now, substitute \(y = 9\) back into the expression for \(x\):
\( \implies x = \frac{21 - 3(9)}{2} \)
\( \implies x = \frac{21 - 27}{2} \)
\( \implies x = \frac{-6}{2} = -3 \)

Thus, the solution is \(x = -3\) and \(y = 9\).
In simple words: Expand and simplify both equations first. Then, express \(x\) in terms of \(y\) and substitute it to solve.

Exam Tip: Expanding the brackets and grouping like terms first is highly recommended to avoid mistakes during substitution.

Question 10. Solve:
\( \frac{2x + 1}{7} + \frac{5y - 3}{3} = 12 \)
\( \frac{3x + 2}{2} - \frac{4y + 3}{9} = 13 \)
Answer:
Let's simplify the first fractional equation by taking the LCM of 7 and 3, which is 21:
\( \implies \frac{3(2x + 1) + 7(5y - 3)}{21} = 12 \)
\( \implies 6x + 3 + 35y - 21 = 252 \)
\( \implies 6x + 35y - 18 = 252 \)
\( \implies 6x + 35y = 270 \)
\( \implies 6x = 270 - 35y \)
\( \implies x = \frac{270 - 35y}{6} \)

Now, let's simplify the second equation by taking the LCM of 2 and 9, which is 18:
\( \implies \frac{9(3x + 2) - 2(4y + 3)}{18} = 13 \)
\( \implies 27x + 18 - 8y - 6 = 234 \)
\( \implies 27x - 8y + 12 = 234 \)
\( \implies 27x - 8y = 222 \) ...(1)

Substituting our expression for \(x\) into equation (1):
\( \implies 27\left(\frac{270 - 35y}{6}\right) - 8y = 222 \)
To clear the fraction, multiply the whole equation by 6:
\( \implies 27(270 - 35y) - 48y = 1332 \)
\( \implies 7290 - 945y - 48y = 1332 \)
\( \implies 7290 - 993y = 1332 \)
\( \implies -993y = 1332 - 7290 \)
\( \implies -993y = -5958 \)
\( \implies y = 6 \)

Substitute this value of \(y\) back into the expression for \(x\):
\( \implies x = \frac{270 - 35(6)}{6} \)
\( \implies x = \frac{270 - 210}{6} \)
\( \implies x = \frac{60}{6} = 10 \)

Thus, the solution is \(x = 10\) and \(y = 6\).
In simple words: First, convert both fractional equations into linear equations without denominators. Then, solve the simplified equations using substitution.

Exam Tip: When clearing denominators, make sure to multiply every term in the equation, including the constant terms on the right-hand side.

Exercise 6(B)

Question 1. Solve:
\( 13 + 2y = 9x \)
\( 3y = 7x \)
Answer:
The system of equations is:
\( 13 + 2y = 9x \) ...(1)
\( 3y = 7x \) ...(2)

Let's use the elimination method. Multiply equation (1) by 3 and equation (2) by 2 to make the \(y\)-coefficients equal:
\( \implies 39 + 6y = 27x \) ...(3)
\( 6y = 14x \) ...(4)

Subtracting equation (4) from equation (3):
\( \implies (39 + 6y) - 6y = 27x - 14x \)
\( \implies 39 = 13x \)
\( \implies x = 3 \)

Substituting this value of \(x\) into equation (2):
\( \implies 3y = 7(3) \)
\( \implies 3y = 21 \)
\( \implies y = 7 \)

Thus, the solution is \(x = 3\) and \(y = 7\).
In simple words: Multiply both equations to make the \(y\) terms equal, subtract one from the other to find \(x\), and then substitute \(x\) to find \(y\).

Exam Tip: Elimination is often faster and less prone to fractions than substitution. Choose the coefficients carefully to minimize calculations.

Question 2. Solve:
\( 3x - y = 23 \)
\( \frac{x}{3} + \frac{y}{4} = 4 \)
Answer:
The given equations are:
\( 3x - y = 23 \) ...(1)
\( \frac{x}{3} + \frac{y}{4} = 4 \) ...(2)

Let's simplify equation (2) by multiplying through by the LCM of 3 and 4, which is 12:
\( \implies 4x + 3y = 48 \) ...(3)

Now, multiply equation (1) by 3 to set up elimination for \(y\):
\( \implies 9x - 3y = 69 \) ...(4)

Adding equation (3) and equation (4):
\( \implies (4x + 3y) + (9x - 3y) = 48 + 69 \)
\( \implies 13x = 117 \)
\( \implies x = 9 \)

Substituting \(x = 9\) into equation (1):
\( \implies 3(9) - y = 23 \)
\( \implies 27 - y = 23 \)
\( \implies y = 4 \)

Therefore, the solution is \(x = 9\) and \(y = 4\).
In simple words: Simplify the fraction equation first. Then multiply the other equation by 3 to eliminate \(y\) by adding them.

Exam Tip: When equations have coefficients with opposite signs (like \(+3y\) and \(-3y\)), adding the equations is the easiest way to eliminate the variable.

Question 3. Solve:
\( \frac{5y}{2} - \frac{x}{3} = 8 \)
\( \frac{y}{2} + \frac{5x}{3} = 12 \)
Answer:
Let's write the given equations with terms ordered consistently:
\( -\frac{x}{3} + \frac{5y}{2} = 8 \) ...(i)
\( \frac{5x}{3} + \frac{y}{2} = 12 \) ...(ii)

Multiply equation (i) by 5 to align the coefficients of \(x\):
\( \implies -\frac{5x}{3} + \frac{25y}{2} = 40 \)

Now, add this new equation to equation (ii) to eliminate the \(x\) term:
\( \implies \left(-\frac{5x}{3} + \frac{25y}{2}\right) + \left(\frac{5x}{3} + \frac{y}{2}\right) = 40 + 12 \)
\( \implies \frac{26y}{2} = 52 \)
\( \implies 13y = 52 \)
\( \implies y = 4 \)

Substituting \(y = 4\) back into equation (i):
\( \implies -\frac{x}{3} + \frac{5(4)}{2} = 8 \)
\( \implies -\frac{x}{3} + 10 = 8 \)
\( \implies -\frac{x}{3} = -2 \)
\( \implies x = 6 \)

Thus, the solution is \(x = 6\) and \(y = 4\).
In simple words: Align the \(x\) terms, multiply the first equation by 5, add the equations to eliminate \(x\), and solve for \(y\) and then \(x\).

Exam Tip: You don't always have to clear fractions entirely. If the fractional coefficients of a variable are multiples of each other, you can eliminate them directly.

Question 4. Solve:
\( \frac{1}{5}(x - 2) = \frac{1}{4}(1 - y) \)
\( 26x + 3y = -4 \)
Answer:
Let's simplify the first equation by cross-multiplying:
\( \implies 4(x - 2) = 5(1 - y) \)
\( \implies 4x - 8 = 5 - 5y \)
\( \implies 4x + 5y = 13 \) ...(1)

The second equation is already simplified:
\( 26x + 3y = -4 \) ...(2)

To eliminate \(y\), let's multiply equation (1) by 3 and equation (2) by 5:
\( \implies 12x + 15y = 39 \) ...(3)
\( \implies 130x + 15y = -20 \) ...(4)

Now, subtract equation (3) from equation (4):
\( \implies (130x + 15y) - (12x + 15y) = -20 - 39 \)
\( \implies 118x = -59 \)
\( \implies x = -\frac{59}{118} \)
\( \implies x = -\frac{1}{2} \)

Substituting \(x = -\frac{1}{2}\) into equation (1):
\( \implies 4\left(-\frac{1}{2}\right) + 5y = 13 \)
\( \implies -2 + 5y = 13 \)
\( \implies 5y = 15 \)
\( \implies y = 3 \)

Thus, the solution is \(x = -\frac{1}{2}\) and \(y = 3\).
In simple words: Cross-multiply to simplify the first equation, multiply both equations to make the \(y\) coefficients equal, subtract to solve for \(x\), and then substitute to find \(y\).

Exam Tip: Take care with arithmetic signs. When subtracting a positive number from a negative number, the result becomes more negative (e.g., \(-20 - 39 = -59\)).

Question 5. Solve:
\( y = 2x - 6 \)
\( y = 0 \)
Answer:
The equations are given as:
\( y = 2x - 6 \) ...(1)
\( y = 0 \) ...(2)

Substituting the value of \(y = 0\) directly into equation (1):
\( \implies 2x - 6 = 0 \)
\( \implies 2x = 6 \)
\( \implies x = 3 \)

Thus, the solution is \(x = 3\) and \(y = 0\).
In simple words: Since \(y\) is already given as 0, simply put 0 in place of \(y\) in the first equation and solve for \(x\).

Exam Tip: If one variable is given as a constant value (like \(y=0\)), substitute it immediately. This is a very easy and quick problem.

Question 6. Solve:
\( \frac{x - y}{6} = 2(4 - x) \)
\( 2x + y = 3(x - 4) \)
Answer:
Let's simplify the first equation:
\( \implies x - y = 12(4 - x) \)
\( \implies x - y = 48 - 12x \)
\( \implies 13x - y = 48 \) ...(1)

Now, let's simplify the second equation:
\( \implies 2x + y = 3x - 12 \)
\( \implies x - y = 12 \) ...(2)

Multiply equation (2) by 13 to prepare for elimination:
\( \implies 13x - 13y = 156 \) ...(3)

Subtracting equation (1) from equation (3):
\( \implies (13x - 13y) - (13x - y) = 156 - 48 \)
\( \implies -12y = 108 \)
\( \implies y = -9 \)

Substituting \(y = -9\) into equation (2):
\( \implies x - (-9) = 12 \)
\( \implies x + 9 = 12 \)
\( \implies x = 3 \)

Thus, the solution is \(x = 3\) and \(y = -9\).
In simple words: Simplify both equations by removing brackets and fractions. Multiply the second equation by 13, and subtract the first equation from it to solve for \(y\).

Exam Tip: Be careful when subtracting a negative term: \(x - (-9)\) simplifies to \(x + 9\).

Question 7. Solve:
\( 3 - (x - 5) = y + 2 \)
\( 2(x + y) = 4 - 3y \)
Answer:
Let's simplify the first equation:
\( \implies 3 - x + 5 = y + 2 \)
\( \implies 8 - x = y + 2 \)
\( \implies -x - y = -6 \)
\( \implies x + y = 6 \) ...(1)

Next, simplify the second equation:
\( \implies 2x + 2y = 4 - 3y \)
\( \implies 2x + 5y = 4 \) ...(2)

Multiply equation (1) by 2 to equate the \(x\)-coefficients:
\( \implies 2x + 2y = 12 \)

Subtracting this equation from equation (2):
\( \implies (2x + 5y) - (2x + 2y) = 4 - 12 \)
\( \implies 3y = -8 \)
\( \implies y = -\frac{8}{3} \)

Now, substitute \(y = -\frac{8}{3}\) back into equation (1):
\( \implies x - \frac{8}{3} = 6 \)
\( \implies x

Question 2. Solve the following system of simultaneous equations using the cross-multiplication method:
\( 3x + 4y = 11 \)
\( 2x + 3y = 8 \)
Answer:
The provided equations are:
\( 3x + 4y - 11 = 0 \)
\( 2x + 3y - 8 = 0 \)

On comparing these with the standard forms \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), we get:
\( a_1 = 3 \), \( b_1 = 4 \), \( c_1 = -11 \)
\( a_2 = 2 \), \( b_2 = 3 \), \( c_2 = -8 \)

By applying the cross-multiplication formula:
\( x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \) and \( y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \)

Substituting the values, we obtain:
\( x = \frac{4(-8) - 3(-11)}{3(3) - 2(4)} \) and \( y = \frac{-11(2) - (-8)(3)}{3(3) - 2(4)} \)

Solving these further:
\( x = \frac{-32 + 33}{9 - 8} \) and \( y = \frac{-22 + 24}{9 - 8} \)

\( x = \frac{1}{1} \) and \( y = \frac{2}{1} \)

Thus, \( x = 1 \) and \( y = 2 \).
In simple words: Write both equations in standard form first. Then, use the cross-multiplication formula to plug in the coefficients and solve for x and y.

Exam Tip: Be very careful with negative signs when substituting values into the cross-multiplication formula, as sign errors are a very common way to lose marks.

Question 3. Solve the following system of simultaneous equations using the cross-multiplication method:
\( 6x + 7y - 11 = 0 \)
\( 5x + 2y = 13 \)
Answer:
The given system of equations is:
\( 6x + 7y - 11 = 0 \)
\( 5x + 2y - 13 = 0 \)

Comparing these with \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), we find:
\( a_1 = 6 \), \( b_1 = 7 \), \( c_1 = -11 \)
\( a_2 = 5 \), \( b_2 = 2 \), \( c_2 = -13 \)

Using the cross-multiplication formula:
\( x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \) and \( y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \)

By putting the values into the formula:
\( x = \frac{7(-13) - 2(-11)}{6(2) - 5(7)} \) and \( y = \frac{-11(5) - (-13)(6)}{6(2) - 5(7)} \)

Simplifying the terms:
\( x = \frac{-91 + 22}{12 - 35} \) and \( y = \frac{-55 + 78}{12 - 35} \)

\( x = \frac{-69}{-23} \) and \( y = \frac{23}{-23} \)

This gives:
\( x = 3 \) and \( y = -1 \).
In simple words: Rearrange the equations so that all terms are on the left side. Identify the constants and use the formula to find the values of x and y.

Exam Tip: Always make sure to write the equations in standard form \( ax + by + c = 0 \) before identifying the coefficients \( a \), \( b \), and \( c \).

Question 4. Solve the following system of simultaneous equations using the cross-multiplication method:
\( 5x + 4y + 14 = 0 \)
\( 3x = -10 - 4y \)
Answer:
The given equations can be written as:
\( 5x + 4y + 14 = 0 \)
\( 3x + 4y + 10 = 0 \)

Comparing these equations with \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), we have:
\( a_1 = 5 \), \( b_1 = 4 \), \( c_1 = 14 \)
\( a_2 = 3 \), \( b_2 = 4 \), \( c_2 = 10 \)

Using the cross-multiplication formula:
\( x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \) and \( y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \)

Substituting the corresponding values:
\( x = \frac{4(10) - 4(14)}{5(4) - 3(4)} \) and \( y = \frac{14(3) - 10(5)}{5(4) - 3(4)} \)

This simplifies to:
\( x = \frac{40 - 56}{20 - 12} \) and \( y = \frac{42 - 50}{20 - 12} \)

\( x = \frac{-16}{8} \) and \( y = \frac{-8}{8} \)

Hence, we get:
\( x = -2 \) and \( y = -1 \).
In simple words: Put both equations in standard form, then find the coefficients and apply the cross-multiplication formula to solve for x and y.

Exam Tip: Remember that both equations must have the constant term on the same side (either both LHS or both RHS) to avoid incorrect sign assignments for \( c_1 \) and \( c_2 \).

Question 5. Solve the following system of simultaneous equations using the cross-multiplication method:
\( x - y + 2 = 0 \)
\( 7x + 9y = 130 \)
Answer:
The system of equations is given by:
\( x - y + 2 = 0 \)
\( 7x + 9y - 130 = 0 \)

On comparing these with \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), we obtain:
\( a_1 = 1 \), \( b_1 = -1 \), \( c_1 = 2 \)
\( a_2 = 7 \), \( b_2 = 9 \), \( c_2 = -130 \)

Applying the formula for cross-multiplication:
\( x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \) and \( y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \)

On putting the values:
\( x = \frac{-1(-130) - 9(2)}{1(9) - 7(-1)} \) and \( y = \frac{2(7) - (-130)(1)}{1(9) - 7(-1)} \)

Simplifying further:
\( x = \frac{130 - 18}{9 + 7} \) and \( y = \frac{14 + 130}{9 + 7} \)

\( x = \frac{112}{16} \) and \( y = \frac{144}{16} \)

This yields:
\( x = 7 \) and \( y = 9 \).
In simple words: Rearrange the terms so that everything is on one side, identify the values of a, b, and c, and plug them into the formula to find x and y.

Exam Tip: Notice how the denominator \( a_1b_2 - a_2b_1 \) is identical for both \( x \) and \( y \); calculating it carefully once saves time and prevents double errors.

Question 6. Solve the following system of simultaneous equations using the cross-multiplication method:
\( 4x - y = 5 \)
\( 5y - 4x = 7 \)
Answer:
Writing the equations in standard form:
\( 4x - y - 5 = 0 \)
\( -4x + 5y - 7 = 0 \)

Comparing with the standard equations \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), we get:
\( a_1 = 4 \), \( b_1 = -1 \), \( c_1 = -5 \)
\( a_2 = -4 \), \( b_2 = 5 \), \( c_2 = -7 \)

Using the cross-multiplication formulas:
\( x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \) and \( y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \)

Substituting these values:
\( x = \frac{-1(-7) - 5(-5)}{4(5) - (-4)(-1)} \) and \( y = \frac{-5(-4) - (-7)(4)}{4(5) - (-4)(-1)} \)

Simplifying the calculations:
\( x = \frac{7 + 25}{20 - 4} \) and \( y = \frac{20 + 28}{20 - 4} \)

\( x = \frac{32}{16} \) and \( y = \frac{48}{16} \)

Thus, the solutions are:
\( x = 2 \) and \( y = 3 \).
In simple words: Rewrite both equations in standard form first, taking care of the order of x and y. Then use the formula to find x and y.

Exam Tip: Be careful with the order of variables in the second equation. Since it was given as \( 5y - 4x = 7 \), it must be rewritten as \( -4x + 5y - 7 = 0 \) so that \( a_2 \) is the coefficient of \( x \).

Question 7. Solve the following system of simultaneous equations using the cross-multiplication method:
\( 4x - 3y = 0 \)
\( 2x + 3y = 18 \)
Answer:
Let us express the given equations as:
\( 4x - 3y + 0 = 0 \)
\( 2x + 3y - 18 = 0 \)

On comparing with \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), we get:
\( a_1 = 4 \), \( b_1 = -3 \), \( c_1 = 0 \)
\( a_2 = 2 \), \( b_2 = 3 \), \( c_2 = -18 \)

Applying the cross-multiplication formulas:
\( x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \) and \( y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \)

By substitution:
\( x = \frac{-3(-18) - 3(0)}{4(3) - 2(-3)} \) and \( y = \frac{0(2) - (-18)(4)}{4(3) - 2(-3)} \)

Solving further:
\( x = \frac{54 - 0}{12 + 6} \) and \( y = \frac{0 + 72}{12 + 6} \)

\( x = \frac{54}{18} \) and \( y = \frac{72}{18} \)

This yields:
\( x = 3 \) and \( y = 4 \).
In simple words: Convert the equations to standard form (using 0 as the constant if it is missing). Substitute the coefficients in the formula to get the final answer.

Exam Tip: If an equation has no constant term, always write \( c = 0 \) explicitly so that you do not leave that parameter out of your cross-multiplication calculations.

Question 8. Solve the following system of simultaneous equations using the cross-multiplication method:
\( 8x + 5y = 9 \)
\( 3x + 2y = 4 \)
Answer:
We can write the equations as:
\( 8x + 5y - 9 = 0 \)
\( 3x + 2y - 4 = 0 \)

Comparing these to the standard equations \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \):
\( a_1 = 8 \), \( b_1 = 5 \), \( c_1 = -9 \)
\( a_2 = 3 \), \( b_2 = 2 \), \( c_2 = -4 \)

Using the formula for cross-multiplication:
\( x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \) and \( y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \)

Substituting the respective values:
\( x = \frac{5(-4) - 2(-9)}{8(2) - 3(5)} \) and \( y = \frac{-9(3) - (-4)(8)}{8(2) - 3(5)} \)

Which simplifies to:
\( x = \frac{-20 + 18}{16 - 15} \) and \( y = \frac{-27 + 32}{16 - 15} \)

\( x = \frac{-2}{1} \) and \( y = \frac{5}{1} \)

Thus, we get:
\( x = -2 \) and \( y = 5 \).
In simple words: Rearrange both equations to standard form first. Plug the constants into the cross-multiplication formula to solve for x and y.

Exam Tip: Since the denominator is \( 1 \), the values of \( x \) and \( y \) simplify directly to their numerators, which makes checking your calculations much simpler.

Question 9. Solve the following system of simultaneous equations using the cross-multiplication method:
\( 4x - 3y - 11 = 0 \)
\( 6x + 7y - 5 = 0 \)
Answer:
The standard equations are:
\( 4x - 3y - 11 = 0 \)
\( 6x + 7y - 5 = 0 \)

On comparing with the general forms, we get:
\( a_1 = 4 \), \( b_1 = -3 \), \( c_1 = -11 \)
\( a_2 = 6 \), \( b_2 = 7 \), \( c_2 = -5 \)

Using the cross-multiplication formula:
\( x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \) and \( y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \)

Substituting the values:
\( x = \frac{-3(-5) - 7(-11)}{4(7) - 6(-3)} \) and \( y = \frac{-11(6) - (-5)(4)}{4(7) - 6(-3)} \)

Simplifying the terms:
\( x = \frac{15 + 77}{28 + 18} \) and \( y = \frac{-66 + 20}{28 + 18} \)

\( x = \frac{92}{46} \) and \( y = \frac{-46}{46} \)

This yields:
\( x = 2 \) and \( y = -1 \).
In simple words: Set the equations to standard form and use the standard formula for cross-multiplication to calculate x and y.

Exam Tip: Be extremely careful with signs in the denominator when dealing with negative coefficients (e.g., \( -6 \times (-3) = +18 \)), to prevent fraction calculations from going wrong.

Question 10. Solve the following system of simultaneous equations using the cross-multiplication method:
\( 4x + 6y = 15 \)
\( 3x - 4y = 7 \)
Answer:
We can express the given equations as:
\( 4x + 6y - 15 = 0 \)
\( 3x - 4y - 7 = 0 \)

Comparing with the standard forms, we have:
\( a_1 = 4 \), \( b_1 = 6 \), \( c_1 = -15 \)
\( a_2 = 3 \), \( b_2 = -4 \), \( c_2 = -7 \)

Using the cross-multiplication formulas:
\( x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \) and \( y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \)

By putting in the respective values:
\( x = \frac{6(-7) - (-4)(-15)}{4(-4) - 3(6)} \) and \( y = \frac{-15(3) - (-7)(4)}{4(-4) - 3(6)} \)

On simplifying, we get:
\( x = \frac{-42 - 60}{-16 - 18} \) and \( y = \frac{-45 + 28}{-16 - 18} \)

\( x = \frac{-102}{-34} \) and \( y = \frac{-17}{-34} \)

This yields:
\( x = 3 \) and \( y = \frac{1}{2} \).
In simple words: Rearrange the equations into standard form and use the cross-multiplication formula to find x and y.

Exam Tip: Remember to reduce the fraction to its lowest terms; leaving \( y = \frac{-17}{-34} \) as-is instead of \( \frac{1}{2} \) could result in a deduction of marks.

Exercise 6(D)

Question 1. Solve the following pair of equations:
\( \frac{9}{x} - \frac{4}{y} = 8 \)
\( \frac{13}{x} + \frac{7}{y} = 101 \)
Answer:
Let the given equations be:
\( \frac{9}{x} - \frac{4}{y} = 8 \) ...(1)
\( \frac{13}{x} + \frac{7}{y} = 101 \) ...(2)

Let us multiply equation (1) by 7 and equation (2) by 4:
\( \frac{63}{x} - \frac{28}{y} = 56 \) ...(3)
\( \frac{52}{x} + \frac{28}{y} = 404 \) ...(4)

Adding equations (3) and (4):
\( \frac{115}{x} = 460 \)

\( \implies x = \frac{115}{460} = \frac{1}{4} \)

Now, substituting \( x = \frac{1}{4} \) in equation (1):
\( 9\left(\frac{4}{1}\right) - \frac{4}{y} = 8 \)

\( \implies 36 - \frac{4}{y} = 8 \)

\( \implies -\frac{4}{y} = -28 \)

\( \implies y = \frac{1}{7} \)

Hence, \( x = \frac{1}{4} \) and \( y = \frac{1}{7} \).
In simple words: Multiply the first equation by 7 and the second by 4 to make the y-terms equal but opposite. Add them together to find x, then use x to solve for y.

Exam Tip: Since \( \frac{1}{x} \) acts as a linear variable, you do not always need to substitute \( u = \frac{1}{x} \) explicitly; eliminating \( \frac{1}{y} \) directly is often faster and less prone to notation confusion.

Question 2. Solve the following pair of equations:
\( \frac{3}{x} + \frac{2}{y} = 10 \)
\( \frac{9}{x} - \frac{7}{y} = 10.5 \)
Answer:
The equations given are:
\( \frac{3}{x} + \frac{2}{y} = 10 \) ...(i)
\( \frac{9}{x} - \frac{7}{y} = 10.5 \) ...(ii)

Multiplying equation (i) by 3, we get:
\( \frac{9}{x} + \frac{6}{y} = 30 \) ...(iii)

Subtracting equation (ii) from equation (iii):
\( \frac{13}{y} = 19.5 \)

\( \implies y = \frac{13}{19.5} = \frac{2}{3} \)

Substituting the value of \( y \) into equation (i):
\( \frac{3}{x} + \frac{2 \times 3}{2} = 10 \)

\( \implies \frac{3}{x} + 3 = 10 \)

\( \implies \frac{3}{x} = 7 \)

\( \implies x = \frac{3}{7} \)

Thus, the solution is \( x = \frac{3}{7} \) and \( y = \frac{2}{3} \).
In simple words: Multiply the first equation by 3 to match the x-terms. Subtract the equations to solve for y, and then use that value to find x.

Exam Tip: When dividing by decimals (like \( \frac{13}{19.5} \)), convert the decimal to a fraction (e.g., \( 19.5 = \frac{39}{2} \)) to easily simplify the complex fraction into \( \frac{2}{3} \).

Question 3. Solve the following pair of equations:
\( 5x + \frac{8}{y} = 19 \)
\( 3x - \frac{4}{y} = 7 \)
Answer:
Let the given equations be:
\( 5x + \frac{8}{y} = 19 \) ...(i)
\( 3x - \frac{4}{y} = 7 \) ...(ii)

Multiplying equation (ii) by 2 gives:
\( 6x - \frac{8}{y} = 14 \) ...(iii)

Adding equations (i) and (iii):
\( 11x = 33 \)

\( \implies x = 3 \)

Substituting \( x = 3 \) back into equation (i):
\( 5(3) + \frac{8}{y} = 19 \)

\( \implies 15 + \frac{8}{y} = 19 \)

\( \implies \frac{8}{y} = 4 \)

\( \implies y = 2 \)

So, the solutions are \( x = 3 \) and \( y = 2 \).
In simple words: Multiply the second equation by 2 so the y-terms can be eliminated by adding the equations. Solve for x, then substitute it back to get y.

Exam Tip: Substitution is very straightforward here since the equations have linear terms in \( x \). Choose the simplest equation to substitute the value of \( x \) to find \( y \).

Question 4. Solve the following pair of equations:
\( 4x + \frac{6}{y} = 15 \)
\( 3x - \frac{4}{y} = 7 \)
Hence, find \( a \) if \( y = ax - 2 \).
Answer:
The equations are given as:
\( 4x + \frac{6}{y} = 15 \) ...(i)
\( 3x - \frac{4}{y} = 7 \) ...(ii)

Multiplying equation (i) by 4 and equation (ii) by 6, we get:
\( 16x + \frac{24}{y} = 60 \) ...(iii)
\( 18x - \frac{24}{y} = 42 \) ...(iv)

Adding equations (iii) and (iv):
\( 34x = 102 \)

\( \implies x = 3 \)

Substituting \( x = 3 \) into equation (i):
\( 4(3) + \frac{6}{y} = 15 \)

\( \implies 12 + \frac{6}{y} = 15 \)

\( \implies \frac{6}{y} = 3 \)

\( \implies y = 2 \)

Now, we use \( y = ax - 2 \) to find the value of \( a \):
\( 2 = a(3) - 2 \)

\( \implies 3a = 4 \)

\( \implies a = \frac{4}{3} = 1\frac{1}{3} \)

Therefore, we have \( x = 3 \), \( y = 2 \), and \( a = 1\frac{1}{3} \).
In simple words: Multiply the first equation by 4 and the second by 6 to eliminate y. After finding x and y, plug their values into the given linear relation to calculate a.

Exam Tip: Be sure to complete the second part of the question. Finding \( x \) and \( y \) is only the intermediate step; always check if there is an additional variable like \( a \) to calculate.

Question 5. Solve the following pair of equations:
\( \frac{3}{x} - \frac{2}{y} = 0 \)
\( \frac{2}{x} + \frac{5}{y} = 19 \)
Hence, find \( a \) if \( y = ax + 3 \).
Answer:
Let us designate the given equations as:
\( \frac{3}{x} - \frac{2}{y} = 0 \) ...(1)
\( \frac{2}{x} + \frac{5}{y} = 19 \) ...(2)

Multiplying equation (1) by 5 and equation (2) by 2 yields:
\( \frac{15}{x} - \frac{10}{y} = 0 \) ...(3)
\( \frac{4}{x} + \frac{10}{y} = 38 \) ...(4)

Adding equations (3) and (4):
\( \frac{19}{x} = 38 \)

\( \implies x = \frac{19}{38} = \frac{1}{2} \)

Substituting \( x = \frac{1}{2} \) into equation (1):
\( 3\left(\frac{2}{1}\right) - \frac{2}{y} = 0 \)

\( \implies 6 - \frac{2}{y} = 0 \)

\( \implies \frac{2}{y} = 6 \)

\( \implies y = \frac{1}{3} \)

Using the relation \( y = ax + 3 \) to find \( a \):
\( \frac{1}{3} = a\left(\frac{1}{2}\right) + 3 \)

\( \implies \frac{a}{2} = \frac{1}{3} - 3 \)

\( \implies \frac{a}{2} = \frac{-8}{3} \)

\( \implies a = \frac{-16}{3} \)

So, we get \( x = \frac{1}{2} \), \( y = \frac{1}{3} \), and \( a = -\frac{16}{3} \).
In simple words: Multiply the first equation by 5 and the second by 2 to eliminate the y-term. Find x and y, and then substitute them into the given equation to find a.

Exam Tip: Fraction arithmetic can be tricky. When working out \( \frac{a}{2} = -\frac{8}{3} \), cross-multiply carefully to ensure you do not make sign or multiplication errors.

Question 6. Solve:
(i) \( \frac{20}{x+y} + \frac{3}{x-y} = 7 \) and \( \frac{8}{x-y} - \frac{15}{x+y} = 5 \)
(ii) \( \frac{34}{3x+4y} + \frac{15}{3x-2y} = 5 \) and \( \frac{25}{3x-2y} - \frac{8.50}{3x+4y} = 4.5 \)
Answer:
(i)
We have the equations:
\( \frac{20}{x+y} + \frac{3}{x-y} = 7 \) ...(1)
\( -\frac{15}{x+y} + \frac{8}{x-y} = 5 \) ...(2)

Let \( \frac{1}{x+y} = u \) and \( \frac{1}{x-y} = v \). The equations become:
\( 20u + 3v = 7 \) ...(3)
\( -15u + 8v = 5 \) ...(4)

Multiplying equation (3) by 8 and equation (4) by 3:
\( 160u + 24v = 56 \) ...(5)
\( -45u + 24v = 15 \) ...(6)

Subtracting equation (6) from (5):
\( 205u = 41 \)

\( \implies u = \frac{41}{205} = \frac{1}{5} \)

Substitute \( u = \frac{1}{5} \) in equation (3):
\( 20\left(\frac{1}{5}\right) + 3v = 7 \)

\( \implies 4 + 3v = 7 \)

\( \implies 3v = 3 \)

\( \implies v = 1 \)

Since \( u = \frac{1}{x+y} = \frac{1}{5} \implies x+y = 5 \) ...(7)
Since \( v = \frac{1}{x-y} = 1 \implies x-y = 1 \) ...(8)

Adding equations (7) and (8):
\( 2x = 6 \)

\( \implies x = 3 \)

Substitute \( x = 3 \) in (7):
\( 3+y = 5 \)

\( \implies y = 2 \)

Therefore, the solution is \( x = 3, y = 2 \).

(ii)
The given equations are:
\( \frac{34}{3x+4y} + \frac{15}{3x-2y} = 5 \) ...(1)
\( -\frac{8.50}{3x+4y} + \frac{25}{3x-2y} = 4.5 \) ...(2)

Let \( 3x + 4y = a \) and \( 3x - 2y = b \). The equations become:
\( \frac{34}{a} + \frac{15}{b} = 5 \) ...(3)
\( -\frac{8.50}{a} + \frac{25}{b} = 4.5 \) ...(4)

Multiplying equation (4) by 4:
\( -\frac{34}{a} + \frac{100}{b} = 18 \) ...(5)

Adding equations (3) and (5):
\( \frac{115}{b} = 23 \)

\( \implies b = 5 \)

Therefore, \( 3x - 2y = 5 \) ...(6)

Substituting \( b = 5 \) into equation (3):
\( \frac{34}{a} + \frac{15}{5} = 5 \)

\( \implies \frac{34}{a} + 3 = 5 \)

\( \implies \frac{34}{a} = 2 \)

\( \implies a = 17 \)

Therefore, \( 3x + 4y = 17 \) ...(7)

Subtracting equation (6) from equation (7):
\( (3x + 4y) - (3x - 2y) = 17 - 5 \)

\( \implies 6y = 12 \)

\( \implies y = 2 \)

Substituting \( y = 2 \) into equation (6):
\( 3x - 2(2) = 5 \)

\( \implies 3x - 4 = 5 \)

\( \implies 3x = 9 \)

\( \implies x = 3 \)

Hence, the solution is \( x = 3 \) and \( y = 2 \).
In simple words: For equations with expressions like (x+y) or (3x+4y) in the denominator, substitute them with dummy variables first. Solve the simpler linear system, then substitute back to get the final values of x and y.

Exam Tip: Be very methodical with two-stage substitutions. Ensure you don't make sign errors when performing subtraction to eliminate the secondary variable at the end.

Question 7. Solve:
(i) \( x+y = 2xy \) and \( x-y = 6xy \)
(ii) \( x+y = 7xy \) and \( 2x-3y = -xy \)
Answer:
(i)
We are given the following equations:
\( x + y = 2xy \) ...(1)
\( x - y = 6xy \) ...(2)

Adding equations (1) and (2):
\( 2x = 8xy \)

Since \( x \neq 0 \), we divide both sides by \( x \):
\( 2 = 8y \)

\( \implies y = \frac{1}{4} \)

Substituting \( y = \frac{1}{4} \) back into equation (1):
\( x + \frac{1}{4} = 2x\left(\frac{1}{4}\right) \)

\( \implies x + \frac{1}{4} = \frac{x}{2} \)

\( \implies x - \frac{x}{2} = -\frac{1}{4} \)

\( \implies \frac{x}{2} = -\frac{1}{4} \)

\( \implies x = -\frac{1}{2} \)

So, \( x = -\frac{1}{2} \) and \( y = \frac{1}{4} \).

(ii)
The given system of equations is:
\( x + y = 7xy \) ...(1)
\( 2x - 3y = -xy \) ...(2)

Multiplying equation (1) by 3, we get:
\( 3x + 3y = 21xy \) ...(3)

Adding equations (2) and (3):
\( 5x = 20xy \)

Assuming \( x \neq 0 \), dividing both sides by \( x \) gives:
\( 5 = 20y \)

\( \implies y = \frac{1}{4} \)

Substituting \( y = \frac{1}{4} \) into equation (1):
\( x + \frac{1}{4} = 7x\left(\frac{1}{4}\right) \)

\( \implies x + \frac{1}{4} = \frac{7x}{4} \)

\( \implies \frac{7x}{4} - x = \frac{1}{4} \)

\( \implies \frac{3x}{4} = \frac{1}{4} \)

\( \implies 3x = 1 \)

\( \implies x = \frac{1}{3} \)

So, \( x = \frac{1}{3} \) and \( y = \frac{1}{4} \).
In simple words: When equations have an xy term on one side, add or multiply the equations to eliminate one of the variables (x or y) entirely, then solve for the other.

Exam Tip: Remember that division by \( x \) is only valid if \( x \neq 0 \). If \( x = 0 \) is also a trivial solution, you should verify if it fits the original equations (here \( (0,0) \) is indeed a trivial solution, but standard textbook questions usually seek non-zero solutions).

Question 8. Solve:
\( \frac{a}{x} - \frac{b}{y} = 0 \)
\( \frac{ab^2}{x} + \frac{a^2b}{y} = a^2 + b^2 \)
Answer:
We are given the equations:
\( \frac{a}{x} - \frac{b}{y} = 0 \)
\( \frac{ab^2}{x} + \frac{a^2b}{y} = a^2 + b^2 \)

Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). The equations become:
\( au - bv + 0 = 0 \) ...(1)
\( ab^2u + a^2bv - (a^2+b^2) = 0 \) ...(2)

Using the method of cross-multiplication:
\( \frac{u}{-b[-(a^2+b^2)] - a^2b(0)} = \frac{-v}{a[-(a^2+b^2)] - ab^2(0)} = \frac{1}{a(a^2b) - (-b)(ab^2)} \)

\( \implies \frac{u}{b(a^2+b^2)} = \frac{v}{a(a^2+b^2)} = \frac{1}{a^3b + ab^3} \)

\( \implies \frac{u}{b(a^2+b^2)} = \frac{v}{a(a^2+b^2)} = \frac{1}{ab(a^2+b^2)} \)

Now, solving for \( u \) and \( v \):
\( u = \frac{b(a^2+b^2)}{ab(a^2+b^2)} = \frac{1}{a} \)
\( v = \frac{a(a^2+b^2)}{ab(a^2+b^2)} = \frac{1}{b} \)

Since \( \frac{1}{x} = u = \frac{1}{a} \implies x = a \)
And \( \frac{1}{y} = v = \frac{1}{b} \implies y = b \)

Hence, \( x = a \) and \( y = b \).
In simple words: Substitute u for 1/x and v for 1/y. Use the cross-multiplication formula with the literal coefficients a and b, then simplify to find x and y.

Exam Tip: Do not be intimidated by algebraic terms like \( a \) and \( b \) as coefficients. Treat them exactly like numbers and factor out \( (a^2 + b^2) \) to simplify the fractions.

Question 9. Solve:
\( \frac{2xy}{x+y} = \frac{3}{2} \)
\( \frac{xy}{2x-y} = -\frac{3}{10} \)
Answer:
The given equations are:
\( \frac{2xy}{x+y} = \frac{3}{2} \) and \( \frac{xy}{2x-y} = -\frac{3}{10} \)

Taking reciprocal on both sides for both equations:
\( \frac{x+y}{2xy} = \frac{2}{3} \)
\( \implies \frac{x}{2xy} + \frac{y}{2xy} = \frac{2}{3} \)
\( \implies \frac{1}{2y} + \frac{1}{2x} = \frac{2}{3} \)
Multiplying by 2:
\( \frac{1}{y} + \frac{1}{x} = \frac{4}{3} \) ...(1)

And for the second equation:
\( \frac{2x-y}{xy} = -\frac{10}{3} \)
\( \implies \frac{2x}{xy} - \frac{y}{xy} = -\frac{10}{3} \)
\( \implies \frac{2}{y} - \frac{1}{x} = -\frac{10}{3} \) ...(2)

Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Equations (1) and (2) become:
\( u + v = \frac{4}{3} \implies 3u + 3v = 4 \) ...(3)
\( -u + 2v = -\frac{10}{3} \implies -3u + 6v = -10 \) ...(4)

Adding equations (3) and (4):
\( 9v = -6 \)

\( \implies v = -\frac{6}{9} = -\frac{2}{3} \)

Substitute \( v = -\frac{2}{3} \) into equation (1):
\( u - \frac{2}{3} = \frac{4}{3} \)

\( \implies u = \frac{4}{3} + \frac{2}{3} = 2 \)

Therefore:
\( x = \frac{1}{u} = \frac{1}{2} \)
\( y = \frac{1}{v} = -\frac{3}{2} \)

Hence, \( x = \frac{1}{2} \) and \( y = -\frac{3}{2} \).
In simple words: Flip the fractions first so that the denominators are simpler. Split the terms, substitute dummy variables, and solve the system to find x and y.

Exam Tip: Taking reciprocals is a very powerful technique when you have products of variables like \( xy \) in the numerator and sums in the denominator.

Question 10. Solve:
\( \frac{3}{2x} + \frac{2}{3y} = -\frac{1}{3} \)
\( \frac{3}{4x} + \frac{1}{2y} = -\frac{1}{8} \)
Answer:
Let the given equations be:
\( \frac{3}{2x} + \frac{2}{3y} = -\frac{1}{3} \) and \( \frac{3}{4x} + \frac{1}{2y} = -\frac{1}{8} \)

Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). The equations become:
\( \frac{3}{2}u + \frac{2}{3}v = -\frac{1}{3} \)
Multiplying by 6 to clear fractions:
\( 9u + 4v = -2 \implies 27u + 12v + 6 = 0 \) ...(1)

And for the second equation:
\( \frac{3}{4}u + \frac{1}{2}v = -\frac{1}{8} \)
Multiplying by 32 to clear fractions:
\( 24u + 16v = -4 \implies 24u + 16v + 4 = 0 \) ...(2)

Using cross-multiplication to solve (1) and (2):
\( \frac{u}{12(4) - 16(6)} = \frac{-v}{27(4) - 24(6)} = \frac{1}{27(16) - 24(12)} \)

\( \implies \frac{u}{48 - 96} = \frac{-v}{108 - 144} = \frac{1}{432 - 288} \)

\( \implies \frac{u}{-48} = \frac{v}{36} = \frac{1}{144} \)

Solving for \( u \) and \( v \):
\( u = \frac{-48}{144} = -\frac{1}{3} \)
\( v = \frac{36}{144} = \frac{1}{4} \)

Since \( \frac{1}{x} = u = -\frac{1}{3} \implies x = -3 \)
And \( \frac{1}{y} = v = \frac{1}{4} \implies y = 4 \)

Hence, the solutions are \( x = -3 \) and \( y = 4 \).
In simple words: Substitute u for 1/x and v for 1/y. Multiply by the LCM to clear the fractions, then use cross-multiplication to solve for u and v, and take their reciprocals.

Exam Tip: Double-check your LCM multiplication step carefully. A small slip-up while clearing fraction denominators can lead to completely wrong coefficient values.

Exercise 6(E)

Question 1. The ratio of two numbers is \( \frac{2}{3} \). If 2 is subtracted from the first and 8 from the second, the ratio becomes the reciprocal of the original ratio. Find the numbers.
Answer:
Let the two required numbers be \( x \) and \( y \).
According to the given condition, we have:
\( \frac{x}{y} = \frac{2}{3} \)
\( \implies 3x - 2y = 0 \) ...(1)

Also, when 2 is subtracted from the first and 8 from the second, the ratio becomes the reciprocal of \( \frac{2}{3} \) (which is \( \frac{3}{2} \)):
\( \frac{x-2}{y-8} = \frac{3}{2} \)
\( \implies 2(x - 2) = 3(y - 8) \)
\( \implies 2x - 4 = 3y - 24 \)
\( \implies 2x - 3y = -20 \) ...(2)

Multiplying equation (1) by 2 and equation (2) by 3:
\( 6x - 4y = 0 \) ...(3)
\( 6x - 9y = -60 \) ...(4)

Subtracting equation (4) from equation (3):
\( (6x - 4y) - (6x - 9y) = 0 - (-60) \)
\( \implies 5y = 60 \)
\( \implies y = 12 \)

Substituting \( y = 12 \) back into equation (1):
\( 3x - 2(12) = 0 \)
\( \implies 3x = 24 \)
\( \implies x = 8 \)

Thus, the two numbers are 8 and 12.
In simple words: Set up the two equations based on the ratio clues. Solve the resulting system of linear equations using elimination to find the two numbers.

Exam Tip: Read the phrase "reciprocal of the original ratio" carefully. If the original ratio is \( \frac{2}{3} \), its reciprocal is \( \frac{3}{2} \).

Question 2. Two numbers are in the ratio \( 4 : 7 \). If thrice the larger be added to twice the smaller, the sum is 59. Find the numbers.
Answer:
Let the smaller number be \( x \) and the larger number be \( y \).
Based on the given ratio, we can write:
\( \frac{x}{y} = \frac{4}{7} \)
\( \implies 7x - 4y = 0 \) ...(1)

Also, thrice the larger number added to twice the smaller number is 59:
\( 2x + 3y = 59 \) ...(2)

To solve this, let us multiply equation (1) by 3 and equation (2) by 4:
\( 21x - 12y = 0 \) ...(3)
\( 8x + 12y = 236 \) ...(4)

Adding equations (3) and (4):
\( 29x = 236 \)
\( \implies x = \frac{236}{29} \)

From equation (1), we can find \( y \):
\( 7\left(\frac{236}{29}\right) = 4y \)
\( \implies y = \frac{7 \times 59}{29} \)
\( \implies y = \frac{413}{29} \)

Therefore, the required numbers are \( \frac{236}{29} \) and \( \frac{413}{29} \).
In simple words: Form two equations based on the statements. Solve using elimination to get the two fractional values.

Exam Tip: Do not be alarmed if you get fraction answers like \( \frac{236}{29} \). Always double-check your arithmetic steps to be confident in your final result.

Question 3. When the greater of two numbers increased by 1 divides the sum of the numbers, the result is \( \frac{3}{2} \). When the difference of these numbers is divided by the smaller, the result is \( \frac{1}{2} \). Find the numbers.
Answer:
Let the greater number be \( x \) and the smaller number be \( y \).
According to the first statement, when \( x+1 \) divides the sum \( x+y \), the quotient is \( \frac{3}{2} \):
\( \frac{x+y}{x+1} = \frac{3}{2} \)
\( \implies 2(x + y) = 3(x + 1) \)
\( \implies 2x + 2y = 3x + 3 \)
\( \implies x - 2y = -3 \) ...(1)

From the second statement, when the difference \( x-y \) is divided by the smaller number \( y \), the quotient is \( \frac{1}{2} \):
\( \frac{x-y}{y} = \frac{1}{2} \)
\( \implies 2(x - y) = y \)
\( \implies 2x - 2y = y \)
\( \implies 2x - 3y = 0 \) ...(2)

Multiplying equation (1) by 2, we get:
\( 2x - 4y = -6 \) ...(3)

Subtracting equation (2) from equation (3):
\( -y = -6 \)
\( \implies y = 6 \)

Substituting \( y = 6 \) back into equation (1):
\( x - 2(6) = -3 \)
\( \implies x - 12 = -3 \)
\( \implies x = 9 \)

Hence, the greater number is 9 and the smaller number is 6.
In simple words: Translate both statements into mathematical expressions to form two equations. Solve using the elimination method to find both numbers.

Exam Tip: Clearly define which variable is the "greater" and which is the "smaller" at the beginning so that you do not swap their places when translating "difference of these numbers" (which must be greater minus smaller, \( x - y \)).

Question 4. Two numbers are in the ratio \( 4 : 5 \). If 30 is subtracted from each of the numbers, the ratio becomes \( 1 : 2 \). Find the numbers.
Answer:
Let the common multiplier for both numbers be \( x \).
Thus, the two numbers are \( 4x \) and \( 5x \).
According to the given condition, subtracting 30 from both gives a ratio of \( 1 : 2 \):
\( \frac{4x - 30}{5x - 30} = \frac{1}{2} \)

Cross-multiplying to solve for \( x \):
\( 2(4x - 30) = 5x - 30 \)
\( \implies 8x - 60 = 5x - 30 \)
\( \implies 3x = 30 \)
\( \implies x = 10 \)

Now, we calculate the numbers:
First number = \( 4x = 4(10) = 40 \)
Second number = \( 5x = 5(10) = 50 \)

Hence, the required numbers are 40 and 50.
In simple words: Use a common variable to write the numbers. Set up the ratio after subtracting 30 from both, and cross-multiply to solve.

Exam Tip: Introducing a single common variable \( x \) is highly efficient for ratio problems compared to using a system of two variables, saving valuable exam time.

Question 5. If the numerator of a fraction is increased by 2 and the denominator is decreased by 1, it becomes \( \frac{2}{3} \). If the numerator is increased by 1 and the denominator is increased by 2, it becomes \( \frac{1}{3} \). Find the fraction.
Answer:
Let the numerator be \( x \) and the denominator be \( y \).
First condition:
\( \frac{x+2}{y-1} = \frac{2}{3} \)
\( \implies 3(x + 2) = 2(y - 1) \)
\( \implies 3x + 6 = 2y - 2 \)
\( \implies 3x - 2y = -8 \) ...(1)

Second condition:
\( \frac{x+1}{y+2} = \frac{1}{3} \)
\( \implies 3(x + 1) = y + 2 \)
\( \implies 3x + 3 = y + 2 \)
\( \implies 3x - y = -1 \) ...(2)

Subtracting equation (2) from equation (1):
\( (3x - 2y) - (3x - y) = -8 - (-1) \)
\( \implies -y = -7 \)
\( \implies y = 7 \)

Substituting \( y = 7 \) into equation (2):
\( 3x - 7 = -1 \)
\( \implies 3x = 6 \)
\( \implies x = 2 \)

Therefore, the required fraction is \( \frac{2}{7} \).
In simple words: Represent the fraction as x/y. Translate both fraction adjustment conditions into two simple equations, then subtract them to find the numerator and denominator.

Exam Tip: Always state the final answer as a fraction \( \frac{x}{y} \) rather than just writing \( x = 2 \) and \( y = 7 \) separately, as the question specifically asks for "the fraction".

Question 6. The sum of the numerator and the denominator of a fraction is equal to 7. Four times the numerator is 8 less than 5 times the denominator. Find the fraction.
Answer:
Let us assume the numerator is \( x \) and the denominator is \( y \).
The sum of both is 7:
\( x + y = 7 \) ...(1)

Also, 4 times the numerator is 8 less than 5 times the denominator:
\( 4x = 5y - 8 \)
\( \implies 5y - 4x = 8 \) ...(2)

Multiplying equation (1) by 4, we get:
\( 4x + 4y = 28 \) ...(3)

Adding equations (2) and (3):
\( 9y = 36 \)
\( \implies y = 4 \)

Substituting \( y = 4 \) back into equation (1):
\( x + 4 = 7 \)
\( \implies x = 3 \)

Hence, the required fraction is \( \frac{3}{4} \).
In simple words: Form two equations based on the sum and relation statements. Use elimination to find x and y, and then write them as a fraction.

Exam Tip: Be precise when translating "8 less than 5 times". It must be written as \( 5y - 8 \), not \( 8 - 5y \).

Question 7. If the numerator of a fraction is doubled and its denominator is increased by 1, its value becomes 1. If its numerator is increased by 4 and its denominator is doubled, its value becomes \( \frac{1}{2} \). Find the fraction.
Answer:
Let the fraction be represented by \( \frac{x}{y} \).
According to the first condition, doubling the numerator and adding 1 to the denominator gives 1:
\( \frac{2x}{y+1} = 1 \)
\( \implies 2x = y + 1 \)
\( \implies 2x - y = 1 \) ...(1)

According to the second condition, adding 4 to the numerator and doubling the denominator gives \( \frac{1}{2} \):
\( \frac{x+4}{2y} = \frac{1}{2} \)
\( \implies 2(x + 4) = 2y \)
\( \implies x + 4 = y \)
\( \implies x - y = -4 \) ...(2)

Subtracting equation (2) from equation (1):
\( (2x - y) - (x - y) = 1 - (-4) \)
\( \implies x = 5 \)

Substituting \( x = 5 \) back into equation (1):
\( 2(5) - y = 1 \)
\( \implies 10 - y = 1 \)
\( \implies y = 9 \)

Thus, the fraction is \( \frac{5}{9} \).
In simple words: Set up the two relationships given as fractional equations. Simplify them to linear equations, subtract to find the numerator, and solve for the denominator.

Exam Tip: Always check your final fraction with the original conditions. For example, if the fraction is \( \frac{5}{9} \), doubling the numerator and adding 1 to the denominator gives \( \frac{10}{10} = 1 \), which verifies your answer.

Exercise 6(E)

Question 8. A fraction becomes \( \frac{1}{2} \) when \( 5 \) is subtracted from its numerator and \( 3 \) is subtracted from its denominator. If the denominator of this fraction is \( 5 \) more than its numerator, find the fraction.
Answer: Let the numerator of the fraction be \( x \) and its denominator be \( y \).
Therefore, the fraction can be represented as \( \frac{x}{y} \).
According to the first given condition:
\( \frac{x - 5}{y - 3} = \frac{1}{2} \)

\( \implies 2(x - 5) = y - 3 \)

\( \implies 2x - 10 = y - 3 \)

\( \implies 2x - y = 7 \) ----(i)
According to the second given condition:
\( x + 5 = y \)

\( \implies x - y = -5 \) ----(ii)
By subtracting equation (ii) from equation (i), we get:
\( x = 12 \)
Substituting \( x = 12 \) in equation (ii):
\( y = 12 + 5 = 17 \)
Thus, the required fraction is \( \frac{12}{17} \).
In simple words: If you subtract 5 from the top of the fraction and 3 from the bottom, it becomes 1/2. Since the bottom is also 5 more than the top, we solve these conditions to find the fraction is 12/17.
Exam Tip: Clearly define the variables at the start and always write the final answer as a fraction rather than just leaving the values of x and y.

Question 9. A fraction becomes \( \frac{1}{2} \) when \( 5 \) is subtracted from its numerator and \( 3 \) is subtracted from its denominator. If the denominator of this fraction is \( 5 \) more than its numerator, find the fraction.
Answer: Let the top value (numerator) of the fraction be \( x \) and the bottom value (denominator) be \( y \).
This makes the fraction \( \frac{x}{y} \).
Based on the first rule in the problem:
\( \frac{x - 5}{y - 3} = \frac{1}{2} \)

\( \implies 2(x - 5) = y - 3 \)

\( \implies 2x - 10 = y - 3 \)

\( \implies 2x - y = 7 \) ----(i)
Based on the second rule:
\( x + 5 = y \)

\( \implies x - y = -5 \) ----(ii)
Subtracting equation (ii) from equation (i) gives:
\( x = 12 \)
Substituting this value into equation (ii):
\( y = 12 + 5 = 17 \)
Hence, the fraction is \( \frac{12}{17} \).
In simple words: We can set up two linear equations using the given conditions of the fraction. Solving these equations tells us the fraction is 12/17.
Exam Tip: When subtracting equations, be extremely careful with the negative signs to avoid calculation errors.

Question 10. The sum of the digits of a two-digit number is \( 7 \). If the digits are reversed, the new number decreased by \( 2 \) is equal to twice the original number. Find the number.
Answer: Let the digit at the unit's place be \( x \) and the digit at the ten's place be \( y \).
So, the original number is \( 10y + x \).
On reversing the digits, the new number is \( 10x + y \).
According to the first condition:
\( x + y = 7 \) ----(1)
According to the second condition:
\( 10x + y - 2 = 2(10y + x) \)

\( \implies 10x + y - 2 = 20y + 2x \)

\( \implies 8x - 19y = 2 \) ----(2)
Multiplying equation (1) by \( 19 \):
\( 19x + 19y = 133 \) ----(3)
Adding equation (2) and equation (3):
\( 27x = 135 \)

\( \implies x = 5 \)
From equation (1):
\( 5 + y = 7 \)

\( \implies y = 2 \)
Thus, the required number is:
\( 10(2) + 5 = 25 \).
In simple words: Representing the two digits as x and y lets us write equations for their sum and their reversed values. Solving these reveals the number is 25.
Exam Tip: Remember that a two-digit number with tens digit y and units digit x is written as 10y + x, not yx.

Question 11. In a two-digit number, the ten's digit is \( 3 \) times the unit's digit. If the sum of the number and its unit's digit is \( 32 \), find the number.
Answer: Let the unit's digit be \( x \) and the ten's digit be \( y \).
The original number is \( 10y + x \).
According to the first condition:
\( y = 3x \)

\( \implies 3x - y = 0 \) ----(1)
According to the second condition:
\( (10y + x) + x = 32 \)

\( \implies 10y + 2x = 32 \) ----(2)
Multiply equation (1) by \( 10 \):
\( 30x - 10y = 0 \) ----(3)
Now, adding equation (3) and equation (2):
\( 32x = 32 \)

\( \implies x = 1 \)
From equation (1):
\( y = 3(1) = 3 \)
Thus, the required number is:
\( 10(3) + 1 = 31 \).
In simple words: Setting up equations based on the relationship of the digits and the sum of the number with its units digit gives the digits 3 and 1, so the number is 31.
Exam Tip: Be careful to read whether the sum is with the units digit specifically, rather than the sum of the digits.

Question 12. The ten's digit of a two-digit number exceeds twice its unit's digit by \( 2 \). If the number itself is \( 5 \) more than \( 3 \) times the sum of its digits, find the number.
Answer: Let the unit's digit be \( x \) and the ten's digit be \( y \).
The original number is \( 10y + x \).
According to the first condition:
\( y - 2x = 2 \)

\( \implies -2x + y = 2 \) ----(1)
According to the second condition:
\( (10y + x) - 3(y + x) = 5 \)

\( \implies 10y + x - 3y - 3x = 5 \)

\( \implies 7x - 2y = 5 \) ----(2)
Multiplying equation (1) by \( 2 \):
\( -4x + 2y = 4 \) ----(3)
Now, adding equation (2) and equation (3):
\( 3x = 9 \)

\( \implies x = 3 \)
From equation (1):
\( -2(3) + y = 2 \)

\( \implies y = 8 \)
So, the required number is:
\( 10(8) + 3 = 83 \).
In simple words: By turning the two hints into linear equations, we find the units digit is 3 and the tens digit is 8, which forms the number 83.
Exam Tip: In digit problems, always check your final digits by substituting them back into the original word problem to verify.

Question 13. Four times a certain two-digit number is seven times the number obtained on interchanging its digits. If the difference between the digits is \( 4 \), find the number.
Answer: Let \( x \) be the digit at the ten's place and \( y \) be the digit at the unit's place.
Thus, the number is \( 10x + y \).
According to the first condition:
\( 4(10x + y) = 7(10y + x) \)

\( \implies 40x + 4y = 70y + 7x \)

\( \implies 33x - 66y = 0 \)

\( \implies x - 2y = 0 \) ----(i)
According to the second condition:
\( x - y = 4 \) ----(ii)
Subtracting equation (i) from equation (ii), we get:
\( y = 4 \)
Substituting \( y = 4 \) in equation (i):
\( x - 2(4) = 0 \)

\( \implies x = 8 \)
Therefore, the required number is \( 10(8) + 4 = 84 \).
In simple words: The problem states that 4 times the number is equal to 7 times its reverse, and the difference of the digits is 4. Solving this gives the number 84.
Exam Tip: Dividing equations by common factors (like dividing by 33 in this case) keeps numbers small and simplifies calculations.

Question 14. The sum of a two-digit number and the number obtained by interchanging its digits is \( 121 \). If the difference between the digits is \( 3 \), find the number.
Answer: Let the digit at the ten's place be \( x \) and the digit at the unit's place be \( y \).
So, the original number is \( 10x + y \).
The number with interchanged digits is \( 10y + x \).
According to the first condition:
\( 10x + y + 10y + x = 121 \)

\( \implies 11x + 11y = 121 \)

\( \implies 11(x + y) = 121 \)

\( \implies x + y = 11 \) ----(i)
According to the second condition:
\( x - y = 3 \) ----(ii)
Adding equation (i) and equation (ii):
\( 2x = 14 \)

\( \implies x = 7 \)
Substituting \( x = 7 \) in (i):
\( 7 + y = 11 \)

\( \implies y = 4 \)
Hence, the required number is \( 74 \).
In simple words: Adding a number and its reverse gives 121, and its digits differ by 3. Solving these together reveals the number is 74.
Exam Tip: For digit sum questions where the sum of a number and its reverse is given, dividing by 11 directly yields the sum of the digits.

Question 15. A two-digit number is \( 8 \) times the sum of its digits. It is also equal to \( 14 \) times the difference of its digits plus \( 2 \). Find the number.
Answer: Let the ten's digit be \( x \) and the unit's digit be \( y \).
The original number is \( 10x + y \).
According to the first condition:
\( 10x + y = 8(x + y) \)

\( \implies 2x = 7y \) ----(i)
According to the second condition:
\( 10x + y = 14(x - y) + 2 \)

\( \implies 4x - 15y = -2 \) ----(ii)
Or,
\( 10x + y = 14(y - x) + 2 \)

\( \implies 24x - 13y = 2 \) ----(iii)
Solving equations (i) and (ii), we find:
\( y = 2 \) and \( x = 7 \)
Solving equations (i) and (iii), we get:
\( y = \frac{2}{71} \)
Since the units digit \( y \) must be a single-digit integer, the second case is not possible.
Thus, the required number is \( 72 \).
In simple words: The number is 8 times its digit sum, and also 14 times the digit difference plus 2. Working through the cases gives the valid digit values as 7 and 2, making the number 72.
Exam Tip: Always discard fractional or negative values when solving for digits of a number, as they must be positive single-digit integers.

Exercise 6(F)

Question 1. Five years ago, the age of A was four times the age of B. Five years hence, the age of A will be twice the age of B. Find their present ages.
Answer: Let A's present age be \( x \) years and B's present age be \( y \) years.
Based on the condition five years ago:
\( x - 5 = 4(y - 5) \)

\( \implies x - 4y = -15 \) ----(1)
Based on the condition five years from now:
\( x + 5 = 2(y + 5) \)

\( \implies x - 2y = 5 \) ----(2)
Subtracting equation (1) from equation (2), we get:
\( 2y = 20 \)

\( \implies y = 10 \)
Substituting \( y = 10 \) in equation (1):
\( x - 4(10) = -15 \)

\( \implies x = 25 \)
Therefore, the present ages of A and B are \( 25 \) years and \( 10 \) years respectively.
In simple words: We look at how old A and B were five years ago and how old they will be in five years. Solving these relations shows A is 25 and B is 10.
Exam Tip: For age problems, remember to add or subtract years from both individuals' present ages when moving forward or backward in time.

Question 2. A's present age is \( 20 \) years more than B's present age. Five years ago, A's age was three times B's age. Find their present ages.
Answer: Let the present age of A be \( x \) years and the present age of B be \( y \) years.
According to the first condition:
\( x = y + 20 \)

\( \implies x - y = 20 \) ----(1)
According to the second condition (five years ago):
\( x - 5 = 3(y - 5) \)

\( \implies x - 3y = -10 \) ----(2)
Subtracting equation (1) from equation (2):
\( -2y = -30 \)

\( \implies y = 15 \)
Substitute \( y = 15 \) into equation (1):
\( x = 15 + 20 = 35 \)
Thus, A's present age is \( 35 \) years and B's present age is \( 15 \) years.
In simple words: Since A is 20 years older than B and was three times B's age five years ago, we can write and solve equations to find A is 35 and B is 15.
Exam Tip: Be sure to write units (years) in the final answer to secure full marks.

Question 3. Four years ago, a mother's age was four times her daughter's age. Six years hence, her age will be two and a half times her daughter's age. Find their present ages.
Answer: Let the present age of the mother be \( x \) years and that of her daughter be \( y \) years.
According to the first condition:
\( x - 4 = 4(y - 4) \)

\( \implies x - 4y = -12 \) ----(i)
According to the second condition:
\( x + 6 = 2.5(y + 6) \)

\( \implies x + 6 = \frac{5}{2}(y + 6) \)

\( \implies 2x + 12 = 5y + 30 \)

\( \implies x - \frac{5}{2}y = 9 \) ----(ii)
Solving equations (i) and (ii), we get:
\( y = 14 \) and \( x = 44 \)
Hence, the mother's present age is \( 44 \) years and the daughter's present age is \( 14 \) years.
In simple words: Translating the relationships of their ages four years ago and six years from now into equations gives the mother's age as 44 and the daughter's as 14.
Exam Tip: Convert fractional or decimal multipliers like "two and a half times" (\( 2.5 \)) to fractions (\( \frac{5}{2} \)) to make cross-multiplication simpler.

Question 4. A man's present age is twice the sum of the ages of his two children. In \( 20 \) years, his age will be equal to the sum of their ages at that time. Find the present age of the man.
Answer: Let the present age of the man be \( x \) years and the sum of the present ages of his two children be \( y \) years.
According to the first condition:
\( x = 2y \) ----(i)
In \( 20 \) years, the man's age will be \( x + 20 \). Since there are two children, each will age by \( 20 \) years, meaning the sum of their ages will increase by \( 40 \) years:
\( x + 20 = y + 40 \) ----(ii)
Substituting equation (i) into (ii):
\( 2y + 20 = y + 40 \)

\( \implies y = 20 \)
So,
\( x = 2(20) = 40 \)
Therefore, the present age of the man is \( 40 \) years.
In simple words: Today, the father is twice as old as the sum of his two kids' ages. In 20 years, since both kids grow older, their combined age increases by 40 years, helping us find the father's age is 40.
Exam Tip: A common mistake is adding only 20 instead of 40 to the children's sum. Remember, for \( n \) people, the sum of ages increases by \( n \times \text{years} \) over time.

Question 5. The annual incomes of A and B are in the ratio \( 3:4 \). If each of them saves Rs. \( 5,000 \) per year, and their expenditures are in the ratio \( 5:7 \), find their annual incomes.
Answer: Let A's annual income be Rs. \( x \) and B's annual income be Rs. \( y \).
According to the income ratio:
\( \frac{x}{y} = \frac{3}{4} \)

\( \implies 4x - 3y = 0 \) ----(1)
Since each saves Rs. \( 5,000 \), their expenditures are \( (x - 5000) \) and \( (y - 5000) \).
According to the expenditure ratio:
\( \frac{x - 5000}{y - 5000} = \frac{5}{7} \)

\( \implies 7(x - 5000) = 5(y - 5000) \)

\( \implies 7x - 35000 = 5y - 25000 \)

\( \implies 7x - 5y = 10000 \) ----(2)
Multiplying equation (1) by \( 7 \) and equation (2) by \( 4 \):
\( 28x - 21y = 0 \) ----(3)
\( 28x - 20y = 40000 \) ----(4)
Subtracting equation (3) from equation (4):
\( y = 40000 \)
From equation (1):
\( 4x - 3(40000) = 0 \)

\( \implies 4x = 120000 \)

\( \implies x = 30000 \)
Thus, A's annual income is Rs. \( 30,000 \) and B's annual income is Rs. \( 40,000 \).
In simple words: Representing their incomes as variables and subtracting savings gives their expenditures. Solving the ratio of expenditures shows A earns Rs. 30,000 and B earns Rs. 40,000.
Exam Tip: Remember the basic formula: Income - Savings = Expenditure, and use it to construct the equations.

Question 6. In an examination, the ratio of pass candidates to fail candidates was \( 4:1 \). If \( 20 \) fewer candidates had appeared and \( 10 \) fewer had failed, the ratio of pass to fail candidates would have been \( 5:1 \). Find the total number of candidates who appeared for the examination.
Answer: Let the number of passing candidates be \( x \) and failing candidates be \( y \).
According to the given ratio:
\( \frac{x}{y} = \frac{4}{1} \)

\( \implies x - 4y = 0 \) ----(1)
If \( 20 \) fewer students appeared, the new total is \( x + y - 20 \). If \( 10 \) fewer failed, the new number of failed students is \( y - 10 \).
The new number of passing students is \( (x + y - 20) - (y - 10) = x - 10 \).
Based on the given condition:
\( \frac{x - 20}{y - 10} = \frac{5}{1} \)

\( \implies x - 20 = 5y - 50 \)

\( \implies x - 5y = -30 \) ----(2)
Subtracting equation (2) from equation (1):
\( y = 30 \)
From equation (1):
\( x - 4(30) = 0 \)

\( \implies x = 120 \)
The total number of candidates who appeared is:
\( x + y = 120 + 30 = 150 \).
In simple words: Setting up equations using the original and hypothetical passing-to-failing ratios tells us there were 120 passing and 30 failing students, giving a total of 150 candidates.
Exam Tip: Make sure to answer the specific question asked — here, the total number of candidates (\( x + y \)), not just the individual values of \( x \) or \( y \).

Question 7. A and B have some pencils. If A gives \( 10 \) pencils to B, B will have twice as many pencils as A. If B gives \( 10 \) pencils to A, they will both have an equal number of pencils. Find the number of pencils each has.
Answer: Let the number of pencils with A be \( x \) and with B be \( y \).
If A gives \( 10 \) pencils to B:
\( y + 10 = 2(x - 10) \)

\( \implies y + 10 = 2x - 20 \)

\( \implies 2x - y = 30 \) ----(1)
If B gives \( 10 \) pencils to A:
\( y - 10 = x + 10 \)

\( \implies x - y = -20 \) ----(2)
Subtracting equation (2) from equation (1):
\( x = 50 \)
From equation (2):
\( 50 - y = -20 \)

\( \implies y = 70 \)
Thus, A has \( 50 \) pencils and B has \( 70 \) pencils.
In simple words: By writing equations to model the sharing of pencils back and forth, we solve to find A started with 50 pencils and B had 70.
Exam Tip: When one person gives items to another, remember to deduct the items from the giver and add them to the receiver.

Question 8. In a group of \( 1250 \) people consisting of adults and children, the ticket for an adult is Rs. \( 75 \) and for a child is Rs. \( 25 \). If the total amount collected is Rs. \( 61250 \), find the number of adults and children in the group.
Answer: Let the number of adults be \( x \) and children be \( y \).
According to the first condition:
\( x + y = 1250 \) ----(1)
According to the second condition:
\( 75x + 25y = 61250 \)
Dividing this equation by \( 25 \):
\( 3x + y = 2450 \) ----(2)
Subtracting equation (1) from equation (2):
\( 2x = 1200 \)

\( \implies x = 600 \)
From equation (1):
\( 600 + y = 1250 \)

\( \implies y = 650 \)
Therefore, there are \( 600 \) adults and \( 650 \) children.
In simple words: Using the total number of visitors and the total cost of their tickets, we solve two linear equations to find that there are 600 adults and 650 children.
Exam Tip: Simplify large coefficients by dividing the entire equation by a common factor (like 25 in this problem) to make calculations much faster.

Question 9. By selling article A at \( 5\% \) profit and article B at \( 7\% \) profit, the total selling price of both is Rs. \( 1167 \). If article A is sold at \( 7\% \) profit and article B is sold at \( 5\% \) profit, the total selling price of both is Rs. \( 1165 \). Find the cost price of each article.
Answer: Let the cost price of article A be Rs. \( x \) and B be Rs. \( y \).
Based on the first condition:
\( \left(x + \frac{5}{100}x\right) + \left(y + \frac{7}{100}y\right) = 1167 \)

\( \implies \frac{105x}{100} + \frac{107y}{100} = 1167 \)

\( \implies 105x + 107y = 116700 \) ----(1)
Based on the second condition:
\( \frac{107x}{100} + \frac{105y}{100} = 1165 \)

\( \implies 107x + 105y = 116500 \) ----(2)
Adding equations (1) and (2):
\( 212x + 212y = 233200 \)

\( \implies x + y = 1100 \) ----(3)
Subtracting equation (1) from equation (2):
\( 2x - 2y = -200 \)

\( \implies -x + y = 100 \) ----(4)
Adding equations (3) and (4):
\( 2y = 1200 \)

\( \implies y = 600 \)
From equation (3):
\( x + 600 = 1100 \)

\( \implies x = 500 \)
So, the cost price of article A is Rs. \( 500 \) and article B is Rs. \( 600 \).
In simple words: Expressing the cost prices and profits as equations lets us solve for the value of each item. This shows article A costs Rs. 500 and article B costs Rs. 600.
Exam Tip: When equations have symmetric coefficients (like 105 and 107 here), adding and subtracting the equations is the easiest path to a solution.

Question 10. Pooja and Ritu can together complete a piece of work in \( \frac{120}{7} \) days. If Pooja works at three-fourths of Ritu's rate, find the number of days each would take to complete the work alone.
Answer: Let the time taken by Pooja to complete the work be \( x \) days and by Ritu be \( y \) days.
Thus, Pooja's 1-day work is \( \frac{1}{x} \) and Ritu's 1-day work is \( \frac{1}{y} \).
Together, their 1-day work is \( \frac{7}{120} \):
\( \frac{1}{x} + \frac{1}{y} = \frac{7}{120} \) ----(1)
We are given that Pooja works at \( \frac{3}{4} \) of Ritu's rate:
\( \frac{1}{x} = \frac{3}{4}\left(\frac{1}{y}\right) \)

\( \implies y = \frac{3}{4}x \) ----(2)
Using the value of \( y \) from (2) in (1):
\( \frac{1}{x} + \frac{4}{3x} = \frac{7}{120} \)

\( \implies \frac{7}{3x} = \frac{7}{120} \)

\( \implies 3x = 120 \)

\( \implies x = 40 \)
From (2):
\( y = \frac{3}{4}(40) = 30 \)
Thus, Pooja will take \( 40 \) days and Ritu will take \( 30 \) days to complete the work individually.
In simple words: Setting up equations for how much work they each do in one day helps us find that Pooja takes 40 days and Ritu takes 30 days to finish alone.
Exam Tip: In work-time problems, always work with the reciprocal of time (the work done in a single day) to add rates correctly.

Exercise 6(G)

Question 1. Rohit has some money and Ajay has some money. If Ajay gives Rs. \( 100 \) to Rohit, Rohit will have twice as much money as Ajay. If Rohit gives Rs. \( 10 \) to Ajay, Ajay will have six times as much money as Rohit. Find how much money each has.
Answer: Let Rohit have Rs. \( x \) and Ajay have Rs. \( y \).
If Ajay gives Rs. \( 100 \) to Rohit:
\( x + 100 = 2(y - 100) \)

\( \implies x - 2y = -300 \) ----(1)
If Rohit gives Rs. \( 10 \) to Ajay:
\( 6(x - 10) = y + 10 \)

\( \implies 6x - y = 70 \) ----(2)
Multiplying equation (2) by \( 2 \):
\( 12x - 2y = 140 \) ----(3)
Subtracting equation (1) from (3):
\( 11x = 440 \)

\( \implies x = 40 \)
From equation (1):
\( 40 - 2y = -300 \)

\( \implies -2y = -340 \)

\( \implies y = 170 \)
Hence, Rohit has Rs. \( 40 \) and Ajay has Rs. \( 170 \).
In simple words: By building equations for both cash-sharing scenarios, we can solve them to find that Rohit has Rs. 40 and Ajay has Rs. 170.
Exam Tip: When setting up equations, make sure the multiplier (like "six times") is applied to the correct person's side.

Question 2. The sum of a two-digit number and the number obtained by reversing its digits is \( 99 \). If the difference between the digits of the number is \( 3 \), find the number.
Answer: Let the tens place digit be \( x \) and the units place digit be \( y \).
The number is \( 10x + y \) and the reversed number is \( 10y + x \).
From the first condition:
\( 10x + y + 10y + x = 99 \)

\( \implies 11x + 11y = 99 \)

\( \implies x + y = 9 \) ----(i)
From the second condition, we have two possible cases:
Case 1:
\( x - y = 3 \) ----(ii)
Solving (i) and (ii):
\( 2x = 12 \)

\( \implies x = 6 \)

\( \implies y = 3 \)
This gives the number \( 63 \).
Case 2:
\( y - x = 3 \) ----(iii)
Solving (i) and (iii):
\( 2y = 12 \)

\( \implies y = 6 \)

\( \implies x = 3 \)
This gives the number \( 36 \).
Thus, the required number is \( 63 \) or \( 36 \).
In simple words: Since we don't know which digit is larger, the difference yields two valid cases. The answer can be either 63 or 36.
Exam Tip: Unless specified which digit is larger, always check both possibilities (\( x - y \) and \( y - x \)) to avoid losing marks.

Question 3. Seven times a two-digit number is equal to four times the number obtained by reversing its digits. If the difference between the digits is \( 3 \), find the number.
Answer: Let the digit at the tens place be \( x \) and the units place be \( y \).
The number is \( 10x + y \) and the reversed number is \( 10y + x \).
According to the first condition:
\( 7(10x + y) = 4(10y + x) \)

\( \implies 70x + 7y = 40y + 4x \)

\( \implies 66x = 33y \)

\( \implies 2x - y = 0 \) ----(1)
Since \( 2x = y \), the units digit \( y \) must be larger than \( x \). Thus, our second equation is:
\( y - x = 3 \) ----(2)
Adding equation (1) and equation (2):
\( x = 3 \)
From equation (1):
\( 2(3) - y = 0 \)

\( \implies y = 6 \)
Hence, the required number is \( 36 \).
In simple words: The problem says seven times the number equals four times its reverse. This tells us the units digit is double the tens digit, which leads to the answer 36.
Exam Tip: Analyzing which digit is larger first (e.g., \( 2x = y \)) tells us which subtraction order to use for the difference, saving steps.

Question 4. The cost of \( 2 \) tickets for station A and \( 3 \) tickets for station B is Rs. \( 77 \). The cost of \( 3 \) tickets for station A and \( 5 \) tickets for station B is Rs. \( 124 \). Find the fare of a ticket for station A and for station B.
Answer: Let the ticket fare for station A be Rs. \( x \) and for station B be Rs. \( y \).
According to the given information:
\( 2x + 3y = 77 \) ----(1)
\( 3x + 5y = 124 \) ----(2)
Multiplying equation (1) by \( 3 \) and equation (2) by \( 2 \):
\( 6x + 9y = 231 \) ----(3)
\( 6x + 10y = 248 \) ----(4)
Subtracting equation (3) from equation (4):
\( y = 17 \)
From equation (1):
\( 2x + 3(17) = 77 \)

\( \implies 2x + 51 = 77 \)

\( \implies 2x = 26 \)

\( \implies x = 13 \)
Thus, the fare for station A is Rs. \( 13 \) and for station B is Rs. \( 17 \).
In simple words: Setting up equations for the two ticket purchase options lets us find that a ticket for A is Rs. 13 and for B is Rs. 17.
Exam Tip: Use the elimination method to solve the simultaneous equations by matching the coefficients of one variable.

Question 5. The sum of the digits of a two-digit number is \( 11 \). If the digit at the tens place is increased by \( 5 \) and the digit at the units place is decreased by \( 5 \), the digits of the number are reversed. Find the number.
Answer: Let the tens digit be \( x \) and the units digit be \( y \).
So, the number is \( 10x + y \).
The sum of the digits is:
\( x + y = 11 \) ----(i)
Based on the second condition:
\( 10(x + 5) + (y - 5) = 10y + x \)

\( \implies 10x + 50 + y - 5 = 10y + x \)

\( \implies 9x - 9y = -45 \)

\( \implies x - y = -5 \) ----(ii)
Subtracting equation (i) from equation (ii):
\( -2y = -16 \)

\( \implies y = 8 \)
Substituting \( y = 8 \) in equation (i):
\( x + 8 = 11 \)

\( \implies x = 3 \)
Therefore, the required number is \( 10(3) + 8 = 38 \).
In simple words: The sum of the digits is 11, and modifying them as stated reverses the number. Solving this reveals the digits are 3 and 8, making the number 38.
Exam Tip: Be careful when simplifying terms like \( 10(x + 5) + (y - 5) \). Expand brackets first to keep the algebra tidy.

Question 6. A chemist has one solution of \( 90\% \) acid and another of \( 97\% \) acid. How many litres of each must be mixed to obtain \( 21 \) litres of a \( 95\% \) acid solution?
Answer: Let the volume of the \( 90\% \) acid solution be \( x \) litres and the \( 97\% \) solution be \( y \) litres.
From the total required volume:
\( x + y = 21 \) ----(1)
From the acid content:
\( 90\% \text{ of } x + 97\% \text{ of } y = 95\% \text{ of } 21 \)

\( \implies \frac{90x}{100} + \frac{97y}{100} = \frac{95 \times 21}{100} \)

\( \implies 90x + 97y = 1995 \) ----(2)
Multiplying equation (1) by \( 90 \):
\( 90x + 90y = 1890 \) ----(3)
Subtracting equation (3) from equation (2):
\( 7y = 105 \)

\( \implies y = 15 \)
From equation (1):
\( x + 15 = 21 \)

\( \implies x = 6 \)
Thus, the chemist needs \( 6 \) litres of the \( 90\% \) solution and \( 15 \) litres of the \( 97\% \) solution.
In simple words: By making equations for the total volume and the total amount of acid, we can solve to see we need 6 litres of the weaker acid and 15 litres of the stronger acid.
Exam Tip: When working with percentages in mixing problems, multiplying the entire equation by 100 gets rid of the denominators quickly.

Question 7. A person bought two types of sweets. The first kind cost Rs. \( 250 \) per kg and the second kind cost Rs. \( 350 \) per kg. If the total quantity of sweets bought was \( 40 \) kg and the total cost was Rs. \( 11,800 \), find the quantity of each kind of sweets purchased.
Answer: Let the quantity of the first kind of sweet be \( x \) kg and the second kind be \( y \) kg.
The total quantity of sweets is:
\( x + y = 40 \) ----(i)
The total cost of the sweets is:
\( 250x + 350y = 11800 \) ----(ii)
Multiplying equation (i) by \( 250 \):
\( 250x + 250y = 10000 \)
Subtracting this from equation (ii):
\( 100y = 1800 \)

\( \implies y = 18 \)
From equation (i):
\( x + 18 = 40 \)

\( \implies x = 22 \)
Therefore, \( 22 \) kg of the first kind of sweet and \( 18 \) kg of the second kind of sweet were bought.
In simple words: Using the total weight and the total cost, we write two equations to find that 22 kg of the cheaper sweets and 18 kg of the more expensive ones were purchased.
Exam Tip: Simplify the cost equation by dividing by 10 first to make the coefficients easier to handle.

Question 8. Mr. Ahuja weighs \( 5 \) kg more than Mrs. Ahuja. If Mr. Ahuja loses \( 5 \) kg by dieting, his weight will equal Mrs. Ahuja's weight. If Mrs. Ahuja also loses \( 4 \) kg, her weight will be \( \frac{7}{8} \) of Mr. Ahuja's original weight. Find their original weights.
Answer: Let Mr. Ahuja's original weight be \( x \) kg and Mrs. Ahuja's original weight be \( y \) kg.
According to the first condition:
\( x - 5 = y \)

\( \implies x - y = 5 \) ----(1)
According to the second condition:
\( y - 4 = \frac{7}{8}x \)

\( \implies 8(y - 4) = 7x \)

\( \implies 7x - 8y = -32 \) ----(2)
Multiplying equation (1) by \( 7 \):
\( 7x - 7y = 35 \) ----(3)
Subtracting equation (2) from (3):
\( y = 67 \)
From equation (1):
\( x - 67 = 5 \)

\( \implies x = 72 \)
Thus, Mr. Ahuja originally weighed \( 72 \) kg and Mrs. Ahuja originally weighed \( 67 \) kg.
In simple words: By writing equations to model their weights before and after weight loss, we find Mr. Ahuja's original weight is 72 kg and Mrs. Ahuja's is 67 kg.
Exam Tip: Be careful with the placement of fractional multipliers. "Mrs. Ahuja's weight is 7/8 of Mr. Ahuja's weight" means Mrs. Ahuja's weight = 7/8 * Mr. Ahuja's weight.

Question 9. The monthly expense of a family has a constant part and a variable part which depends on the number of members in the family. For a family of \( 4 \) people, the total monthly expense is Rs. \( 10,400 \), and for a family of \( 7 \) people, it is Rs. \( 15,800 \). Find the constant expense per month and the expense per member.
Answer: Let the constant monthly expense be Rs. \( x \) and the variable monthly expense per member be Rs. \( y \).
For a family of \( 4 \) people:
\( x + 4y = 10400 \) ----(i)
For a family of \( 7 \) people:
\( x + 7y = 15800 \) ----(ii)
Subtracting equation (i) from equation (ii):
\( 3y = 5400 \)

\( \implies y = 1800 \)
Substituting \( y = 1800 \) into equation (i):
\( x + 4(1800) = 10400 \)

\( \implies x + 7200 = 10400 \)

\( \implies x = 3200 \)
Therefore, the constant monthly expense is Rs. \( 3,200 \) and the expense per member is Rs. \( 1,800 \).
In simple words: The family has fixed costs plus a cost per member. Solving the linear equations shows the fixed monthly cost is Rs. 3,200 and the cost per person is Rs. 1,800.
Exam Tip: In fixed-and-variable charge problems, the variable charge is multiplied by the number of units/members, while the fixed charge remains unmultiplied.

Question 10. The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of \( 10 \) km, the charge paid is Rs. \( 315 \) and for a journey of \( 15 \) km, the charge paid is Rs. \( 465 \). What will a person have to pay for travelling a distance of \( 32 \) km?
Answer: Let the fixed charge be Rs. \( x \) and the rate per kilometer be Rs. \( y \).
According to the given conditions:
\( x + 10y = 315 \) ----(i)
\( x + 15y = 465 \) ----(ii)
Subtracting equation (i) from equation (ii):
\( 5y = 150 \)

\( \implies y = 30 \)
Substituting \( y = 30 \) in equation (i):
\( x + 10(30) = 315 \)

\( \implies x = 15 \)
So, the fixed charge is Rs. \( 15 \) and the rate per km is Rs. \( 30 \).
To travel \( 32 \) km, a person must pay:
\( \text{Fare} = x + 32y = 15 + 32(30) = 15 + 960 = 975 \)
Thus, the total fare is Rs. \( 975 \).
In simple words: Taxi fares have a fixed start cost and a cost per kilometer. We find these are Rs. 15 and Rs. 30, so a 32 km trip costs Rs. 975 in total.
Exam Tip: Be sure to perform the final step of calculating the fare for the requested distance (32 km) rather than just finding the values of x and y.

Question 11. A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs. \( 27 \) for a book kept for seven days, while Susy paid Rs. \( 21 \) for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Answer: Let the fixed charge for the first \( 3 \) days be Rs. \( x \) and the daily rate for each extra day be Rs. \( y \).
For a book kept for \( 7 \) days (\( 3 \) fixed days and \( 4 \) extra days):
\( x + 4y = 27 \) ----(i)
For a book kept for \( 5 \) days (\( 3 \) fixed days and \( 2 \) extra days):
\( x + 2y = 21 \) ----(ii)
Subtracting equation (ii) from equation (i):
\( 2y = 6 \)

\( \implies y = 3 \)
Substituting \( y = 3 \) in equation (ii):
\( x + 2(3) = 21 \)

\( \implies x = 15 \)
Thus, the fixed charge is Rs. \( 15 \) and the charge for each extra day is Rs. \( 3 \).
In simple words: The library charges Rs. 15 for the first 3 days, and then Rs. 3 for each extra day. We find this by comparing what two different people paid.
Exam Tip: In library problems, subtract the number of fixed days from the total days to find the coefficient of the extra-day rate.

Question 12. The area of a rectangle is reduced by \( 9 \) square units if its length is reduced by \( 5 \) units and breadth is increased by \( 3 \) units. If the length is increased by \( 3 \) units and breadth is increased by \( 2 \) units, the area increases by \( 67 \) square units. Find the length and breadth of the rectangle.
Answer: Let the length of the rectangle be \( x \) units and its breadth be \( y \) units.
Its original area is \( xy \) square units.
According to the first condition:
\( (x - 5)(y + 3) = xy - 9 \)

\( \implies xy + 3x - 5y - 15 = xy - 9 \)

\( \implies 3x - 5y = 6 \) ----(i)
According to the second condition:
\( (x + 3)(y + 2) = xy + 67 \)

\( \implies xy + 2x + 3y + 6 = xy + 67 \)

\( \implies 2x + 3y = 61 \) ----(ii)
Multiply equation (i) by \( 2 \) and equation (ii) by \( 3 \):
\( 6x - 10y = 12 \) ----(iii)
\( 6x + 9y = 183 \) ----(iv)
Subtracting equation (iii) from (iv):
\( 19y = 171 \)

\( \implies y = 9 \)
Substituting \( y = 9 \) in equation (i):
\( 3x - 5(9) = 6 \)

\( \implies 3x = 51 \)

\( \implies x = 17 \)
Therefore, the length of the rectangle is \( 17 \) units and its breadth is \( 9 \) units.
In simple words: Expressing the modifications to the rectangle's dimensions and their impact on the area yields a system of equations. Solving this tells us the length is 17 units and the breadth is 9 units.
Exam Tip: Make sure to cancel out the \( xy \) term from both sides of the expanded equation to leave a simple linear form.

Question 13. Two pipes of different diameters can together fill a swimming pool in \( 12 \) hours. If the pipe with the larger diameter is used for \( 4 \) hours and the pipe with the smaller diameter for \( 9 \) hours, only half of the pool is filled. How long would it take for each pipe alone to fill the pool?
Answer: Let the filling rate of the larger pipe (pipe A) be \( x \) pool per hour and that of the smaller pipe (pipe B) be \( y \) pool per hour.
Together, they fill the pool in \( 12 \) hours, so:
\( x + y = \frac{1}{12} \)

\( \implies 12x + 12y = 1 \) ----(i)
If the larger pipe works for \( 4 \) hours and the smaller for \( 9 \) hours, they fill half the pool:
\( 4x + 9y = \frac{1}{2} \)

\( \implies 8x + 18y = 1 \) ----(ii)
Multiply equation (i) by \( 2 \) and equation (ii) by \( 3 \):
\( 24x + 24y = 2 \) ----(iii)
\( 24x + 54y = 3 \) ----(iv)
Subtracting equation (iii) from equation (iv):
\( 30y = 1 \)

\( \implies y = \frac{1}{30} \)
Substituting \( y = \frac{1}{30} \) in equation (i):
\( 12x + 12\left(\frac{1}{30}\right) = 1 \)

\( \implies 12x + \frac{2}{5} = 1 \)

\( \implies 12x = \frac{3}{5} \)

\( \implies x = \frac{1}{20} \)
Thus, the larger pipe will take \( 20 \) hours and the smaller pipe will take \( 30 \) hours to fill the swimming pool individually.
In simple words: By comparing how much work they do individually in an hour, we solve equations for when they run together and when they run for different durations. This tells us the larger pipe needs 20 hours and the smaller pipe needs 30 hours.
Exam Tip: Define variables as work rates (e.g., fraction of pool filled per hour) to make the addition of joint rates straightforward.