CBSE Class 10 Mathematics Linear Equations Assignment Set 03

Important Concepts

Take a Look:

• Equation: A mathematical statement asserting that two algebraic expressions are equal.
• Linear Equation in One Variable: An equation of the form \( ax + b = 0 \), where \( a \) and \( b \) are real numbers and \( a \neq 0 \).
• Linear Equation in Two Variables: An equation of the form \( ax + by + c = 0 \), where \( a \), \( b \), and \( c \) are real numbers such that \( a \neq 0 \) and \( b \neq 0 \).
• Consistent System of Linear Equations: A pair of linear equations in two variables is consistent if it has at least one common solution.
• Inconsistent System of Linear Equations: A pair of linear equations in two variables is inconsistent if it has no common solution.

For a given system of linear equations in two variables:

\[ a_1x + b_1y + c_1 = 0 \]

\[ a_2x + b_2y + c_2 = 0 \]

• (i) Inconsistent (No Solution):
  This occurs if the lines are parallel. The condition is:
  \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
• (ii) Infinitely Many Solutions (Consistent):
  This occurs if the lines are coincident (lie on top of each other). The condition is:
  \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
• (iii) Unique Solution (Consistent):
  This occurs if the lines intersect at exactly one point. The condition is:
  \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)

Algebraic Methods for Solving a Pair of Linear Equations:

1. Method of Substitution: Expressing one variable in terms of the other from one equation and substituting it into the second equation.
2. Method of Elimination: Multiplying one or both equations by suitable non-zero constants so that the coefficients of one variable become equal, then adding or subtracting the equations to eliminate that variable.
3. Method of Cross-Multiplication: For the system:
   \[ a_1x + b_1y + c_1 = 0 \]
   \[ a_2x + b_2y + c_2 = 0 \]
   We can write:
   \[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
   Which gives:
   \( x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \) and \( y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \)
   This can be visualised using the following diagram:

Level- I
 

Question. The pair 2x=3y-5 and 2y= 5x-4 of linear equations represents two lines which are
(a) Parallel
(b) coincident
(c) intersecting
(d) either parallel or coincident
Answer: (c) intersecting
Rewriting the given equations in standard form \( ax + by + c = 0 \):
First equation: \( 2x - 3y + 5 = 0 \), which gives \( a_1 = 2, b_1 = -3, c_1 = 5 \).
Second equation: \( 5x - 2y - 4 = 0 \), which gives \( a_2 = 5, b_2 = -2, c_2 = -4 \).
Comparing the coefficients, we get:
\( a_1/a_2 = 2/5 \)
\( b_1/b_2 = -3/-2 = 3/2 \)
Since \( a_1/a_2 \neq b_1/b_2 \), the given pair represents intersecting lines.
In simple words: When we compare the ratio of the coefficients of x and y, they are not equal. This means the two lines will cross each other at a single point, so they are intersecting lines.
Exam Tip: Always arrange the linear equations in the standard form \( ax + by + c = 0 \) before comparing coefficients to avoid sign errors.

Question. The pair x=p and y=q of the linear equations in two variables x and y graphically represents two lines which are
(a) Parallel
(b) coincident
(c) intersecting at(p,q)
(d) intersecting at(q,p)
Answer: (c) intersecting at(p,q)
The line \( x = p \) is parallel to the y-axis and passes through all points where the x-coordinate is \( p \). The line \( y = q \) is parallel to the x-axis and passes through all points where the y-coordinate is \( q \). These two perpendicular lines will intersect each other at the point \( (p, q) \).
In simple words: The line \( x = p \) is a straight vertical line, and \( y = q \) is a straight horizontal line. They cross each other exactly at the point \( (p, q) \).
Exam Tip: Remember that any line of the form \( x = \text{constant} \) is vertical, and \( y = \text{constant} \) is horizontal. They always intersect at \( (\text{constant}_x, \text{constant}_y) \).

Question. If the lines represented by the pair of linear equations 2x+5y=3 and (k+1)x +2(k+2)y=2k are coincident, then the value of k is
(a) -3
(b) 3
(c) 1
(d) -2
Answer: (b) 3
For two linear equations to represent coincident lines, the ratio of their coefficients must be equal:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
Here, \( a_1 = 2, b_1 = 5, c_1 = -3 \) and \( a_2 = k+1, b_2 = 2(k+2), c_2 = -2k \).
Substituting these values, we get:
\( \frac{2}{k+1} = \frac{5}{2(k+2)} = \frac{-3}{-2k} \)
Taking the first two ratios:
\( \frac{2}{k+1} = \frac{5}{2k+4} \)
\( \implies 2(2k+4) = 5(k+1) \)
\( \implies 4k + 8 = 5k + 5 \)
\( \implies k = 3 \)
Checking with the third ratio, \( \frac{2}{3+1} = \frac{1}{2} \) and \( \frac{-3}{-2(3)} = \frac{1}{2} \). Since both equations give the same value, \( k = 3 \).
In simple words: When two lines lie on top of each other, the ratios of their x-coefficients, y-coefficients, and constant numbers are all equal. Solving this ratio gives us \( k = 3 \).
Exam Tip: Always double-check your value of \( k \) by substituting it back into all three ratios to make sure they are fully equal.

Question. If the pair of linear equations (3k+1)x+3y-2=0 and (k2 +1 )x+(k-2)y-5=0 inconsistent, then The value of k is
(a) 1
(b) -1
(c) 2
(d) -2
Answer: (b) -1
For the system to be inconsistent, the lines must be parallel, which requires:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
Given:
\( a_1 = 3k+1, b_1 = 3, c_1 = -2 \)
\( a_2 = k^2+1, b_2 = k-2, c_2 = -5 \)
Substituting these into the condition:
\( \frac{3k+1}{k^2+1} = \frac{3}{k-2} \neq \frac{2}{5} \)
Solving the equality:
\( (3k+1)(k-2) = 3(k^2+1) \)
\( \implies 3k^2 - 5k - 2 = 3k^2 + 3 \)
\( \implies -5k = 5 \)
\( \implies k = -1 \)
For \( k = -1 \), the ratio \( \frac{b_1}{b_2} = \frac{3}{-3} = -1 \), which is indeed not equal to \( \frac{2}{5} \). Thus, \( k = -1 \).
In simple words: For the lines to be parallel (no solution), the ratio of the x-parts must equal the ratio of the y-parts, but not the constant parts. Cross-multiplying and solving shows that \( k \) must be \( -1 \).
Exam Tip: Don't forget to verify that the final ratio is indeed unequal to \( \frac{c_1}{c_2} \) to ensure the lines are parallel and not coincident.

Question. If the pair of linear equations 2x+3y=11 and 2px+(p+q)y=p+5q has infinitely many solution Then (a) p=2q (b)q=2p (c)p=-2q (d) q= -2p
(a) p=2q
(b) q=2p
(c) p=-2q
(d) q= -2p
Answer: (b) q=2p
For a system to have infinitely many solutions, the lines must be coincident:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
Here:
\( a_1 = 2, b_1 = 3, c_1 = 11 \)
\( a_2 = 2p, b_2 = p+q, c_2 = p+5q \)
Equating the ratios:
\( \frac{2}{2p} = \frac{3}{p+q} = \frac{11}{p+5q} \)
Using the first two parts:
\( \frac{1}{p} = \frac{3}{p+q} \)
\( \implies p + q = 3p \)
\( \implies q = 2p \)
We can verify this relation by substituting \( q = 2p \) into the third ratio, which simplifies to \( \frac{11}{p+5(2p)} = \frac{11}{11p} = \frac{1}{p} \). This matches the first ratio, confirming \( q = 2p \).
In simple words: Since there are infinite solutions, all three coefficient ratios are equal. Setting the first ratio equal to the second ratio simplifies directly to \( q = 2p \).
Exam Tip: Simplify ratios first (like writing \( \frac{2}{2p} \) as \( \frac{1}{p} \)) to make cross-multiplication much quicker and less prone to errors.

Level- II
 

Question. Find the value of k for which the given system of equations has unique solution: 2x+3y-5=0, kx-6y-8=0
Answer: For the given pair of linear equations to have a unique solution, the lines must intersect at one point. The condition for this is:
\( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
From the equations, we have:
\( a_1 = 2, b_1 = 3 \)
\( a_2 = k, b_2 = -6 \)
Substituting these values:
\( \frac{2}{k} \neq \frac{3}{-6} \)
\( \implies \frac{2}{k} \neq -\frac{1}{2} \)
\( \implies k \neq -4 \)
Thus, the system will have a unique solution for all real values of \( k \) except \( -4 \).
In simple words: For a unique solution, the lines must not be parallel. Comparing the ratios of \( x \) and \( y \) coefficients shows that \( k \) can be any number except \( -4 \).
Exam Tip: When writing the final answer for unique solutions, clearly specify that \( k \) can take "any real value except" the calculated number, rather than just writing \( k \neq -4 \).

Question. For what value of k will the following system of linear equations have infinite number of solution. 2x+3y=2; (k+2)x+(2k+1)y=2(k-1)
Answer: To have infinitely many solutions, the given system of linear equations must satisfy the following condition:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
For the equations \( 2x + 3y - 2 = 0 \) and \( (k+2)x + (2k+1)y - 2(k-1) = 0 \):
\( a_1 = 2, b_1 = 3, c_1 = -2 \)
\( a_2 = k+2, b_2 = 2k+1, c_2 = -2(k-1) \)
Substituting these in the condition, we get:
\( \frac{2}{k+2} = \frac{3}{2k+1} = \frac{-2}{-2(k-1)} \)
Let us solve using the first two ratios:
\( \frac{2}{k+2} = \frac{3}{2k+1} \)
\( \implies 2(2k+1) = 3(k+2) \)
\( \implies 4k + 2 = 3k + 6 \)
\( \implies k = 4 \)
Let us verify by putting \( k = 4 \) into the third ratio:
\( \frac{-2}{-2(4-1)} = \frac{1}{3} \)
Since \( \frac{2}{4+2} = \frac{1}{3} \) and \( \frac{3}{2(4)+1} = \frac{1}{3} \), the value of \( k = 4 \) satisfies all ratios. Therefore, \( k = 4 \).
In simple words: When there are infinitely many solutions, the lines are completely identical, so their coefficient ratios are equal. Solving the relation between the x and y ratios gives us \( k = 4 \).
Exam Tip: Be careful with signs when writing down \( c_2 = -2(k-1) \). Remember to simplify negative signs in ratios early in your calculation to avoid confusion.

Question. Find two numbers whose sum is 18 and difference is6.
Answer: Let the two required numbers be represented by \( x \) and \( y \).
According to the problem, we can set up the following equations:
1) \( x + y = 18 \)
2) \( x - y = 6 \)
Let us add the two equations to eliminate \( y \):
\( (x + y) + (x - y) = 18 + 6 \)
\( \implies 2x = 24 \)
\( \implies x = 12 \)
Now, substitute \( x = 12 \) into the first equation:
\( 12 + y = 18 \)
\( \implies y = 6 \)
Thus, the two numbers are 12 and 6.
In simple words: We can set up two simple equations using the sum and the difference of the two numbers. Adding them together lets us easily find that the numbers are 12 and 6.
Exam Tip: For simple word problems, always define your variables clearly at the start of your answer to secure step-marks.

Question. Solve for x and y. X+6/y=6, 3x-8/y=5.
Answer: We are given the system of equations:
1) \( x + \frac{6}{y} = 6 \)
2) \( 3x - \frac{8}{y} = 5 \)
To make these equations easier to solve, let us substitute \( \frac{1}{y} = u \). This gives:
3) \( x + 6u = 6 \)
4) \( 3x - 8u = 5 \)
Multiplying equation (3) by 3 to match the \( x \) coefficients:
5) \( 3x + 18u = 18 \)
Subtracting equation (4) from equation (5):
\( (3x + 18u) - (3x - 8u) = 18 - 5 \)
\( \implies 26u = 13 \)
\( \implies u = \frac{1}{2} \)
Substituting this back, since \( u = \frac{1}{y} \):
\( \frac{1}{y} = \frac{1}{2} \implies y = 2 \)
Now, substitute \( u = \frac{1}{2} \) into equation (3):
\( x + 6\left(\frac{1}{2}\right) = 6 \)
\( \implies x + 3 = 6 \)
\( \implies x = 3 \)
Thus, the solution is \( x = 3 \) and \( y = 2 \).
In simple words: Since \( y \) is in the denominator, we can substitute \( \frac{1}{y} \) with a new letter, say \( u \), to make the equations look simpler. After finding \( u \) and \( x \), we can easily flip \( u \) back to find \( y \).
Exam Tip: Substituting a variable in the denominator with a new variable like \( u \) or \( v \) is a standard technique to convert non-linear equations into linear ones.

Question. The sum of the numerator and the denominator of a fraction is 20 if we subtract 5 from the numerator and 5 from denominator, then the ratio of the numerator and the denominator will be 1:4 .Find the fraction.
Answer: Let the numerator of the fraction be \( x \) and its denominator be \( y \). Thus, the fraction is \( \frac{x}{y} \).
From the first condition, the sum of the numerator and denominator is 20:
1) \( x + y = 20 \)
From the second condition, if we subtract 5 from both, the ratio becomes 1:4:
\( \frac{x-5}{y-5} = \frac{1}{4} \)
\( \implies 4(x-5) = 1(y-5) \)
\( \implies 4x - 20 = y - 5 \)
2) \( 4x - y = 15 \)
Now, add equation (1) and equation (2) to eliminate \( y \):
\( (x + y) + (4x - y) = 20 + 15 \)
\( \implies 5x = 35 \)
\( \implies x = 7 \)
Substitute \( x = 7 \) into equation (1):
\( 7 + y = 20 \)
\( \implies y = 13 \)
Hence, the required fraction is \( \frac{7}{13} \).
In simple words: We can represent the top and bottom of the fraction as \( x \) and \( y \). Using the clues, we write two equations and solve them to find that the top is 7 and the bottom is 13.
Exam Tip: Always write your final answer as a fraction \( \frac{x}{y} \) rather than leaving it as separate values of \( x \) and \( y \).
 

Level- III
 

Question. Solve the following system of equations by using the method of elimination by equating the coefficients: x/10+y/5+1=15, x/8+y/6=15.
Answer: We are given the following system of equations:
1) \( \frac{x}{10} + \frac{y}{5} + 1 = 15 \)
2) \( \frac{x}{8} + \frac{y}{6} = 15 \)
First, let us simplify equation (1):
\( \frac{x}{10} + \frac{y}{5} = 14 \)
Multiplying the entire equation by 10 (LCM of 10 and 5):
3) \( x + 2y = 140 \)
Next, simplify equation (2) by multiplying by 24 (LCM of 8 and 6):
4) \( 3x + 4y = 360 \)
To use the method of elimination by equating coefficients, we multiply equation (3) by 2 to get matching \( y \) coefficients:
5) \( 2x + 4y = 280 \)
Subtracting equation (5) from equation (4):
\( (3x + 4y) - (2x + 4y) = 360 - 280 \)
\( \implies x = 80 \)
Now, substitute \( x = 80 \) into equation (3):
\( 80 + 2y = 140 \)
\( \implies 2y = 60 \)
\( \implies y = 30 \)
Thus, the solution is \( x = 80 \) and \( y = 30 \).
In simple words: First, we clear the fractions by multiplying by the common denominators. Then, we make the \( y \)-parts equal in both equations so we can subtract them and find \( x \) and \( y \) easily.
Exam Tip: Simplifying fractional equations into standard linear form is always the safest first step to prevent arithmetic errors during elimination.

Question. If two digit number is four times the sum of its digits and twice the product of digits. Find the number.
Answer: Let the digit at the tens place be \( x \) and the digit at the units place be \( y \).
Therefore, the two-digit number is represented by \( 10x + y \).
According to the first condition, the number is four times the sum of its digits:
\( 10x + y = 4(x + y) \)
\( \implies 10x + y = 4x + 4y \)
\( \implies 6x = 3y \)
1) \( y = 2x \)
According to the second condition, the number is twice the product of its digits:
2) \( 10x + y = 2xy \)
Now, substitute the value of \( y \) from equation (1) into equation (2):
\( 10x + 2x = 2x(2x) \)
\( \implies 12x = 4x^2 \)
Dividing both sides by \( 4x \) (since \( x \neq 0 \) for a two-digit number):
\( 3 = x \)
Substitute \( x = 3 \) back into equation (1):
\( y = 2(3) = 6 \)
Thus, the required number is \( 10(3) + 6 = 36 \).
In simple words: We can represent a two-digit number as \( 10x + y \). Using the given relations, we find a simple link between the digits, which helps us solve for them to get the number 36.
Exam Tip: Remember that a two-digit number is always written in the expanded form \( 10x + y \), where \( x \) is the tens digit and \( y \) is the units digit.

Question. Solve the following system of equations. bx/a- ay/b +a +b=0 bx –ay +2ab=0
Answer: We are given the system of equations:
1) \( \frac{bx}{a} - \frac{ay}{b} + a + b = 0 \)
2) \( bx - ay + 2ab = 0 \)
First, let us multiply equation (1) by \( ab \) to clear the denominators:
\( b^2x - a^2y + a^2b + ab^2 = 0 \) --- (3)
Next, let us multiply equation (2) by \( b \) to equate the coefficients of \( x \):
\( b^2x - aby + 2ab^2 = 0 \) --- (4)
Subtracting equation (4) from equation (3):
\( (b^2x - a^2y + a^2b + ab^2) - (b^2x - aby + 2ab^2) = 0 \)
\( \implies -a^2y + aby + a^2b - ab^2 = 0 \)
Let us rearrange and factorise:
\( aby - a^2y + a^2b - ab^2 = 0 \)
\( \implies ay(b - a) - ab(b - a) = 0 \)
\( \implies a(b - a)(y - b) = 0 \)
Since \( a \neq 0 \) and \( a \neq b \):
\( y - b = 0 \)
\( \implies y = b \)
Now, substitute \( y = b \) into equation (2):
\( bx - a(b) + 2ab = 0 \)
\( \implies bx + ab = 0 \)
\( \implies bx = -ab \)
Since \( b \neq 0 \):
\( x = -a \)
Thus, the correct solution is \( x = -a \) and \( y = b \).
In simple words: We can clear the fraction by multiplying the first equation by \( ab \). Then, we subtract the two equations to eliminate \( x \) and find \( y = b \) and \( x = -a \).
Exam Tip: Be very careful when factoring algebraic expressions like \( a(b-a)(y-b) = 0 \). Keeping your steps clear helps you identify the solutions for \( x \) and \( y \) without getting lost in the variables.

Question. Solve graphically the system of linear equations. 4x-3y+4=0, 4x +3y=20 also find the area of the region bounded by the lines and x-axis.
Answer: To solve the system of linear equations graphically, we find at least two points for each line:
For the first line \( 4x - 3y + 4 = 0 \):
\( y = \frac{4x + 4}{3} \)
- When \( x = -1 \), \( y = 0 \). Point \( A(-1, 0) \)
- When \( x = 2 \), \( y = 4 \). Point \( P(2, 4) \)

For the second line \( 4x + 3y = 20 \):
\( y = \frac{20 - 4x}{3} \)
- When \( x = 5 \), \( y = 0 \). Point \( C(5, 0) \)
- When \( x = 2 \), \( y = 4 \). Point \( P(2, 4) \)

Plotting these points on a graph paper and drawing the straight lines, we observe that the two lines intersect each other at \( P(2, 4) \). Therefore, the solution to the system is \( x = 2 \) and \( y = 4 \).

Finding the Area of the Bounded Region:
The region bounded by the two lines and the x-axis forms a triangle \( APC \).
The vertices of this triangle are \( A(-1, 0) \), \( C(5, 0) \), and the intersection point \( P(2, 4) \).
- The base \( AC \) of the triangle is along the x-axis:
\( \text{Base} = 5 - (-1) = 6 \) units
- The height of the triangle is the perpendicular distance from \( P \) to the x-axis, which is the y-coordinate of \( P \):
\( \text{Height} = 4 \) units
- The area of the triangle is:
\( \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \)
\( \implies \text{Area} = \frac{1}{2} \times 6 \times 4 = 12 \) square units.
In simple words: We plot both lines on a graph and find where they cross, which is at \( (2, 4) \). The shape formed between these lines and the x-axis is a triangle. Using the basic formula for the area of a triangle, we get 12 square units.
Exam Tip: Label the intersection point and the x-intercepts clearly on your graph, and write the final area with "square units" or "sq. units" to score full marks.

Question. The sum of two naturals number is 8 and sum of their reciprocals is 8/15. Find the numbers
Answer: Let the two natural numbers be denoted by \( x \) and \( y \).
We are given:
1) \( x + y = 8 \)
2) \( \frac{1}{x} + \frac{1}{y} = \frac{8}{15} \)
Let us simplify equation (2) by finding a common denominator:
\( \frac{x+y}{xy} = \frac{8}{15} \)
Substitute the value \( x + y = 8 \) from equation (1) into this:
\( \frac{8}{xy} = \frac{8}{15} \)
\( \implies xy = 15 \)
From equation (1), we can express \( y \) as \( 8 - x \). Substituting this into \( xy = 15 \):
\( x(8 - x) = 15 \)
\( \implies 8x - x^2 = 15 \)
\( \implies x^2 - 8x + 15 = 0 \)
Factoring this quadratic equation:
\( x^2 - 5x - 3x + 15 = 0 \)
\( \implies x(x - 5) - 3(x - 5) = 0 \)
\( \implies (x - 5)(x - 3) = 0 \)
This gives \( x = 5 \) or \( x = 3 \).
If \( x = 5 \), then \( y = 3 \). If \( x = 3 \), then \( y = 5 \).
Thus, the two required natural numbers are 5 and 3.
In simple words: Using the sum of the numbers and the sum of their reciprocals, we can find that their product must be 15. The only two natural numbers that add up to 8 and multiply to 15 are 5 and 3.
Exam Tip: Since the question specifies "natural numbers," check that your final values are indeed positive integers before finalizing.

Level- IV
 

Question. Solve for x and y: 2/2x+y – 1/x-2y +5/9 =0 9/2x+y – 6/x-2y +4 =0
Answer: We are given the system:
1) \( \frac{2}{2x+y} - \frac{1}{x-2y} + \frac{5}{9} = 0 \)
2) \( \frac{9}{2x+y} - \frac{6}{x-2y} + 4 = 0 \)
Let us substitute \( \frac{1}{2x+y} = u \) and \( \frac{1}{x-2y} = v \). This gives the linear system:
3) \( 2u - v = -\frac{5}{9} \)
4) \( 9u - 6v = -4 \)
Multiplying equation (3) by 6 to equate the coefficients of \( v \):
5) \( 12u - 6v = -\frac{10}{3} \)
Subtracting equation (4) from equation (5):
\( (12u - 6v) - (9u - 6v) = -\frac{10}{3} - (-4) \)
\( \implies 3u = \frac{2}{3} \)
\( \implies u = \frac{2}{9} \)
Substitute \( u = \frac{2}{9} \) into equation (3):
\( 2\left(\frac{2}{9}\right) - v = -\frac{5}{9} \)
\( \implies \frac{4}{9} - v = -\frac{5}{9} \)
\( \implies v = 1 \)
Now substitute back \( u \) and \( v \) to get equations in \( x \) and \( y \):
\( 2x + y = \frac{9}{2} \implies 4x + 2y = 9 \) --- (6)
\( x - 2y = 1 \) --- (7)
Adding equations (6) and (7) to eliminate \( y \):
\( (4x + 2y) + (x - 2y) = 9 + 1 \)
\( \implies 5x = 10 \)
\( \implies x = 2 \)
Substitute \( x = 2 \) into equation (7):
\( 2 - 2y = 1 \)
\( \implies 2y = 1 \)
\( \implies y = \frac{1}{2} \)
Thus, the solution is \( x = 2 \) and \( y = \frac{1}{2} \).
In simple words: By substituting the complex denominators with simpler variables \( u \) and \( v \), we solve for them first. Then, we substitute back to get simpler linear equations which we easily solve to find \( x = 2 \) and \( y = 1/2 \).
Exam Tip: Substituting variables inside complex denominators is highly likely to appear in long-answer questions. Solve each stage with high precision as a single sign mistake can make the final system of equations unsolvable.

Question. Draw the graph the following equations: 2x+3y-12=0 and 7x-3y-15=0.Determine the coordinates of the vertices of the triangle formed by the lines and the y-axis.
Answer: We can determine points to plot each line as follows:
For the first line \( 2x + 3y - 12 = 0 \):
\( y = \frac{12 - 2x}{3} \)
- When \( x = 0 \), \( y = 4 \). Point: \( (0, 4) \)
- When \( x = 3 \), \( y = 2 \). Point: \( (3, 2) \)

For the second line \( 7x - 3y - 15 = 0 \):
\( y = \frac{7x - 15}{3} \)
- When \( x = 0 \), \( y = -5 \). Point: \( (0, -5) \)
- When \( x = 3 \), \( y = 2 \). Point: \( (3, 2) \)

Plotting these on graph paper, we see that:
1) The two lines intersect at the point \( (3, 2) \).
2) The first line cuts the y-axis at \( (0, 4) \).
3) The second line cuts the y-axis at \( (0, -5) \).

Thus, the triangle formed by these two lines and the y-axis has vertices at \( (0, 4) \), \( (3, 2) \), and \( (0, -5) \).
In simple words: We find where each line crosses the y-axis (by setting \( x = 0 \)) and where they cross each other. These three points form the corners of our triangle, which are \( (0,4) \), \( (3,2) \), and \( (0,-5) \).
Exam Tip: Pay close attention to whether the question asks for the region bounded by the "x-axis" or "y-axis," as confusing them is a very common student mistake.

Question. The sum of the digits of a two- digit number is 12 the number obtained by interchanging the two digits exceed the given number by 18. Find the number.
Answer: Let the tens digit of the two-digit number be \( x \) and the units digit be \( y \).
Thus, the original number is \( 10x + y \).
From the first condition, the sum of the digits is 12:
1) \( x + y = 12 \)
The number obtained by interchanging the digits is \( 10y + x \).
According to the second condition, the interchanged number exceeds the original number by 18:
\( (10y + x) - (10x + y) = 18 \)
\( \implies 9y - 9x = 18 \)
Dividing the entire equation by 9:
2) \( y - x = 2 \)
Now, add equation (1) and equation (2):
\( (x + y) + (y - x) = 12 + 2 \)
\( \implies 2y = 14 \)
\( \implies y = 7 \)
Substitute \( y = 7 \) into equation (1):
\( x + 7 = 12 \)
\( \implies x = 5 \)
Hence, the original two-digit number is \( 10(5) + 7 = 57 \).
In simple words: Representing the digits as \( x \) and \( y \), we get two equations: one for the sum of the digits, and one from the difference between the flipped number and the original. Solving them gives 5 and 7, making the number 57.
Exam Tip: Remember that interchanging digits of \( 10x+y \) always yields \( 10y+x \), and their difference is always a multiple of 9, specifically \( 9(y-x) \).

Question. Abdul traveled 300km by train and 200km by taxi, it took him 5 hours 30 minutes. But if he travels 260 km by train and 240 km by bus he takes 6 minutes longer. Find the speed of the train and of the taxi.
Answer: Let the speed of the train be \( x \) km/hr and the speed of the taxi be \( y \) km/hr.
We know that:
\( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)

Case 1: Abdul travels 300 km by train and 200 km by taxi.
Total time = 5 hours 30 minutes = \( 5\frac{1}{2} = \frac{11}{2} \) hours.
Therefore:
1) \( \frac{300}{x} + \frac{200}{y} = \frac{11}{2} \)

Case 2: He travels 260 km by train and 240 km by taxi (where "bus" is a typo in the text).
Total time = 5 hours 36 minutes = \( 5\frac{36}{60} = 5\frac{3}{5} = \frac{28}{5} \) hours.
Therefore:
2) \( \frac{260}{x} + \frac{240}{y} = \frac{28}{5} \)

Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). The equations become:
\( 300u + 200v = \frac{11}{2} \implies 600u + 400v = 11 \) --- (3)
\( 260u + 240v = \frac{28}{5} \implies 1300u + 1200v = 28 \) --- (4)

Multiplying equation (3) by 3 to equate coefficients of \( v \):
5) \( 1800u + 1200v = 33 \)

Subtracting equation (4) from equation (5):
\( (1800u + 1200v) - (1300u + 1200v) = 33 - 28 \)
\( \implies 500u = 5 \)
\( \implies u = \frac{1}{100} \)

Since \( u = \frac{1}{x} \), the speed of the train is \( x = 100 \) km/hr.

Substituting \( u = \frac{1}{100} \) in equation (3):
\( 600\left(\frac{1}{100}\right) + 400v = 11 \)
\( \implies 6 + 400v = 11 \)
\( \implies 400v = 5 \)
\( \implies v = \frac{1}{80} \)

Since \( v = \frac{1}{y} \), the speed of the taxi is \( y = 80 \) km/hr.

Thus, the speed of the train is 100 km/hr and the speed of the taxi is 80 km/hr.
In simple words: We set up equations based on distance and speed to represent the total travel time for both situations. Solving these equations reveals that the train travels at 100 km/hr and the taxi travels at 80 km/hr.
Exam Tip: Be sure to convert time given in hours and minutes fully into hours (e.g., 5 hours 36 minutes as \( \frac{28}{5} \) hours) before setting up equations.

Question. Solve the following pairs of equation for x and y. 15/x-y +22/x+y=5, 40/x-y +55/x+y =13.
Answer: We are given the system:
1) \( \frac{15}{x-y} + \frac{22}{x+y} = 5 \)
2) \( \frac{40}{x-y} + \frac{55}{x+y} = 13 \)
Let us substitute \( \frac{1}{x-y} = u \) and \( \frac{1}{x+y} = v \) to linearize the equations:
3) \( 15u + 22v = 5 \)
4) \( 40u + 55v = 13 \)
Multiplying equation (3) by 5 and equation (4) by 2:
5) \( 75u + 110v = 25 \)
6) \( 80u + 110v = 26 \)
Subtracting equation (5) from equation (6):
\( (80u + 110v) - (75u + 110v) = 26 - 25 \)
\( \implies 5u = 1 \)
\( \implies u = \frac{1}{5} \)
Substituting \( u = \frac{1}{5} \) into equation (3):
\( 15\left(\frac{1}{5}\right) + 22v = 5 \)
\( \implies 3 + 22v = 5 \)
\( \implies 22v = 2 \)
\( \implies v = \frac{1}{11} \)
Substituting back the values of \( u \) and \( v \):
\( x - y = 5 \) --- (7)
\( x + y = 11 \) --- (8)
Adding equations (7) and (8):
\( 2x = 16 \)
\( \implies x = 8 \)
Substitute \( x = 8 \) into equation (8):
\( 8 + y = 11 \)
\( \implies y = 3 \)
Thus, the solution is \( x = 8 \) and \( y = 3 \).
In simple words: We can simplify these equations by setting \( \frac{1}{x-y} \) and \( \frac{1}{x+y} \) as new single letters. After solving for these letters, we get two simple equations to find that \( x = 8 \) and \( y = 3 \).
Exam Tip: Be comfortable with this "double substitution" method. It is very common in board exams for questions carrying 4 or 5 marks.

Self Evaluation
 

Question. Find the value of 'p' if(-3,p) lies on 7x+2y=14.
Answer: Since the point \( (-3, p) \) lies on the line represented by the equation \( 7x + 2y = 14 \), its coordinates must satisfy this equation.
Substituting \( x = -3 \) and \( y = p \) into the equation:
\( 7(-3) + 2p = 14 \)
\( \implies -21 + 2p = 14 \)
\( \implies 2p = 14 + 21 \)
\( \implies 2p = 35 \)
\( \implies p = \frac{35}{2} \)
Hence, the value of \( p \) is \( \frac{35}{2} \) (or \( 17.5 \)).
In simple words: When a point lies on a line, putting its coordinates into the equation must work out perfectly. Plugging in \( x = -3 \) and \( y = p \) allows us to solve and find \( p = 35/2 \).
Exam Tip: Whenever a point is said to "lie on a line," your immediate step should be to substitute the point's coordinates into the line's equation.

Question. Solve the following system of linear equations using the method of cross-multiplication: ax +by =1 bx +ay = (a+b)2 / a2+b2 =1
Answer: To solve the given system using the cross-multiplication method, we first write both equations in standard form \( Ax + By + C = 0 \):
1) \( ax + by - 1 = 0 \)
2) \( bx + ay - \frac{2ab}{a^2+b^2} = 0 \) [where the constant in the second equation is simplified as \( \frac{(a+b)^2}{a^2+b^2} - 1 = \frac{2ab}{a^2+b^2} \)]

Comparing with the standard forms \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \):
\( a_1 = a, b_1 = b, c_1 = -1 \)
\( a_2 = b, b_2 = a, c_2 = -\frac{2ab}{a^2+b^2} \)

By the cross-multiplication rule:
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]

Substituting the coefficients:
\[ \frac{x}{b\left(-\frac{2ab}{a^2+b^2}\right) - a(-1)} = \frac{y}{-1(b) - \left(-\frac{2ab}{a^2+b^2}\right)a} = \frac{1}{a(a) - b(b)} \]

Let us simplify each denominator:
For \( x \):
\( b\left(-\frac{2ab}{a^2+b^2}\right) + a = a - \frac{2ab^2}{a^2+b^2} \)
\( \implies \frac{a(a^2+b^2) - 2ab^2}{a^2+b^2} = \frac{a^3 - ab^2}{a^2+b^2} = \frac{a(a^2-b^2)}{a^2+b^2} \)

For \( y \):
\( -b + \frac{2a^2b}{a^2+b^2} = \frac{-b(a^2+b^2) + 2a^2b}{a^2+b^2} \)
\( \implies \frac{-a^2b - b^3 + 2a^2b}{a^2+b^2} = \frac{a^2b - b^3}{a^2+b^2} = \frac{b(a^2-b^2)}{a^2+b^2} \)

For the constant term:
\( a^2 - b^2 \)

So the equations become:
\[ \frac{x}{\frac{a(a^2-b^2)}{a^2+b^2}} = \frac{y}{\frac{b(a^2-b^2)}{a^2+b^2}} = \frac{1}{a^2 - b^2} \]

Now, finding \( x \):
\( x = \frac{a(a^2-b^2)}{(a^2+b^2)(a^2-b^2)} = \frac{a}{a^2+b^2} \)

Finding \( y \):
\( y = \frac{b(a^2-b^2)}{(a^2+b^2)(a^2-b^2)} = \frac{b}{a^2+b^2} \)

Hence, the solution is \( x = \frac{a}{a^2+b^2} \) and \( y = \frac{b}{a^2+b^2} \).
In simple words: First, we write both equations in the standard form. Then, we apply the cross-multiplication formula, simplify the expressions in the denominators, and find the values for \( x \) and \( y \) as \( \frac{a}{a^2+b^2} \) and \( \frac{b}{a^2+b^2} \).
Exam Tip: In cross-multiplication questions involving variables as coefficients, keep the terms factored (such as leaving \( a(a^2-b^2) \)) instead of expanding them completely, as this makes cancellation much easier in the final step.

Question. Solve for x and y. bx +ay = a+b. ax[1/a-b -1/a+b]+ by [1/b-a -1/b+a]=2
Answer: We are given:
1) \( bx + ay = a + b \)
2) \( ax \left[ \frac{1}{a-b} - \frac{1}{a+b} \right] + by \left[ \frac{1}{b-a} - \frac{1}{b+a} \right] = 2 \)

Let us first simplify the bracketed terms in equation (2):
\( \frac{1}{a-b} - \frac{1}{a+b} = \frac{(a+b) - (a-b)}{(a-b)(a+b)} = \frac{2b}{a^2-b^2} \)
\( \frac{1}{b-a} - \frac{1}{b+a} = \frac{(b+a) - (b-a)}{(b-a)(b+a)} = \frac{2a}{b^2-a^2} = -\frac{2a}{a^2-b^2} \)

Substituting these simplified fractions back into equation (2):
\( ax \left( \frac{2b}{a^2-b^2} \right) + by \left( -\frac{2a}{a^2-b^2} \right) = 2 \)
\( \implies \frac{2abx}{a^2-b^2} - \frac{2aby}{a^2-b^2} = 2 \)
\( \implies 2ab(x - y) = 2(a^2 - b^2) \)
Dividing both sides by \( 2ab \):
3) \( x - y = \frac{a^2 - b^2}{ab} \)

From equation (3), we can write:
4) \( x = y + \frac{a^2 - b^2}{ab} \)

Now, substitute this value of \( x \) into equation (1):
\( b \left( y + \frac{a^2 - b^2}{ab} \right) + ay = a + b \)
\( \implies by + \frac{a^2 - b^2}{a} + ay = a + b \)
\( \implies y(a + b) + a - \frac{b^2}{a} = a + b \)
Subtracting \( a \) from both sides:
\( \implies y(a + b) - \frac{b^2}{a} = b \)
\( \implies y(a + b) = b + \frac{b^2}{a} \)
\( \implies y(a + b) = \frac{ab + b^2}{a} \)
\( \implies y(a + b) = \frac{b(a + b)}{a} \)
Dividing both sides by \( (a+b) \):
\( y = \frac{b}{a} \)

Now, substitute \( y = \frac{b}{a} \) into equation (4) to find \( x \):
\( x = \frac{b}{a} + \frac{a^2 - b^2}{ab} \)
\( \implies x = \frac{b^2 + a^2 - b^2}{ab} \)
\( \implies x = \frac{a^2}{ab} = \frac{a}{b} \)

Thus, the solution is \( x = \frac{a}{b} \) and \( y = \frac{b}{a} \).
In simple words: First, we simplify the fractions inside the square brackets. This turns the second equation into a much simpler form: \( x - y = \frac{a^2 - b^2}{ab} \). Solving this with our first equation gives us \( x = a/b \) and \( y = b/a \).
Exam Tip: When you see complex algebraic brackets in long equations, always simplify those brackets separately before substituting them back into the main equation to keep the calculation clean.