CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Worksheet Set 08

Question. The area of the triangle formed by \( 2x - y + 6 = 0 \), \( 4x + 5y - 16 = 0 \) and the x – axis is 
(a) 15 sq. units
(b) 16 sq. units
(c) 14 sq. units
(d) 12 sq. units
Answer: (c) 14 sq. units

Question. The system of equations \( 2x + 3y - 7 = 0 \) and \( 6x + 5y - 11 = 0 \) has  
(a) unique solution
(b) infinite many solutions
(c) no solution
(d) non zero solution
Answer: (a) unique solution

Question. The pair of linear equations \( 5x - 3y = 11 \) and \( -10x + 6y = -22 \) are 
(a) dependent(consistent)
(b) None of the options
(c) consistent
(d) inconsistent
Answer: (a) dependent(consistent)

Question. The system of linear equations \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) has no solution if (1)
(a) \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
(b) \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
(c) \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
(d) None of the options
Answer: (a) \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)

Question. Ten students of class X took part in Mathematics quiz. The number of girls is 4 more than that of the boys. The algebraic representation of the above situation is 
(a) None of the options
(b) \( x - y = 10 \) and \( x + y = 4 \)
(c) \( x = y - 12 \) and \( y = 6 + x \)
(d) \( y = x + 4 \) and \( x = 10 - y \)
Answer: (d) \( y = x + 4 \) and \( x = 10 - y \)

Question. Given the linear equation \( 3x + 4y - 8 = 0 \), write another linear equation in two variables such that the geometrical representation of the pair so formed is parallel lines. 
Answer: Given equation \( 3x + 4y - 8 = 0 \).
Lines are parallel when \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
One of the linear equation in two variables can be \( 6x + 8y + k = 0 \) where \( k \) is constant not equal to -16.

Question. Write whether the following pair of linear equations is consistent or not.  
\( x + y = 14 \)
\( x - y = 4 \)
Answer: Given \( x + y = 14 \) and \( x - y = 4 \) such that \( a_1 = 1, b_1 = 1, c_1 = -14 \) and \( a_2 = 1, b_2 = -1, c_2 = -4 \).
Now, \( \frac{a_1}{a_2} = 1 \) and \( \frac{b_1}{b_2} = -1 \).
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the equations have unique solution. Therefore pair of linear equations is consistent.

Question. In a deer park, the number of heads and the number of legs of deer and human visitors were counted and it was found that there were 39 heads and 132 legs. Find the number of deer and human visitors in the park.  
Answer: Let a number of humans be \( x \) and deer be \( y \).
According to the condition, \( x + y = 39 ....(i) \)
and \( 2x + 4y = 132 \implies x + 2y = 66 .....(ii) \)
On solving (i) and (ii), we get
\( (x + 2y) - (x + y) = 66 - 39 \)
\( \implies y = 27 \)
Putting \( y = 27 \) in (i), we get \( x + 27 = 39 \implies x = 12 \).
So, \( y = 27, x = 12 \).

Question. Write the value of k for which the system of equations \( x + y - 4 = 0 \) and \( 2x + ky - 3 = 0 \) has no solution. 
Answer: Given system of equations \( x + y - 4 = 0 \) and \( 2x + ky - 3 = 0 \). With \( a_1=1, b_1=1, c_1=-4 \) and \( a_2=2, b_2=k, c_2=-3 \).
For no solution \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \implies \frac{1}{2} = \frac{1}{k} \neq \frac{-4}{-3} ...... (1) \)
Then \( \frac{1}{2} = \frac{1}{k} \implies k = 2 \).
Substitute \( k = 2 \) in (1) we get \( \frac{1}{2} = \frac{1}{2} \neq \frac{4}{3} \).
Hence, \( k = 2 \) for no solution.

Question. Determine k for which the system of equations has infinite solutions: (1) \( 4x + y = 3 \) and \( 8x + 2y = 5k \)
Answer: For infinite many solutions \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
\( \frac{4}{8} = \frac{1}{2} = \frac{-3}{-5k} \)
\( \implies \frac{1}{2} = \frac{3}{5k} \)
\( \implies 5k = 6 \implies k = \frac{6}{5} \)

Question. Solve the following system of equations: (2) \( 2x - \frac{3}{y} = 9, 3x + \frac{7}{y} = 2, y \neq 0 \)
Answer: The given system of equations is
\( 2x - \frac{3}{y} = 9 .......... (i) \)
\( 3x + \frac{7}{y} = 2, y \neq 0 ............ (ii) \)
Taking \( \frac{1}{y} = u \), the given equations become,
\( 2x - 3u = 9 ........... (iii) \)
\( 3x + 7u = 2 ........... (iv) \)
From (iii), we get \( 2x = 9 + 3u \implies x = \frac{9+3u}{2} \)
Substituting \( x = \frac{9+3u}{2} \) in (iv), we get \( 3 \left( \frac{9+3u}{2} \right) + 7u = 2 \)
\( \implies \frac{27 + 9u + 14u}{2} = 2 \)
\( \implies 27 + 23u = 4 \)
\( \implies 23u = 4 - 27 \implies 23u = -23 \implies u = -1 \).
Hence, \( y = \frac{1}{u} = \frac{1}{-1} = -1 \).
Putting \( u = -1 \) in \( x = \frac{9+3u}{2} \), we get \( x = \frac{9+3(-1)}{2} = \frac{9-3}{2} = \frac{6}{2} = 3 \).
Hence, Solution of the given system of equation is \( x = 3, y = -1 \).

Question. Determine the values of a and b for which the following system of linear equations has infinite solutions: (2) \( 2x - (a - 4)y = 2b + 1, 4x - (a - 1)y = 5b - 1 \)
Answer: \( 2x - (a - 4)y = 2b + 1 ........ (i) \)
\( 4x - (a - 1)y = 5b - 1 ....... (ii) \)
Compare the equations with form \( a_1x + b_1y = c_1 \) and \( a_2x + b_2y = c_2 \).
We get, \( a_1 = 2, b_1 = -(a - 4), c_1 = 2b + 1 \) and \( a_2 = 4, b_2 = -(a - 1), c_2 = 5b - 1 \).
Equations has infinite number of solutions, if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
Therefore, \( \frac{2}{4} = \frac{-(a-4)}{-(a-1)} = \frac{2b+1}{5b-1} \)
\( \implies \frac{1}{2} = \frac{a-4}{a-1} \) and \( \frac{1}{2} = \frac{2b+1}{5b-1} \)
\( \implies a - 1 = 2a - 8 \implies a = 7 \)
And \( 5b - 1 = 4b + 2 \implies b = 3 \).
Hence \( a = 7 \) and \( b = 3 \).

Question. Solve the following system of linear equation by substitution method: (2) \( 2x - y = 2 ...(i), x + 3y = 15 ....(ii) \)
Answer: Given, \( 2x - y = 2 ..(i) \) and \( x + 3y = 15 ..(ii) \).
From eqn. (i), we get \( y = 2x - 2 ... (iii) \)
Substituting the value of \( y \) in eqn. (ii), we get \( x + 3(2x - 2) = 15 \)
\( \implies x + 6x - 6 = 15 \implies 7x = 21 \implies x = 3 \).
Substituting this value of \( x \) in (iii), we get \( y = 2(3) - 2 \implies y = 6 - 2 = 4 \).
Hence the value of \( x \) and \( y \) of given equations are 3 and 4 respectively.

Question. Solve for x and y: \( \frac{3}{x+y} + \frac{2}{x-y} = 2, \frac{9}{x+y} - \frac{4}{x-y} = 1 \).  
Answer: According to the question,
\( \frac{3}{x+y} + \frac{2}{x-y} = 2 \)
\( \frac{9}{x+y} - \frac{4}{x-y} = 1 \)
Putting \( \frac{1}{x+y} = u \) and \( \frac{1}{x-y} = v \),
\( 3u + 2v = 2 .......... (i) \)
\( 9u - 4v = 1 .......... (ii) \)
Multiplying (i) by 2 and (ii) by 1, we get
\( 6u + 4v = 4 ........ (iii) \)
\( 9u - 4v = 1 ......... (iv) \)
Adding (iii) and (iv), \( 15u = 5 \implies u = \frac{1}{3} \).
Putting \( u = \frac{1}{3} \) in (i), \( 3 \times \frac{1}{3} + 2v = 2 \implies 1 + 2v = 2 \implies 2v = 1 \implies v = \frac{1}{2} \).
Now, \( u = \frac{1}{3} \implies x + y = 3 ..........(v) \)
And \( v = \frac{1}{2} \implies x - y = 2 ............(vi) \)
Adding (v) and (vi), \( 2x = 5 \implies x = \frac{5}{2} \).
Putting \( x = \frac{5}{2} \) in (v), \( \frac{5}{2} + y = 3 \implies y = 3 - \frac{5}{2} = \frac{1}{2} \).
The solution is \( x = \frac{5}{2}, y = \frac{1}{2} \).

Question. Draw the graphs of the equations \( 4x - y - 8 = 0 \) and \( 2x - 3y + 6 = 0 \). Also, determine the vertices of the triangle formed by the lines and x-axis. 
Answer: \( 4x - y - 8 = 0 \implies y = 4x - 8 \)
Solution table for \( 4x - y - 8 = 0 \) is:
x: 0, 1, 2
y: -8, -4, 0
\( 2x - 3y + 6 = 0 \implies 3y = 2x + 6 \)
Solution table for \( 2x - 3y + 6 = 0 \) is:
x: 0, 3, -3
y: 2, 4, 0
Vertices of the triangle formed by lines and x-axis are \( (2, 0), (3, 4) \) and \( (-3, 0) \).

Question. Solve the following pairs of equations by reducing them to a pair of linear equations: (3) \( \frac{1}{(3x+y)} + \frac{1}{(3x-y)} = \frac{3}{4} \) and \( \frac{1}{2(3x+y)} - \frac{1}{2(3x-y)} = -\frac{1}{8} \)
Answer: The given equations are:
\( \frac{1}{3x+y} + \frac{1}{3x-y} = \frac{3}{4} ........... (1) \)
\( \frac{1}{2(3x+y)} - \frac{1}{2(3x-y)} = -\frac{1}{8} .......... (2) \)
Put \( \frac{1}{3x+y} = u ......... (3) \) and \( \frac{1}{3x-y} = v .......... (4) \).
Then, \( u + v = \frac{3}{4} .......(5) \)
\( \frac{1}{2}u - \frac{1}{2}v = -\frac{1}{8} \implies u - v = -\frac{1}{4} .......(6) \)
Adding (5) and (6), \( 2u = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \implies u = \frac{1}{4} \).
Subtracting (6) from (5), \( 2v = \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1 \implies v = \frac{1}{2} \).
From (3), \( 3x + y = 4 ....(7) \) and from (4), \( 3x - y = 2 ....(8) \).
Adding (7) and (8), \( 6x = 6 \implies x = 1 \).
Substituting \( x = 1 \) in (7), \( 3(1) + y = 4 \implies y = 1 \).
Hence, the solution is \( x = 1, y = 1 \).

Question. 2 men and 5 boys can finish a piece of work in 4 days, while 3 men and 6 boys can finish it in 3 days. Find the time taken by one man alone to finish the work and that taken by one boy alone to finish the work. (3)
Answer: Suppose man's 1 day's work be \( \frac{1}{x} \) and boy's 1 day's work be \( \frac{1}{y} \).
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \).
By first condition, \( 2u + 5v = \frac{1}{4} ....... (i) \)
By second condition, \( 3u + 6v = \frac{1}{3} ........ (ii) \)
Multiplying (i) by 6 and (ii) by 5, we get
\( 12u + 30v = \frac{6}{4} ..........(iii) \)
\( 15u + 30v = \frac{5}{3} ...........(iv) \)
Subtracting (iii) from (iv), \( 3u = \frac{5}{3} - \frac{3}{2} = \frac{10-9}{6} = \frac{1}{6} \implies u = \frac{1}{18} \).
Putting \( u = \frac{1}{18} \) in (i), \( \frac{2}{18} + 5v = \frac{1}{4} \implies 5v = \frac{1}{4} - \frac{1}{9} = \frac{5}{36} \implies v = \frac{1}{36} \).
Now, \( u = \frac{1}{18} \implies x = 18 \) and \( v = \frac{1}{36} \implies y = 36 \).
The man will complete the work in 18 days and the boy in 36 days when they work alone.

Question. A motor boat takes 6 hours to cover 100 km downstream and 30 km upstream. If the boat goes 75 km downstream and returns back to the starting point in 8 hours. Find the speed of the boat in still water and the speed of the stream.  
Answer: Let the speed of the motor boat in still water be \( x \) km / hour and the speed of the stream be \( y \) km/hour.
Speed downstream = \( (x + y) \) km / hour and Speed upstream = \( (x - y) \) km / hour.
In the first case, \( \frac{100}{x+y} + \frac{30}{x-y} = 6 ...(1) \)
In the second case, \( \frac{75}{x+y} + \frac{75}{x-y} = 8 ...(2) \)
Put \( \frac{1}{x+y} = u \) and \( \frac{1}{x-y} = v \).
\( 100u + 30v = 6 \implies 50u + 15v = 3 ...(3) \)
\( 75u + 75v = 8 \implies 15u + 15v = \frac{8}{5} ...(4) \)
Subtracting (4) from (3), \( 35u = 3 - \frac{8}{5} = \frac{7}{5} \implies u = \frac{1}{25} \).
Putting \( u = \frac{1}{25} \) in (3), \( 50 \times \frac{1}{25} + 15v = 3 \implies 2 + 15v = 3 \implies v = \frac{1}{15} \).
Now, \( x + y = 25 \) and \( x - y = 15 \).
Adding these, \( 2x = 40 \implies x = 20 \). Subtracting them, \( 2y = 10 \implies y = 5 \).
Speed of boat in still water is 20 km/hour and speed of stream is 5 km/hour.

Question. Solve the following system of linear equation graphically \( 4x - 5y - 20 = 0 \) and \( 3x + 5y - 15 = 0 \). Also, find the coordinates of the vertices of the Triangle formed by these two lines and the Y-axis.  
Answer: Given equations are \( 4x - 5y - 20 = 0 \) and \( 3x + 5y - 15 = 0 \).
For \( 4x - 5y - 20 = 0 \): When \( y = 0, x = 5 \); when \( x = 0, y = -4 \).
For \( 3x + 5y - 15 = 0 \): When \( y = 0, x = 5 \); when \( x = 0, y = 3 \).
Graphing these, the two lines intersect at \( A(5, 0) \).
The lines meet y-axis at \( B(0, -4) \) and \( C(0, 3) \) respectively.
Therefore, the vertices of the triangle are \( (5, 0), (0, -4) \) and \( (0, 3) \).

Question. Solve the following system of linear equations graphically and shade the region between the two lines and x-axis: (4) \( 3x + 2y - 4 = 0, 2x - 3y - 7 = 0 \)
Answer: Given system of linear equations is \( 3x + 2y - 4 = 0 .....(1) \) and \( 2x - 3y - 7 = 0 .....(2) \).
For eqn (1), \( y = \frac{4-3x}{2} \). If \( x = 0, y = 2 \); if \( x = 2, y = -1 \).
For eqn (2), \( y = \frac{2x-7}{3} \). If \( x = 2, y = -1 \); if \( x = 5, y = 1 \).
Plotting these points on graph, we find the lines intersect at \( (2, -1) \). The region between the lines and x-axis is shaded.
 

Short Answer Type questions
 

Question. Represent the linear equation in two variables in its standard form. 2y - 3 + 2x =5
Answer: The standard form of a linear equation in two variables is \( ax + by + c = 0 \). Let us rearrange the terms of the given equation:
\( 2x + 2y - 3 = 5 \)
Subtract 5 from both sides to set the equation to zero:
\( 2x + 2y - 3 - 5 = 0 \)
\( 2x + 2y - 8 = 0 \)
We can simplify this by dividing the entire equation by 2:
\( x + y - 4 = 0 \)
Both equations represent the standard form, with the simplified version being \( x + y - 4 = 0 \).
In simple words: Rearrange the terms so that everything is on the left-hand side, equal to zero, following the order of x first, then y, then the constant.
Exam Tip: While \( 2x + 2y - 8 = 0 \) is correct, dividing by the common factor to write \( x + y - 4 = 0 \) is the most mathematically elegant way to present your standard form.

Question. What is the value of a, b, c in the given equation 3x - 5= 2y
Answer: First, we write the equation in the standard form \( ax + by + c = 0 \):
\( 3x - 2y - 5 = 0 \)
Now, comparing this with \( ax + by + c = 0 \), we identify the coefficients:
- \( a = 3 \)
- \( b = -2 \)
- \( c = -5 \)
In simple words: Bring the 2y to the left side to get 3x - 2y - 5 = 0. The numbers in front of x, y, and the leftover constant are your values for a, b, and c.
Exam Tip: Do not forget the negative sign when writing the coefficients. Here, \( b \) is \( -2 \) and \( c \) is \( -5 \), not \( 2 \) and \( 5 \).

Question. Find the value of b and c in the equation 3x= 15
Answer: Let us rewrite the equation \( 3x = 15 \) in the standard form \( ax + by + c = 0 \). Since there is no \( y \) term, its coefficient is \( 0 \):
\( 3x + 0y - 15 = 0 \)
Comparing this to \( ax + by + c = 0 \), we get:
- \( a = 3 \)
- \( b = 0 \)
- \( c = -15 \)
Therefore, the value of \( b \) is \( 0 \) and the value of \( c \) is \( -15 \).
In simple words: Because there is no y in the equation, its coefficient is zero. Bring 15 to the other side to get the constant value as -15.
Exam Tip: If a variable is completely missing from a linear equation, its standard coefficient is always written as zero.

Question. Check whether (4,0) is a solution of the equation2x+3y =8
Answer: To check if \( (4, 0) \) is a solution, we substitute \( x = 4 \) and \( y = 0 \) into the Left-Hand Side (LHS) of the equation \( 2x + 3y = 8 \):
LHS \( = 2x + 3y \)
\( = 2(4) + 3(0) \)
\( = 8 + 0 \)
\( = 8 \)
Since the LHS equals the Right-Hand Side (RHS), which is \( 8 \), the given coordinates satisfy the equation.
Therefore, \( (4, 0) \) is indeed a solution of the equation.
In simple words: Replace x with 4 and y with 0 in the equation. Since the left side works out to equal 8, it is a valid solution.
Exam Tip: Always show the step-by-step substitution of values into the LHS and state clearly that LHS = RHS to get full marks on verification questions.

Question. Find if 2, -1 is a solution of the equation x + 3y =1
Answer: We substitute \( x = 2 \) and \( y = -1 \) into the Left-Hand Side (LHS) of the equation \( x + 3y = 1 \):
LHS \( = x + 3y \)
\( = 2 + 3(-1) \)
\( = 2 - 3 \)
\( = -1 \)
Now, comparing LHS with RHS:
LHS \( = -1 \) and RHS \( = 1 \).
Since LHS \( \neq \) RHS, the coordinates do not satisfy the equation.
Therefore, \( (2, -1) \) is not a solution of the equation.
In simple words: Put 2 for x and -1 for y. The left side equals -1, which does not equal the right side of 1, so it is not a solution.
Exam Tip: Always write a final concluding sentence stating whether the point is a solution or not, based on your LHS and RHS comparison.

Question. Form an equation in two variables with the given information: the number of ducks is three more than three times number of hens, and the total of all hens and ducks is 156.
Answer: Let the number of hens be \( x \) and the number of ducks be \( y \).
From the first statement, the number of ducks is three more than three times the number of hens:
\( y = 3x + 3 \)
In standard form, we can write this as:
\( 3x - y + 3 = 0 \)
From the second statement, the total number of hens and ducks is 156:
\( x + y = 156 \)
In standard form, we write this as:
\( x + y - 156 = 0 \)
These two linear equations represent the given situation mathematically.
In simple words: Let hens be x and ducks be y. Translate the words directly into two separate equations: y = 3x + 3 and x + y = 156.
Exam Tip: Be sure to define your variables at the very beginning of your solution, specifying clearly what \( x \) and \( y \) represent.

Question. Find four solutions of the given equations 3x -y=4
Answer: Let us rewrite the equation to express \( y \) in terms of \( x \):
\( y = 3x - 4 \)
We can find four different solutions by substituting four different values for \( x \):
- For \( x = 0 \): \( y = 3(0) - 4 = -4 \). Solution: \( (0, -4) \)
- For \( x = 1 \): \( y = 3(1) - 4 = -1 \). Solution: \( (1, -1) \)
- For \( x = 2 \): \( y = 3(2) - 4 = 2 \). Solution: \( (2, 2) \)
- For \( x = 3 \): \( y = 3(3) - 4 = 5 \). Solution: \( (3, 5) \)
These four coordinates are solutions to the given linear equation.
In simple words: Change the equation to y = 3x - 4. Pick any four values for x to get four corresponding values for y.
Exam Tip: Arrange your final solutions in a neat table to make it easy for the examiner to read and grade your work.

Question. Find the value of k, when x = -1 and y =2 in the equation 3x -7y = 3k .
Answer: We substitute the values \( x = -1 \) and \( y = 2 \) into the given equation \( 3x - 7y = 3k \):
\( 3(-1) - 7(2) = 3k \)
\( -3 - 14 = 3k \)
\( -17 = 3k \)
Divide both sides by 3 to find the value of \( k \):
\( k = -\frac{17}{3} \)
Thus, the value of \( k \) is \( -\frac{17}{3} \).
In simple words: Put -1 for x and 2 for y in the equation. Simplify the left side to get -17, and then divide by 3 to find k.
Exam Tip: Do not panic if your value for a variable like \( k \) is a fraction; fractional values are common in coordinate geometry problems.

Question. Find two solutions for each of the following equations:
(i) 4x + 3y =12
(ii) 2x + 5y = 0
Answer: Let us find two solutions for each equation:

(i) For \( 4x + 3y = 12 \):
- Substitute \( x = 0 \): \( 4(0) + 3y = 12 \implies 3y = 12 \implies y = 4 \). First solution: \( (0, 4) \).
- Substitute \( y = 0 \): \( 4x + 3(0) = 12 \implies 4x = 12 \implies x = 3 \). Second solution: \( (3, 0) \).

(ii) For \( 2x + 5y = 0 \):
- Substitute \( x = 0 \): \( 2(0) + 5y = 0 \implies 5y = 0 \implies y = 0 \). First solution: \( (0, 0) \).
- Substitute \( x = 5 \): \( 2(5) + 5y = 0 \implies 10 + 5y = 0 \implies 5y = -10 \implies y = -2 \). Second solution: \( (5, -2) \).
In simple words: For each equation, try setting x = 0 to get one point, and then choose another simple value to find a second point.
Exam Tip: If an equation has no constant term (like \( 2x + 5y = 0 \)), it will always pass through the origin \( (0, 0) \).

Question. Express the following linear equations in the form ax + by + c = 0 and find the values of a , b, c.In the equation x - \(\frac{1}{5}\)y = -9.35 (with a bar over 35)
Answer: Let us rewrite the equation \( x - \frac{1}{5}y = -9.\overline{35} \) in standard form by bringing the constant to the LHS:
\( x - \frac{1}{5}y + 9.\overline{35} = 0 \)
Now, comparing this with \( ax + by + c = 0 \), we identify the coefficients:
- \( a = 1 \)
- \( b = -\frac{1}{5} \) (or \( -0.2 \))
- \( c = 9.\overline{35} \)
In simple words: Move the constant from the right side to the left side so that the equation equals zero, then read off your coefficients.
Exam Tip: Keep the bar over the repeating decimal parts intact when writing the constant value \( c \).

Question. Write the following as linear equation in two variables.2x = 15 and -3y -4 =0
Answer: To express these single-variable equations in two variables, we insert the missing variable with a coefficient of 0:
- For \( 2x = 15 \):
First rewrite it as \( 2x - 15 = 0 \). Adding the \( y \) term, we get:
\( 2x + 0y - 15 = 0 \)

- For \( -3y - 4 = 0 \):
Adding the \( x \) term, we get:
\( 0x - 3y - 4 = 0 \)
In simple words: To write an equation with two variables when one is missing, just add the missing letter with a zero in front of it.
Exam Tip: Always write the standard form with a \( + \) or \( - \) sign clearly separating all three terms, including the \( 0x \) or \( 0y \) term.

Question. Write five solutions of the given equation: \(\pi\) x + y = 7.
Answer: We express \( y \) in terms of \( x \) as:
\( y = 7 - \pi x \)
We can find five different solutions by choosing five values for \( x \):
- For \( x = 0 \): \( y = 7 - \pi(0) = 7 \). Solution: \( (0, 7) \)
- For \( x = 1 \): \( y = 7 - \pi(1) = 7 - \pi \). Solution: \( (1, 7 - \pi) \)
- For \( x = 2 \): \( y = 7 - \pi(2) = 7 - 2\pi \). Solution: \( (2, 7 - 2\pi) \)
- For \( x = 3 \): \( y = 7 - \pi(3) = 7 - 3\pi \). Solution: \( (3, 7 - 3\pi) \)
- For \( x = 4 \): \( y = 7 - \pi(4) = 7 - 4\pi \). Solution: \( (4, 7 - 4\pi) \)
In simple words: Rewrite the equation as y = 7 - \(\pi\)x. Pick five numbers for x and write down the corresponding coordinate pairs.
Exam Tip: Keep \( \pi \) in its symbol form in your coordinates rather than using \( 3.14 \) or \( \frac{22}{7} \), unless the question asks for a decimal approximation.

Question. Write the four solutions of the equation (2x - 1)/(3y - 5) = 1/3.
Answer: Let us simplify the equation by cross-multiplying first:
\( 3(2x - 1) = 1(3y - 5) \)
\( 6x - 3 = 3y - 5 \)
\( 6x - 3y - 3 + 5 = 0 \)
\( 6x - 3y + 2 = 0 \)
Now, express \( y \) in terms of \( x \):
\( 3y = 6x + 2 \implies y = 2x + \frac{2}{3} \)
Let us find four different solutions by substituting four values for \( x \):
- If \( x = 0 \): \( y = 2(0) + \frac{2}{3} = \frac{2}{3} \). Solution: \( (0, \frac{2}{3}) \)
- If \( x = 1 \): \( y = 2(1) + \frac{2}{3} = \frac{8}{3} \). Solution: \( (1, \frac{8}{3}) \)
- If \( x = -1 \): \( y = 2(-1) + \frac{2}{3} = -\frac{4}{3} \). Solution: \( (-1, -\frac{4}{3}) \)
- If \( x = 2 \): \( y = 2(2) + \frac{2}{3} = \frac{14}{3} \). Solution: \( (2, \frac{14}{3}) \)
In simple words: Cross-multiply to clear the fractions and write the equation simply. Then pick values for x to calculate y.
Exam Tip: Be mindful of restrictions on variables; here, the denominator \( 3y - 5 \neq 0 \) means \( y \neq \frac{5}{3} \), which is satisfied by all our chosen solutions.

Question. Write the solution of the equation x +y =4
Answer: To provide solutions for \( x + y = 4 \), we can find coordinate pairs that add up to 4. Since a linear equation in two variables has infinitely many solutions, any valid pair can be given:
- If we choose \( x = 2 \), then \( y = 4 - 2 = 2 \). Solution: \( (2, 2) \).
- If we choose \( x = 0 \), then \( y = 4 - 0 = 4 \). Solution: \( (0, 4) \).
- If we choose \( x = 4 \), then \( y = 4 - 4 = 0 \). Solution: \( (4, 0) \).
Any of these pairs satisfy the given equation.
In simple words: Any pair of numbers that add up to 4 is a solution to this equation.
Exam Tip: When asked for "the solution" of such an equation, provide multiple simple examples to demonstrate your understanding of infinite solution sets.

Question. Lata and Gautami together contributed Rs. 100 for a donation camp. Represent this situation graphically.
Answer: Let Lata's contribution be Rs. \( x \) and Gautami's contribution be Rs. \( y \).
According to the given statement:
\( x + y = 100 \)
Let us find some coordinates to plot this on a graph:
- If \( x = 0 \), then \( y = 100 \). Point: \( (0, 100) \)
- If \( x = 50 \), then \( y = 50 \). Point: \( (50, 50) \)
- If \( x = 100 \), then \( y = 0 \). Point: \( (100, 0) \)
We plot these points and join them with a straight line in the first quadrant.

In simple words: The sum of money donated by both girls must equal Rs. 100. If one donates Rs. 50, the other must also donate Rs. 50.
Exam Tip: Since money contributions cannot be negative, do not extend the graph line beyond the axes into negative quadrants.

Question. The taxi fare in a city is as follows: For the first km, the fare is Rs.8 and for every subsequent Km it is Rs. 5. Taking the distance traveled as x and the total fare as y, represent the equation graphically.
Answer: Let the total distance traveled be \( x \) km and the total fare be Rs. \( y \).
The fare for the first kilometer is Rs. 8. The remaining distance is \( (x - 1) \) km, charged at Rs. 5 per kilometer.
This gives the equation:
\( y = 8 + 5(x - 1) \)
\( y = 8 + 5x - 5 \)
\( y = 5x + 3 \) (valid for \( x \ge 1 \))
Let us find some coordinate points to plot this:
- For \( x = 1 \): \( y = 5(1) + 3 = 8 \). Point: \( (1, 8) \)
- For \( x = 2 \): \( y = 5(2) + 3 = 13 \). Point: \( (2, 13) \)
- For \( x = 3 \): \( y = 5(3) + 3 = 18 \). Point: \( (3, 18) \)
Plotted on a Cartesian plane, these coordinates define a straight line starting from \( x = 1 \).

In simple words: You pay a base fare of Rs. 8 for the first kilometer, and Rs. 5 for every kilometer after that.
Exam Tip: Since distance traveled must be at least 1 km to start charging, begin your line plot at the coordinate point \( (1, 8) \).

Question. Represent the equation 2 + 3y = 7x graphically.
Answer: Let us rewrite the equation to express \( y \) in terms of \( x \):
\( 3y = 7x - 2 \implies y = \frac{7x - 2}{3} \)
Let us calculate coordinates to plot the graph:
- If \( x = -1 \): \( y = \frac{7(-1) - 2}{3} = -3 \). Point: \( (-1, -3) \)
- If \( x = 2 \): \( y = \frac{7(2) - 2}{3} = 4 \). Point: \( (2, 4) \)
- If \( x = 5 \): \( y = \frac{7(5) - 2}{3} = 11 \). Point: \( (5, 11) \)
Plotting these points on the Cartesian plane and connecting them yields the graph of the equation.

In simple words: Rearrange the equation, calculate a few clean coordinates (like x = 2 giving y = 4), and draw the line passing through them.
Exam Tip: Try to choose integer values for \( x \) that yield integer values for \( y \). This makes plotting on standard graph paper much easier and highly accurate.

Question. Form an equation for the statement: The sum of cost of pens and twice the cost of pencils is Rs. 6 and represent the situation graphically.
Answer: Let the cost of a pen be Rs. \( x \) and the cost of a pencil be Rs. \( y \).
According to the given statement:
\( x + 2y = 6 \)
Let us find some coordinate points to plot this line:
- If \( x = 0 \): \( 0 + 2y = 6 \implies y = 3 \). Point: \( (0, 3) \)
- If \( y = 0 \): \( x + 2(0) = 6 \implies x = 6 \). Point: \( (6, 0) \)
- If \( x = 2 \): \( 2 + 2y = 6 \implies 2y = 4 \implies y = 2 \). Point: \( (2, 2) \)
We plot these points and join them with a straight line in the first quadrant.

In simple words: The price of one pen plus twice the price of a pencil must equal Rs. 6.
Exam Tip: Ensure that your axes are clearly labeled with the names of the quantities being represented (Cost of Pens and Cost of Pencils).

Question. Form the graph of the equation y = 2x
Answer: To construct the graph of the linear equation \( y = 2x \), we determine some coordinate points:
- When \( x = 0 \), \( y = 2(0) = 0 \). Point: \( (0, 0) \)
- When \( x = 1 \), \( y = 2(1) = 2 \). Point: \( (1, 2) \)
- When \( x = 2 \), \( y = 2(2) = 4 \). Point: \( (2, 4) \)
- When \( x = -1 \), \( y = 2(-1) = -2 \). Point: \( (-1, -2) \)
We plot these coordinates on the Cartesian grid and join them with a straight line passing through the origin.

Question. Plot the graph of the equation y – 2x = 4
Answer: Let us rewrite the equation to express \( y \) in terms of \( x \):
\( y = 2x + 4 \)
Now, we calculate three coordinate points to help us plot the line accurately:
- When \( x = 0 \): \( y = 2(0) + 4 = 4 \). Point: \( (0, 4) \)
- When \( x = -2 \): \( y = 2(-2) + 4 = 0 \). Point: \( (-2, 0) \)
- When \( x = 1 \): \( y = 2(1) + 4 = 6 \). Point: \( (1, 6) \)
We plot these coordinates on our coordinate grid and connect them with a straight line.

In simple words: The y-value is equal to twice the x-value plus 4. Plot points where the line crosses the axes: (-2, 0) and (0, 4), then draw a straight line through them.
Exam Tip: Finding the x-intercept (where \( y = 0 \)) and the y-intercept (where \( x = 0 \)) is a standard and effective technique for plotting linear graphs quickly.

Question. Express y =4 as linear equation in two variables.
Answer: To express \( y = 4 \) as a linear equation in two variables, we rewrite it with both variables \( x \) and \( y \). Since \( x \) is missing, we write it with a coefficient of \( 0 \):
\( 0x + 1y = 4 \)
In the standard form \( ax + by + c = 0 \), this is written as:
\( 0x + y - 4 = 0 \)
In simple words: Write y = 4 as 0x + y = 4 to show that x is part of the equation but has no effect on the value.
Exam Tip: Always make sure to write the coefficient of the missing variable as exactly \( 0 \) to correctly show it in two variables.

Question. Give the representation of 2x +9 =0 as an equation in
a) One variable
b) Two variable
Answer: Let us first simplify the equation:
\( 2x + 9 = 0 \implies 2x = -9 \implies x = -4.5 \)

a) **In one variable:**
The representation of \( x = -4.5 \) is a single point on a 1D horizontal number line, located exactly halfway between \( -4 \) and \( -5 \).

b) **In two variables:**
The equation in two variables is represented as:
\( 2x + 0y + 9 = 0 \)
In a 2D Cartesian plane, this is represented as a straight vertical line parallel to the y-axis, crossing the x-axis at the point \( (-4.5, 0) \).
In simple words: On a simple number line, it is just one dot at -4.5. On a 2D grid, it is a vertical line that crosses the horizontal line at -4.5.
Exam Tip: Clearly label your axes and points on both the number line and coordinate plane to ensure maximum marks.

Question. The temperature in degree Celsius is given by the following formula F = (9/5) C + 32
Answer the following questions
a. What will be the temperature in degree Celsius if the temperature is 45 F?
b. If the temperature is 0 C, what is the temperature in Fahrenheit?
c. Is there a temperature, which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
Answer: Let us solve each part step-by-step using the temperature conversion formula:
**a. Find temperature in Celsius when \( F = 45^\circ \text{F} \):**
Substitute \( F = 45 \) in the formula:
\( 45 = \frac{9}{5}C + 32 \)
\( 45 - 32 = \frac{9}{5}C \)
\( 13 = \frac{9}{5}C \)
Multiply both sides by 5:
\( 65 = 9C \)
\( C = \frac{65}{9} \approx 7.22^\circ \text{C} \)

**b. Find temperature in Fahrenheit when \( C = 0^\circ \text{C} \):**
Substitute \( C = 0 \) in the formula:
\( F = \frac{9}{5}(0) + 32 \)
\( F = 0 + 32 = 32^\circ \text{F} \)

**c. Find if there is a temperature numerically equal in both scales:**
Let the numerically equal temperature value be \( x \), so \( F = C = x \). Substitute \( x \) into the formula:
\( x = \frac{9}{5}x + 32 \)
Rearrange to solve for \( x \):
\( x - \frac{9}{5}x = 32 \)
\( -\frac{4}{5}x = 32 \)
Multiply by 5:
\( -4x = 160 \)
\( x = -40 \)
Yes, \( -40^\circ \) is numerically the same temperature on both the Fahrenheit and Celsius scales (\( -40^\circ \text{C} = -40^\circ \text{F} \)).
In simple words: Substitute the given numbers into the formula and solve. For part (c), setting F and C equal to the same letter shows that -40 is the temperature where both scales read exactly the same.
Exam Tip: Memorize the value of \( -40 \) as the crossover point of Celsius and Fahrenheit, as it is a very common exam question.