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Updated 6 August 2026

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Pair of Linear Equations in Two Variables

Madhya Pradesh Board · Class 10 · Mathematics

NCERT Solutions for Pair of Linear Equations in Two Variables — Madhya Pradesh Board Class 10 Mathematics.

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EXERCISE 3.1

1Form the pair of linear equations in the following problems, and find their solutions graphically.Show solution

Let the number of boys be

x

and the number of girls be

y

.

From the problem:
- Total students:

x+y=10

- Girls are 4 more than boys:

y=x+4

So the pair of linear equations is:

x+y=10,\qquad y=x+4

Substitute

y=x+4

into

x+y=10

x+(x+4)=10

2x+4=10

2x=6

x=3

Then

y=x+4=3+4=7

So the graphical solution would be the point

(3,7)

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1(i)10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.Show solution

Let the number of boys be

x

and the number of girls be

y

.

Then:

x+y=10

y=x+4

Substitute:

x+(x+4)=10

2x=6

x=3

So,

y=3+4=7

Hence, there were 3 boys and 7 girls.

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1(ii)5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.Show solution

Let the cost of one pencil be

x

and one pen be

y

.

From the problem:

5x+7y=50

7x+5y=46

Eliminate

y

:
Multiply the first equation by 5 and the second by 7:

25x+35y=250

49x+35y=322

Subtract:

24x=72

x=3

Substitute in

5x+7y=50

5(3)+7y=50

15+7y=50

7y=35

y=5

So the cost is ₹3 per pencil and ₹5 per pen.

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2On comparing the ratios

\frac{a_1}{a_2}

\frac{b_1}{b_2}

and

\frac{c_1}{c_2}

, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:Show solution

To decide whether the lines are intersecting, parallel, or coincident, compare the ratios

\frac{a_1}{a_2}

\frac{b_1}{b_2}

and

\frac{c_1}{c_2}

for each pair.

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2(i)

5x - 4y + 8 = 0

7x + 6y - 9 = 0

Show solution

For

5x-4y+8=0

and

7x+6y-9=0

\frac{a_1}{a_2}=\frac{5}{7},\quad \frac{b_1}{b_2}=\frac{-4}{6}

Since

\frac{5}{7}\ne \frac{-4}{6},

the lines intersect at a point.

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2(ii)

9x + 3y + 12 = 0

18x + 6y + 24 = 0

Show solution

For

9x+3y+12=0

and

18x+6y+24=0

\frac{a_1}{a_2}=\frac{9}{18}=\frac12,\quad \frac{b_1}{b_2}=\frac{3}{6}=\frac12,\quad \frac{c_1}{c_2}=\frac{12}{24}=\frac12

Since

\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2},

the lines are coincident.

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2(iii)

6x - 3y + 10 = 0

2x - y + 9 = 0

Show solution

For

6x-3y+10=0

and

2x-y+9=0

\frac{a_1}{a_2}=\frac{6}{2}=3,\quad \frac{b_1}{b_2}=\frac{-3}{-1}=3,\quad \frac{c_1}{c_2}=\frac{10}{9}

Since

\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2},

the lines are parallel.

The computed relation shows they are parallel, so if the printed options differ, the correct classification is parallel lines.

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3On comparing the ratios

\frac{a_1}{a_2}

\frac{b_1}{b_2}

and

\frac{c_1}{c_2}

, find out whether the following pair of linear equations are consistent, or inconsistent.Show solution

Use the ratio test:
- If

\frac{a_1}{a_2} \ne \frac{b_1}{b_2}

, the pair is consistent.
- If

\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}

, the pair is inconsistent.
- If

\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

, the pair is consistent with infinitely many solutions.

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3(i)

3x + 2y = 5

;

2x - 3y = 7

Show solution

For

3x+2y=5

and

2x-3y=7

\frac{a_1}{a_2}=\frac{3}{2},\quad \frac{b_1}{b_2}=\frac{2}{-3}

Since

\frac{3}{2}\ne \frac{2}{-3},

the pair is consistent.

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3(ii)

2x - 3y = 8

;

4x - 6y = 9

Show solution

For

2x-3y=8

and

4x-6y=9

\frac{a_1}{a_2}=\frac{2}{4}=\frac12,\quad \frac{b_1}{b_2}=\frac{-3}{-6}=\frac12,\quad \frac{c_1}{c_2}=\frac{8}{9}

Since

\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2},

the pair is inconsistent.

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3(iii)

\frac{3}{2}x + \frac{5}{3}y = 7

;

9x - 10y = 14

Show solution

For

\frac{3}{2}x+\frac{5}{3}y=7

and

9x-10y=14

:

Write in standard form:

\frac{3}{2}x+\frac{5}{3}y-7=0,\qquad 9x-10y-14=0

So,

\frac{a_1}{a_2}=\frac{3/2}{9}=\frac{1}{6},\quad \frac{b_1}{b_2}=\frac{5/3}{-10}=-\frac{1}{6}

Since

\frac{a_1}{a_2}\ne \frac{b_1}{b_2},

the pair is consistent.

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3(iv)

5x - 3y = 11

;

-10x + 6y = -22

Show solution

For

5x-3y=11

and

-10x+6y=-22

\frac{a_1}{a_2}=\frac{5}{-10}=-\frac12,\quad \frac{b_1}{b_2}=\frac{-3}{6}=-\frac12,\quad \frac{c_1}{c_2}=\frac{11}{-22}=-\frac12

Since all three ratios are equal, the pair is dependent and consistent.

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3(v)

\frac{4}{3}x + 2y = 8

;

2x + 3y = 12

Show solution

For

\frac{4}{3}x+2y=8

and

2x+3y=12

\frac{a_1}{a_2}=\frac{4/3}{2}=\frac{2}{3},\quad \frac{b_1}{b_2}=\frac{2}{3}

Also,

\frac{c_1}{c_2}=\frac{8}{12}=\frac{2}{3}

Thus,

\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2},

so the pair is dependent and consistent.

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4Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:Show solution

Check each pair using the ratio criterion:
- If

\frac{a_1}{a_2} \ne \frac{b_1}{b_2}

, the pair is consistent.
- If

\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}

, the pair is inconsistent.
- If all three ratios are equal, the pair is consistent with infinitely many solutions.

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4(i)

x + y = 5

2x + 2y = 10

Show solution

For

x+y=5

and

2x+2y=10

\frac{a_1}{a_2}=\frac{1}{2},\quad \frac{b_1}{b_2}=\frac{1}{2},\quad \frac{c_1}{c_2}=\frac{5}{10}=\frac{1}{2}

Since all three ratios are equal, the lines are coincident. Therefore, the pair is consistent and has infinitely many solutions.

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4(ii)

x - y = 8

3x - 3y = 16

Show solution

For

x-y=8

and

3x-3y=16

\frac{a_1}{a_2}=\frac{1}{3},\quad \frac{b_1}{b_2}=\frac{-1}{-3}=\frac{1}{3},\quad \frac{c_1}{c_2}=\frac{8}{16}=\frac{1}{2}

Since

\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2},

the pair is inconsistent.

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4(iii)

2x + y - 6 = 0

4x - 2y - 4 = 0

Show solution

For

2x+y-6=0

and

4x-2y-4=0

\frac{a_1}{a_2}=\frac{2}{4}=\frac12,\quad \frac{b_1}{b_2}=\frac{1}{-2}=-\frac12

Since

\frac{a_1}{a_2}\ne \frac{b_1}{b_2},

the lines intersect at one point, so the pair is consistent with a unique solution.

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4(iv)

2x - 2y - 2 = 0

4x - 4y - 5 = 0

Show solution

For

2x-2y-2=0

and

4x-4y-5=0

\frac{a_1}{a_2}=\frac{2}{4}=\frac12,\quad \frac{b_1}{b_2}=\frac{-2}{-4}=\frac12,\quad \frac{c_1}{c_2}=\frac{-2}{-5}=\frac25

Since

\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2},

the pair is inconsistent.

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5Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.Show solution

Let the width be

x

m and the length be

x+4

m.

Half the perimeter is 36 m, so

\text{length} + \, \text{width} = 36

(x+4)+x=36

2x+4=36

2x=32

x=16

So width

=16

m and length

=16+4=20

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6Given the linear equation

2x + 3y - 8 = 0

, write another linear equation in two variables such that the geometrical representation of the pair so formed is:Show solution

Let the given equation be

2x+3y-8=0

. To get a second equation:
- for intersecting lines, choose any equation whose coefficients do not make

\frac{a_1}{a_2}=\frac{b_1}{b_2}

;
- for parallel lines, choose another equation with

\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}

;
- for coincident lines, choose an equation which is a multiple of the given one.

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6(i)intersecting linesShow solution

For intersecting lines, choose a line whose coefficients do not satisfy the equality of ratios with

2x+3y-8=0

.

One suitable equation is:

x+y-5=0

Since the ratios are not equal, the two lines will intersect at one point.

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6(ii)parallel linesShow solution

For parallel lines, the coefficients of

x

and

y

must be proportional, but the constant term must not be in the same ratio.

A suitable equation is:

4x+6y-16=0

Here,

\frac{2}{4}=\frac{3}{6}=\frac{8}{16}

would make it coincident, so to make it parallel we need a different constant, such as

-16

? Actually for parallel lines we need

\frac{2}{4}=\frac{3}{6}\ne \frac{8}{16}

This is not true because

\frac{8}{16}=\frac12

.

A correct parallel equation is:

4x+6y-20=0

so that

\frac{2}{4}=\frac{3}{6}=\frac12 \ne \frac{8}{20}

Hence the lines are parallel.

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6(iii)coincident linesShow solution

For coincident lines, choose a multiple of the given equation.

Multiplying

2x+3y-8=0

by 2 gives:

4x+6y-16=0

This represents the same line, so the pair is coincident.

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7Draw the graphs of the equations

x - y + 1 = 0

and

3x + 2y - 12 = 0

. Determine the coordinates of the vertices of the triangle formed by these lines and the

x

-axis, and shade the triangular region.Show solution

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EXERCISE 3.2

1Solve the following pair of linear equations by the substitution method.Show solution

Use the substitution method: express one variable from one equation and substitute into the other. Then solve for the second variable and back-substitute to get the first.

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1(i)

x + y = 14

x - y = 4

1(ii)

s - t = 3

\frac{s}{3} + \frac{t}{2} = 6

1(iii)

3x - y = 3

9x - 3y = 9

1(iv)

0.2x + 0.3y = 1.3

0.4x + 0.5y = 2.3

1(v)

\sqrt{2}x + \sqrt{3}y = 0

\sqrt{3}x - \sqrt{8}y = 0

1(vi)

\frac{3x}{2} - \frac{5y}{3} = -2

\frac{x}{3} + \frac{y}{2} = \frac{13}{6}

2Solve

2x + 3y = 11

and

2x - 4y = -24

and hence find the value of 'm' for which

y = mx + 3

3Form the pair of linear equations for the following problems and find their solution by substitution method.

3(i)The difference between two numbers is 26 and one number is three times the other. Find them.

3(ii)The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

3(iii)The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.

3(iv)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 ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

3(v)A fraction becomes

\frac{9}{11}

, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes

\frac{5}{6}

. Find the fraction.

3(vi)Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages?

EXERCISE 3.3

1Solve the following pair of linear equations by the elimination method and the substitution method :

1(i)

x + y = 5

and

2x - 3y = 4

1(ii)

3x + 4y = 10

and

2x - 2y = 2

1(iii)

3x - 5y - 4 = 0

and

9x = 2y + 7

1(iv)

\frac{x}{2} + \frac{2y}{3} = -1

and

x - \frac{y}{3} = 3

2Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :

2(i)If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes

\frac{1}{2}

if we only add 1 to the denominator. What is the fraction?

2(ii)Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

2(iii)The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

2(iv)Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.

2(v)A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

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Key topics in Pair of Linear Equations in Two Variables include Chapter Overview — Pair of Linear Equations, Decision Tree: How Many Solutions Does a Pair of Linear Equations Have?, Pair of Linear Equations — Complete Chapter Overview. These are the concepts Madhya Pradesh Board Class 10 examiners draw on most — study them first, then practise related questions.

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EXERCISE 3.1

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