---
title: "NCERT Solutions Class 10 Maths Pair of Linear Equations in Two Variables"
url: https://www.swavid.com/maths/class/10/chapter/pair-of-linear-equations-in-two-variables/ncert-solutions
dateModified: 2026-10-07T15:50:31+00:00
---

# NCERT Solutions Class 10 Maths Pair of Linear Equations in Two Variables

This chapter covers questions on formulating and solving pairs of linear equations in two variables using graphical, substitution, and elimination methods.

Free PDF (13 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-10/swavid-ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables-67d58008af.pdf

## EXERCISE 3.1

### Question 1

*3 marks · Short answer*

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

**Part (i)**

1. Let the number of boys be $x$ and the number of girls be $y$.
2. The linear equations are $x + y = 10$ and $y = x + 4$.
3. For $x + y = 10$, solutions are $(0, 10)$ and $(5, 5)$.
4. For $y = x + 4$, solutions are $(0, 4)$ and $(3, 7)$.
5. Plotting the lines on a graph paper, they intersect at $(3, 7)$.
6. Therefore, the number of boys is $3$ and the number of girls is $7$.

Answer (i): Number of boys = 3, Number of girls = 7

**Part (ii)**

1. Let the cost of one pencil be ₹ $x$ and one pen be ₹ $y$.
2. The linear equations are $5x + 7y = 50$ and $7x + 5y = 46$.
3. For $5x + 7y = 50$, solutions are $(3, 5)$ and $(-4, 10)$.
4. For $7x + 5y = 46$, solutions are $(3, 5)$ and $(-2, 12)$.
5. Plotting the lines on a graph paper, they intersect at $(3, 5)$.
6. Therefore, the cost of one pencil is ₹ $3$ and one pen is ₹ $5$.

Answer (ii): Cost of one pencil = ₹ 3, Cost of one pen = ₹ 5

**Answer:** For (i) Number of boys = 3, number of girls = 7. For (ii) Cost of one pencil = ₹ 3, cost of one pen = ₹ 5.

> Common mistake: Interchanging the axes or making errors in plotting the intersection point.

### Question 2

*2 marks · Very short answer*

On 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:
(i) $5x - 4y + 8 = 0$, $7x + 6y - 9 = 0$
(ii) $9x + 3y + 12 = 0$, $18x + 6y + 24 = 0$
(iii) $6x - 3y + 10 = 0$, $2x - y + 9 = 0$

**Part (i)**

1. Given equations: $5x - 4y + 8 = 0$ and $7x + 6y - 9 = 0$.
2. Here $\frac{a_1}{a_2} = \frac{5}{7}$, $\frac{b_1}{b_2} = \frac{-4}{6} = \frac{-2}{3}$.
3. Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at a point.

Answer (i): Intersect at a point

**Part (ii)**

1. Given equations: $9x + 3y + 12 = 0$ and $18x + 6y + 24 = 0$.
2. Here $\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$, and $\frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}$.
3. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident.

Answer (ii): Coincident lines

**Part (iii)**

1. Given equations: $6x - 3y + 10 = 0$ and $2x - y + 9 = 0$.
2. Here $\frac{a_1}{a_2} = \frac{6}{2} = 3$, $\frac{b_1}{b_2} = \frac{-3}{-1} = 3$, and $\frac{c_1}{c_2} = \frac{10}{9}$.
3. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel.

Answer (iii): Parallel lines

**Answer:** For (i) Intersect at a point, For (ii) Coincident lines, For (iii) Parallel lines.

> Common mistake: Ignoring the negative signs while comparing ratios.

### Question 3

*3 marks · Short answer*

On 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.
(i) $3x + 2y = 5$; $2x - 3y = 7$
(ii) $2x - 3y = 8$; $4x - 6y = 9$
(iii) $\frac{3}{2}x + \frac{5}{3}y = 7$; $9x - 10y = 14$
(iv) $5x - 3y = 11$; $-10x + 6y = -22$
(v) $\frac{4}{3}x + 2y = 8$; $2x + 3y = 12$

**Part (i)**

1. Here, $a_1 = 3$, $b_1 = 2$, $c_1 = -5$ and $a_2 = 2$, $b_2 = -3$, $c_2 = -7$.
2. The ratios are $\frac{a_1}{a_2} = \frac{3}{2}$, $\frac{b_1}{b_2} = \frac{2}{-3}$, and $\frac{c_1}{c_2} = \frac{5}{7}$.
3. Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the pair of linear equations is consistent.

Answer (i): Consistent

**Part (ii)**

1. Here, $a_1 = 2$, $b_1 = -3$, $c_1 = -8$ and $a_2 = 4$, $b_2 = -6$, $c_2 = -9$.
2. The ratios are $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2}$, and $\frac{c_1}{c_2} = \frac{-8}{-9} = \frac{8}{9}$.
3. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the pair of linear equations is inconsistent.

Answer (ii): Inconsistent

**Part (iii)**

1. Here, $a_1 = \frac{3}{2}$, $b_1 = \frac{5}{3}$, $c_1 = -7$ and $a_2 = 9$, $b_2 = -10$, $c_2 = -14$.
2. The ratios are $\frac{a_1}{a_2} = \frac{3}{2 \times 9} = \frac{1}{6}$, $\frac{b_1}{b_2} = \frac{5}{3 \times (-10)} = -\frac{1}{6}$.
3. Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the pair of linear equations is consistent.

Answer (iii): Consistent

**Part (iv)**

1. Here, $a_1 = 5$, $b_1 = -3$, $c_1 = -11$ and $a_2 = -10$, $b_2 = 6$, $c_2 = 22$.
2. The ratios are $\frac{a_1}{a_2} = \frac{5}{-10} = -\frac{1}{2}$, $\frac{b_1}{b_2} = \frac{-3}{6} = -\frac{1}{2}$, and $\frac{c_1}{c_2} = \frac{-11}{22} = -\frac{1}{2}$.
3. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the pair of linear equations is dependent and consistent.

Answer (iv): Consistent

**Part (v)**

1. Here, $a_1 = \frac{4}{3}$, $b_1 = 2$, $c_1 = -8$ and $a_2 = 2$, $b_2 = 3$, $c_2 = -12$.
2. The ratios are $\frac{a_1}{a_2} = \frac{4}{3 \times 2} = \frac{2}{3}$, $\frac{b_1}{b_2} = \frac{2}{3}$, and $\frac{c_1}{c_2} = \frac{-8}{-12} = \frac{2}{3}$.
3. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the pair of linear equations is dependent and consistent.

Answer (v): Consistent

**Answer:** (i) Consistent, (ii) Inconsistent, (iii) Consistent, (iv) Consistent, (v) Consistent

> Common mistake: Comparing ratios without shifting all terms to one side, leading to incorrect signs for constant terms c1 and c2.

### Question 4

*3 marks · Short answer*

Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) $x + y = 5$, $2x + 2y = 10$
(ii) $x - y = 8$, $3x - 3y = 16$
(iii) $2x + y - 6 = 0$, $4x - 2y - 4 = 0$
(iv) $2x - 2y - 2 = 0$, $4x - 4y - 5 = 0$

**Part (i)**

1. Here $\frac{a_1}{a_2} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{1}{2}$, and $\frac{c_1}{c_2} = \frac{5}{10} = \frac{1}{2}$.
2. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident and the pair is consistent with infinitely many solutions.

Answer (i): Consistent (infinitely many solutions)

**Part (ii)**

1. Here $\frac{a_1}{a_2} = \frac{1}{3}$, $\frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3}$, and $\frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2}$.
2. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel and the pair is inconsistent.

Answer (ii): Inconsistent

**Part (iii)**

1. Here $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$ and $\frac{b_1}{b_2} = \frac{1}{-2}$.
2. Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the pair is consistent.
3. Solving graphically or algebraically, the lines intersect at $(2, 2)$.

Answer (iii): Consistent, Solution: $x = 2, y = 2$

**Part (iv)**

1. Here $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2}$, and $\frac{c_1}{c_2} = \frac{-2}{-5} = \frac{2}{5}$.
2. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel and the pair is inconsistent.

Answer (iv): Inconsistent

**Answer:** For (i) Consistent with infinitely many solutions, For (ii) Inconsistent, For (iii) Consistent with unique solution $(2, 2)$, For (iv) Inconsistent

> Common mistake: Misidentifying consistent pairs when lines are coincident.

### Question 5

*3 marks · Short answer*

Half the perimeter of a rectangular garden, whose length is $4\text{ m}$ more than its width, is $36\text{ m}$. Find the dimensions of the garden.

**Solution**

1. Let the width of the rectangular garden be $x\text{ m}$ and the length be $y\text{ m}$.
2. According to the question, length is $4\text{ m}$ more than its width, so $y = x + 4$.
3. Also, half the perimeter is $36\text{ m}$, so $\frac{1}{2} \times 2(x + y) = 36$, which gives $x + y = 36$.
4. Substitute $y = x + 4$ into the second equation: $x + (x + 4) = 36$, giving $2x = 32$ or $x = 16\text{ m}$.
5. Substitute $x = 16$ into the first equation: $y = 16 + 4 = 20\text{ m}$.
6. Therefore, the dimensions of the garden are length $= 20\text{ m}$ and width $= 16\text{ m}$.

**Answer:** Length = 20 m, Width = 16 m

> Common mistake: Taking half the perimeter formula incorrectly as full perimeter.

### Question 6

*2 marks · Very short answer*

Given 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:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines

**Part (i)**

1. Given equation is $2x + 3y - 8 = 0$.
2. For intersecting lines, $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
3. A valid equation is $3x + 2y - 5 = 0$.

Answer (i): $3x + 2y - 5 = 0$

**Part (ii)**

1. Given equation is $2x + 3y - 8 = 0$.
2. For parallel lines, $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
3. A valid equation is $4x + 6y - 16 = 0$.

Answer (ii): $4x + 6y - 16 = 0$

**Part (iii)**

1. Given equation is $2x + 3y - 8 = 0$.
2. For coincident lines, $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
3. A valid equation is $4x + 6y - 16 = 0$.

Answer (iii): $4x + 6y - 16 = 0$

**Answer:** For (i) $3x + 2y - 5 = 0$, For (ii) $4x + 6y - 16 = 0$, For (iii) $4x + 6y - 16 = 0$

> Common mistake: Not checking all three ratio conditions for parallel and coincident lines.

### Question 7

*3 marks · Short answer*

Draw 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.

**Solution**

1. For the equation $x - y + 1 = 0$, when $x = 0$, $y = 1$; when $x = 2$, $y = 3$.
2. For the equation $3x + 2y - 12 = 0$, when $x = 0$, $y = 6$; when $x = 4$, $y = 0$.
3. Solving the two equations simultaneously, we get the intersection point as $(2, 3)$.
4. The two lines intersect the $x$-axis at $(-1, 0)$ and $(4, 0)$, forming a triangle with the $x$-axis.
5. The coordinates of the vertices of the triangle are $(-1, 0)$, $(4, 0)$, and $(2, 3)$.

**Answer:** The vertices of the triangle are $(-1, 0)$, $(4, 0)$, and $(2, 3)$.

> Common mistake: Incorrectly identifying the points where the lines intersect the x-axis or misreading the intersection point of the two lines.

## EXERCISE 3.2

### Question 1

*3 marks · Short answer*

Solve the following pair of linear equations by the substitution method.
(i) $x + y = 14$, $x - y = 4$
(ii) $s - t = 3$, $\frac{s}{3} + \frac{t}{2} = 6$
(iii) $3x - y = 3$, $9x - 3y = 9$
(iv) $0.2x + 0.3y = 1.3$, $0.4x + 0.5y = 2.3$
(v) $\sqrt{2}x + \sqrt{3}y = 0$, $\sqrt{3}x - \sqrt{8}y = 0$
(vi) $\frac{3x}{2} - \frac{5y}{3} = -2$, $\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$

**Part (i)**

1. Given equations are $x + y = 14$ and $x - y = 4$.
2. From $x - y = 4$, we get $x = y + 4$.
3. Substituting $x$ in the first equation, $(y + 4) + y = 14$, which gives $y = 5$.
4. Putting $y = 5$ in $x = y + 4$, we get $x = 9$.

Answer (i): $x = 9, y = 5$

**Part (ii)**

1. Given equations are $s - t = 3$ and $\frac{s}{3} + \frac{t}{2} = 6$.
2. From $s - t = 3$, we get $s = t + 3$.
3. Substituting $s$ in the second equation, $\frac{t + 3}{3} + \frac{t}{2} = 6$, giving $t = 6$.
4. Putting $t = 6$ in $s = t + 3$, we get $s = 9$.

Answer (ii): $s = 9, t = 6$

**Part (iii)**

1. Given equations are $3x - y = 3$ and $9x - 3y = 9$.
2. From $3x - y = 3$, we get $y = 3x - 3$.
3. Substituting $y$ in the second equation, $9x - 3(3x - 3) = 9$, which simplifies to $9 = 9$.
4. This is a true statement for all values of $x$, hence the pair has infinitely many solutions.

Answer (iii): Infinitely many solutions

**Part (iv)**

1. Given equations are $0.2x + 0.3y = 1.3$ and $0.4x + 0.5y = 2.3$.
2. From the first equation, $x = \frac{1.3 - 0.3y}{0.2}$.
3. Substituting this into the second equation gives $y = 3$.
4. Putting $y = 3$ gives $x = 2$.

Answer (iv): $x = 2, y = 3$

**Part (v)**

1. Given equations are $\sqrt{2}x + \sqrt{3}y = 0$ and $\sqrt{3}x - \sqrt{8}y = 0$.
2. From the first equation, $x = -\frac{\sqrt{3}}{\sqrt{2}}y$.
3. Substituting $x$ in the second equation gives $\sqrt{3}\left(-\frac{\sqrt{3}}{\sqrt{2}}y\right) - 2\sqrt{2}y = 0$, which yields $y = 0$.
4. Putting $y = 0$ gives $x = 0$.

Answer (v): $x = 0, y = 0$

**Part (vi)**

1. Given equations are $\frac{3x}{2} - \frac{5y}{3} = -2$ and $\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$.
2. Simplifying the equations gives $9x - 10y = -12$ and $2x + 3y = 13$.
3. Expressing $x$ from the second equation as $x = \frac{13 - 3y}{2}$ and substituting in the first gives $y = 3$.
4. Substituting $y = 3$ gives $x = 2$.

Answer (vi): $x = 2, y = 3$

**Answer:** Solutions obtained by substitution method.

> Common mistake: Errors in algebraic simplification of fractions and decimals.

### Question 2

*3 marks · Short answer*

Solve $2x + 3y = 11$ and $2x - 4y = -24$ and hence find the value of '$m$' for which $y = mx + 3$.

**Solution**

1. Given equations are $2x + 3y = 11$ and $2x - 4y = -24$.
2. Subtract the second equation from the first equation: $(2x - 2x) + (3y - (-4y)) = 11 - (-24)$, which gives $7y = 35$, so $y = 5$.
3. Substitute $y = 5$ into the first equation: $2x + 3(5) = 11$, leading to $2x + 15 = 11$, so $2x = -4$, and $x = -2$.
4. Substitute $x = -2$ and $y = 5$ into the relation $y = mx + 3$ to find $m$.
5. We get $5 = m(-2) + 3$, which simplifies to $2 = -2m$, so $m = -1$.

**Answer:** $m = -1$

> Common mistake: Substituting the values of x and y incorrectly into the linear relation for m.

### Question 3

*3 marks · Short answer*

Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(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.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of $10\text{ km}$, the charge paid is ₹ 105 and for a journey of $15\text{ km}$, 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 $25\text{ km}$?
(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.
(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?

**Part (i)**

1. Let the two numbers be $x$ and $y$ such that $x > y$.
2. Given $x - y = 26$ and $x = 3y$.
3. Substituting $x = 3y$ into the first equation, $3y - y = 26$, giving $y = 13$.
4. Thus, $x = 3(13) = 39$.

Answer (i): Numbers are 39 and 13.

**Part (ii)**

1. Let the larger angle be $x$ and smaller angle be $y$.
2. Given $x + y = 180^\circ$ and $x - y = 18^\circ$.
3. From $x = y + 18^\circ$, substituting in the first equation gives $(y + 18^\circ) + y = 180^\circ$, so $y = 81^\circ$.
4. Thus, $x = 99^\circ$.

Answer (ii): Angles are $99^\circ$ and $81^\circ$.

**Part (iii)**

1. Let the cost of one bat be $x$ and one ball be $y$.
2. Form equations: $7x + 6y = 3800$ and $3x + 5y = 1750$.
3. Express $x$ from the second equation as $x = \frac{1750 - 5y}{3}$ and substitute into the first.
4. Solving gives $y = 50$ and $x = 500$.

Answer (iii): Cost of a bat is ₹ 500 and of a ball is ₹ 50.

**Part (iv)**

1. Let fixed charge be ₹ $x$ and charge per km be ₹ $y$.
2. Form equations: $x + 10y = 105$ and $x + 15y = 155$.
3. From the first, $x = 105 - 10y$; substituting in the second gives $y = 10$ and $x = 5$.
4. For $25\text{ km}$, total charge is $x + 25y = 5 + 25(10) = 255$.

Answer (iv): Fixed charge ₹ 5, charge per km ₹ 10, total for 25 km is ₹ 255.

**Part (v)**

1. Let the fraction be $\frac{x}{y}$.
2. Given $\frac{x + 2}{y + 2} = \frac{9}{11}$ and $\frac{x + 3}{y + 3} = \frac{5}{6}$.
3. Simplify to form $11x - 9y = -4$ and $6x - 5y = -3$.
4. Solving by substitution yields $x = 7$ and $y = 9$.

Answer (v): The fraction is $\frac{7}{9}$.

**Part (vi)**

1. Let Jacob's present age be $x$ and his son's age be $y$.
2. Given equations are $x + 5 = 3(y + 5)$ and $x - 5 = 7(y - 5)$
3. Simplifying gives $x - 3y = 10$ and $x - 7y = -30$.
4. Solving by substitution gives $y = 10$ and $x = 40$.

Answer (vi): Jacob's age is 40 years and son's age is 10 years.

**Answer:** Solutions to the word problems.

> Common mistake: Forming incorrect linear equations from the word statements.

## EXERCISE 3.3

### Question 1

*3 marks · Short answer*

Solve the following pair of linear equations by the elimination method and the substitution method :
(i) $x + y = 5$ and $2x - 3y = 4$
(ii) $3x + 4y = 10$ and $2x - 2y = 2$
(iii) $3x - 5y - 4 = 0$ and $9x = 2y + 7$
(iv) $\frac{x}{2} + \frac{2y}{3} = -1$ and $x - \frac{y}{3} = 3$

**Part (i)**

1. Given equations are $x + y = 5$ and $2x - 3y = 4$.
2. Multiplying the first equation by $3$, we get $3x + 3y = 15$.
3. Adding this to $2x - 3y = 4$ gives $5x = 19$, so $x = \frac{19}{5}$.
4. Substituting $x = \frac{19}{5}$ in $x + y = 5$ gives $y = 5 - \frac{19}{5} = \frac{6}{5}$.
5. The solution is $x = \frac{19}{5}$ and $y = \frac{6}{5}$.

Answer (i): $x = \frac{19}{5}, y = \frac{6}{5}$

**Part (ii)**

1. Given equations are $3x + 4y = 10$ and $2x - 2y = 2$.
2. Dividing the second equation by $2$, we get $x - y = 1$, or $x = y + 1$.
3. Substituting $x = y + 1$ in the first equation gives $3(y + 1) + 4y = 10$.
4. This simplifies to $7y + 3 = 10$, which gives $7y = 7$, or $y = 1$.
5. Substituting $y = 1$ gives $x = 1 + 1 = 2$.
6. The solution is $x = 2$ and $y = 1$.

Answer (ii): $x = 2, y = 1$

**Part (iii)**

1. Given equations are $3x - 5y - 4 = 0$ (or $3x - 5y = 4$) and $9x - 2y = 7$.
2. Multiplying the first equation by $3$, we get $9x - 15y = 12$.
3. Subtracting this from $9x - 2y = 7$ gives $13y = -5$, so $y = -\frac{5}{13}$.
4. Substituting $y = -\frac{5}{13}$ in $3x - 5y = 4$ gives $3x - 5\left(-\frac{5}{13}\right) = 4$.
5. This simplifies to $3x = 4 - \frac{25}{13} = \frac{27}{13}$, so $x = \frac{9}{13}$.
6. The solution is $x = \frac{9}{13}$ and $y = -\frac{5}{13}$.

Answer (iii): $x = \frac{9}{13}, y = -\frac{5}{13}$

**Part (iv)**

1. Given equations are $\frac{x}{2} + \frac{2y}{3} = -1$ and $x - \frac{y}{3} = 3$.
2. Multiplying the first equation by $6$, we get $3x + 4y = -6$. Multiplying the second by $3$, we get $3x - y = 9$.
3. Subtracting the second equation from the first gives $5y = -15$, so $y = -3$.
4. Substituting $y = -3$ in $3x - y = 9$ gives $3x - (-3) = 9$, which means $3x = 6$, so $x = 2$.
5. The solution is $x = 2$ and $y = -3$.

Answer (iv): $x = 2, y = -3$

**Answer:** Solutions obtained by elimination and substitution methods.

> Common mistake: Making errors with signs when eliminating or substituting fractions.

### Question 2

*5 marks · Case-based*

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

**Part (i)**

1. Let the numerator be $x$ and the denominator be $y$, so the fraction is $\frac{x}{y}$.
2. According to the question, $\frac{x + 1}{y - 1} = 1 \implies x - y = -2$ and $\frac{x}{y + 1} = \frac{1}{2} \implies 2x - y = 1$.
3. Subtracting the first equation from the second gives $x = 3$.
4. Substituting $x = 3$ into $x - y = -2$ gives $y = 5$.
5. Therefore, the required fraction is $\frac{3}{5}$.

Answer (i): The fraction is $\frac{3}{5}$.

**Part (ii)**

1. Let Nuri's present age be $x$ years and Sonu's present age be $y$ years.
2. Five years ago, $x - 5 = 3(y - 5) \implies x - 3y = -10$.
3. Ten years later, $x + 10 = 2(y + 10) \implies x - 2y = 10$.
4. Subtracting the first equation from the second gives $y = 20$.
5. Substituting $y = 20$ into $x - 2y = 10$ gives $x = 50$.
6. Therefore, Nuri's age is $50$ years and Sonu's age is $20$ years.

Answer (ii): Nuri is $50$ years old and Sonu is $20$ years old.

**Part (iii)**

1. Let the ten's digit be $x$ and the unit's digit be $y$. The number is $10x + y$.
2. The sum of the digits is $9$, so $x + y = 9$.
3. Nine times the number is twice the number obtained by reversing the digits: $9(10x + y) = 2(10y + x)$.
4. Simplifying gives $88x - 11y = 0 \implies 8x - y = 0$.
5. Adding $x + y = 9$ and $8x - y = 0$ gives $9x = 9 \implies x = 1$, and $y = 8$.
6. Therefore, the number is $18$.

Answer (iii): The number is $18$.

**Part (iv)**

1. Let the number of ₹ 50 notes be $x$ and ₹ 100 notes be $y$.
2. Total notes: $x + y = 25$.
3. Total amount: $50x + 100y = 2000 \implies x + 2y = 40$.
4. Subtracting the first equation from the second gives $y = 15$.
5. Substituting $y = 15$ into $x + y = 25$ gives $x = 10$.
6. Therefore, Meena received $10$ notes of ₹ 50 and $15$ notes of ₹ 100.

Answer (iv): $10$ notes of ₹ 50 and $15$ notes of ₹ 100.

**Part (v)**

1. Let the fixed charge for the first three days be ₹ $x$ and the charge for each extra day be ₹ $y$.
2. Saritha paid ₹ 27 for 7 days: $x + 4y = 27$.
3. Susy paid ₹ 21 for 5 days: $x + 2y = 21$.
4. Subtracting the second equation from the first gives $2y = 6 \implies y = 3$.
5. Substituting $y = 3$ into $x + 2y = 21$ gives $x = 15$.
6. Therefore, the fixed charge is ₹ 15 and the charge for each extra day is ₹ 3.

Answer (v): Fixed charge is ₹ 15 and extra day charge is ₹ 3.

**Answer:** Solutions to all five parts found using elimination method.

> Common mistake: Forming incorrect algebraic equations from the given word statements or making sign errors during subtraction in the elimination method.

## Frequently asked questions

### How many exercises and questions are there in NCERT Solutions for Class 10 Maths Chapter 3?

This chapter has a total of 3 exercises with 12 questions in all, spread across Exercise 3.1, Exercise 3.2, and Exercise 3.3. You can find SwaVid's free PDF and step-by-step solutions for all these questions on this page only.

### What topics do the questions in this chapter cover?

The questions cover checking consistency and solving graphically, comparison of ratios for intersecting, parallel, and coincident lines, and finding the vertices of a triangle. They also include solving equations using the substitution and elimination methods along with various word problems.

### Which are the hardest question types in this chapter and how should I approach them?

Word problems leading to linear equations and graphical questions involving finding triangle vertices are generally considered the toughest. To approach them, first define your variables clearly, translate the given conditions into algebraic equations like $a_1x + b_1y + c_1 = 0$, and practice plotting accurate graphs.

### How can I write answers to score full marks in board exams for this chapter?

To secure full marks, always write the given statements clearly, show proper algebraic steps for substitution or elimination methods, and state the final condition ratios explicitly. Referring to SwaVid's free PDF and step-by-step solutions available on this page only will help you understand the ideal presentation format.

### Is a free PDF of these solutions available for download?

Yes, SwaVid provides a complete free PDF and step-by-step solutions for this chapter right on this page only. You can easily download or view them online to prepare for the 2026-27 session exams.

## Related pages

- [Exercise 3.1 solutions](https://www.swavid.com/maths/class/10/chapter/pair-of-linear-equations-in-two-variables/ncert-solutions/exercise-3-1)
- [Exercise 3.2 solutions](https://www.swavid.com/maths/class/10/chapter/pair-of-linear-equations-in-two-variables/ncert-solutions/exercise-3-2)
- [Exercise 3.3 solutions](https://www.swavid.com/maths/class/10/chapter/pair-of-linear-equations-in-two-variables/ncert-solutions/exercise-3-3)
- [Pair of Linear Equations in Two Variables: CBSE previous year questions](https://www.swavid.com/cbse/class-10/maths/pyq/pair-of-linear-equations-in-two-variables)
- [Class 10 Maths chapters](https://www.swavid.com/maths/class/10)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student