---
title: "NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.1"
url: https://www.swavid.com/maths/class/10/chapter/pair-of-linear-equations-in-two-variables/ncert-solutions/exercise-3-1
dateModified: 2026-10-07T15:50:31+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.1

Chapter 3: Pair of Linear Equations in Two Variables. Every question from Exercise 3.1, with full working and the final answer.

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## 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.

## Related pages

- [All Chapter 3 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/pair-of-linear-equations-in-two-variables/ncert-solutions)
- [Exercise 3.2](https://www.swavid.com/maths/class/10/chapter/pair-of-linear-equations-in-two-variables/ncert-solutions/exercise-3-2)
- [Exercise 3.3](https://www.swavid.com/maths/class/10/chapter/pair-of-linear-equations-in-two-variables/ncert-solutions/exercise-3-3)

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