NCERT Class 10 Maths – Exercise 3.1 Solutions

NCERT Class 10 Maths

Chapter 3 – Pair of Linear Equations in Two Variables | Exercise 3.1

(Rationalized Syllabus 2025-26)

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.

Let the number of girls be $x$ and the number of boys be $y$.

$$x + y = 10 \quad \text{…(1)}$$

$$x – y = 4 \quad \Rightarrow x = y + 4 \quad \text{…(2)}$$

Coordinate Table:

For $x + y = 10$ ($y = 10-x$)

x	0	10	5
y	10	0	5

For $x – y = 4$ ($y = x-4$)

x	4	0	7
y	0	-4	3

From the graph, the lines intersect at point $(7, 3)$.

✔️ Number of Girls = 7, Number of Boys = 3

(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.

Let cost of 1 pencil be ₹$x$ and 1 pen be ₹$y$.

$$5x + 7y = 50 \quad \text{…(1)}$$

$$7x + 5y = 46 \quad \text{…(2)}$$

Coordinate Table:

For $5x + 7y = 50$

x	3	10	-4
y	5	0	10

For $7x + 5y = 46$

x	3	8	-2
y	5	-2	12

From the graph, the lines intersect at $(3, 5)$.

✔️ Cost of Pencil = ₹3, Cost of Pen = ₹5

On comparing the ratios $\frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2}$, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

(i) $5x – 4y + 8 = 0$ and $7x + 6y – 9 = 0$

$$\frac{a_1}{a_2} = \frac{5}{7}, \quad \frac{b_1}{b_2} = \frac{-4}{6} = \frac{-2}{3}$$

Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$:

✔️ Intersect at a point

(ii) $9x + 3y + 12 = 0$ and $18x + 6y + 24 = 0$

$$\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}$$

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$:

✔️ Coincident lines

(iii) $6x – 3y + 10 = 0$ 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 $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$:

✖️ Parallel lines

On comparing the ratios, find out whether the following pairs of linear equations are consistent or inconsistent:

(i) $3x + 2y = 5; \quad 2x – 3y = 7$

$$\frac{3}{2} \neq \frac{2}{-3}$$

✔️ Consistent (Intersecting)

(ii) $2x – 3y = 8; \quad 4x – 6y = 9$

$$\frac{2}{4} = \frac{-3}{-6} \neq \frac{8}{9} \Rightarrow \frac{1}{2} = \frac{1}{2} \neq \frac{8}{9}$$

✖️ Inconsistent (Parallel)

(iii) $\frac{3}{2}x + \frac{5}{3}y = 7; \quad 9x – 10y = 14$

$$\frac{3/2}{9} = \frac{1}{6}, \quad \frac{5/3}{-10} = \frac{-1}{6}$$

Since $\frac{1}{6} \neq -\frac{1}{6}$:

✔️ Consistent (Intersecting)

(iv) $5x – 3y = 11; \quad -10x + 6y = -22$

$$\frac{5}{-10} = \frac{-3}{6} = \frac{11}{-22} \Rightarrow -\frac{1}{2} = -\frac{1}{2} = -\frac{1}{2}$$

✔️ Consistent (Dependent/Coincident)

(v) $\frac{4}{3}x + 2y = 8; \quad 2x + 3y = 12$

$$\frac{4/3}{2} = \frac{2}{3}, \quad \frac{2}{3} = \frac{2}{3}, \quad \frac{8}{12} = \frac{2}{3}$$

✔️ Consistent (Dependent/Coincident)

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

(i) $x + y = 5, \quad 2x + 2y = 10$

Ratios: $\frac{1}{2} = \frac{1}{2} = \frac{5}{10}$. Coincident lines.

Consistent (Infinitely many solutions)

(ii) $x – y = 8, \quad 3x – 3y = 16$

Ratios: $\frac{1}{3} = \frac{-1}{-3} \neq \frac{8}{16}$. Parallel lines.

Inconsistent

(iii) $2x + y – 6 = 0, \quad 4x – 2y – 4 = 0$

Ratios: $\frac{2}{4} \neq \frac{1}{-2}$. Intersecting lines.

Consistent

Graphing:

$2x + y = 6 \rightarrow$ Points: $(0,6), (3,0)$

$4x – 2y = 4 \rightarrow$ Points: $(1,0), (2,2)$

Intersection Point: (2, 2)

(iv) $2x – 2y – 2 = 0, \quad 4x – 4y – 5 = 0$

Ratios: $\frac{2}{4} = \frac{-2}{-4} \neq \frac{-2}{-5}$.

Inconsistent

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

Let length $= l$ and width $= w$.

According to the question:

$$l = w + 4 \quad \Rightarrow l – w = 4 \quad \text{…(1)}$$

$$\frac{1}{2} \times 2(l + w) = 36 \quad \Rightarrow l + w = 36 \quad \text{…(2)}$$

Adding (1) and (2): $2l = 40 \Rightarrow l = 20$.

Substituting $l$ in (2): $20 + w = 36 \Rightarrow w = 16$.

✔️ Length = 20 m, Width = 16 m

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

Condition: $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$. Example:

$3x + 2y – 7 = 0$

(ii) Parallel lines

Condition: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. Example:

$4x + 6y – 12 = 0$

(iii) Coincident lines

Condition: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. Example:

$6x + 9y – 24 = 0$

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.

Equation 1: $x – y = -1 \Rightarrow y = x + 1$

x	0	-1	2
y	1	0	3

Equation 2: $3x + 2y = 12 \Rightarrow y = \frac{12-3x}{2}$

x	0	4	2
y	6	0	3

From the graph, the lines intersect at $(2, 3)$.

The lines cut the x-axis ($y=0$) at:

Eq 1: $(-1, 0)$ and Eq 2: $(4, 0)$.

✔️ Vertices: (2, 3), (-1, 0), and (4, 0)

🎉 Exercise 3.1 Completed