NCERT Class 10 Maths – Exercise 3.2 Solutions

NCERT Class 10 Maths

Chapter 3 – Pair of Linear Equations in Two Variables | Exercise 3.2

(Rationalized Syllabus 2025-26)

Solve the following pair of linear equations by the substitution method.

(i) $x + y = 14$ and $x – y = 4$

$$x + y = 14 \quad \text{…(1)}$$

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

Substitute value of $x$ from (2) into (1):

$$(y + 4) + y = 14$$

$$2y = 10 \Rightarrow y = 5$$

Substitute $y = 5$ in equation (2):

$$x = 5 + 4 = 9$$

✔️ x = 9, y = 5

(ii) $s – t = 3$ and $\frac{s}{3} + \frac{t}{2} = 6$

From first equation: $s = t + 3 \quad \text{…(1)}$

Substitute $s$ into second equation:

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

Multiply by 6 to clear fractions:

$$2(t + 3) + 3t = 36$$

$$2t + 6 + 3t = 36$$

$$5t = 30 \Rightarrow t = 6$$

Substitute $t = 6$ in equation (1): $s = 6 + 3 = 9$.

✔️ s = 9, t = 6

(iii) $3x – y = 3$ and $9x – 3y = 9$

From first equation: $y = 3x – 3 \quad \text{…(1)}$

Substitute into second equation:

$$9x – 3(3x – 3) = 9$$

$$9x – 9x + 9 = 9 \Rightarrow 9 = 9$$

This statement is true for all values of $x$. This means the pair of linear equations has infinitely many solutions.

✔️ Infinitely many solutions

(iv) $0.2x + 0.3y = 1.3$ and $0.4x + 0.5y = 2.3$

Multiply both equations by 10 to remove decimals:

$$2x + 3y = 13 \quad \text{…(1)}$$

$$4x + 5y = 23 \quad \text{…(2)}$$

From (1): $2x = 13 – 3y \Rightarrow x = \frac{13 – 3y}{2}$

Substitute into (2):

$$4\left(\frac{13 – 3y}{2}\right) + 5y = 23$$

$$2(13 – 3y) + 5y = 23$$

$$26 – 6y + 5y = 23 \Rightarrow -y = -3 \Rightarrow y = 3$$

Find $x$: $x = \frac{13 – 3(3)}{2} = \frac{4}{2} = 2$.

✔️ x = 2, y = 3

(v) $\sqrt{2}x + \sqrt{3}y = 0$ and $\sqrt{3}x – \sqrt{8}y = 0$

From first equation: $x = -\frac{\sqrt{3}}{\sqrt{2}}y$. Substitute into second:

$$\sqrt{3}\left(-\frac{\sqrt{3}}{\sqrt{2}}y\right) – \sqrt{8}y = 0$$

$$-\frac{3}{\sqrt{2}}y – 2\sqrt{2}y = 0$$

$$-y\left(\frac{3}{\sqrt{2}} + 2\sqrt{2}\right) = 0$$

Since the term in bracket is not zero, $y$ must be 0.

If $y = 0$, then $x = 0$.

✔️ x = 0, y = 0

(vi) $\frac{3x}{2} – \frac{5y}{3} = -2$ and $\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$

Simplify equations by multiplying by 6:

$$9x – 10y = -12 \quad \text{…(1)}$$

$$2x + 3y = 13 \quad \Rightarrow x = \frac{13 – 3y}{2} \quad \text{…(2)}$$

Substitute (2) into (1):

$$9\left(\frac{13 – 3y}{2}\right) – 10y = -12$$

$$\frac{117 – 27y – 20y}{2} = -12$$

$$117 – 47y = -24 \Rightarrow 47y = 141 \Rightarrow y = 3$$

Find $x$: $x = \frac{13 – 3(3)}{2} = \frac{4}{2} = 2$.

✔️ x = 2, y = 3

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

$$2x + 3y = 11 \quad \text{…(1)}$$

$$2x – 4y = -24 \quad \text{…(2)}$$

From (2): $2x = 4y – 24 \Rightarrow x = 2y – 12$. Substitute into (1):

$$2(2y – 12) + 3y = 11$$

$$4y – 24 + 3y = 11 \Rightarrow 7y = 35 \Rightarrow y = 5$$

Substitute $y=5$ into $x = 2y – 12$:

$$x = 2(5) – 12 = 10 – 12 = -2$$

Now find $m$ using $y = mx + 3$:

$$5 = m(-2) + 3$$

$$2 = -2m \Rightarrow m = -1$$

✔️ x = -2, y = 5, m = -1

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.

Let numbers be $x$ and $y$ (where $x > y$).

$$x – y = 26 \quad \text{…(1)}$$

$$x = 3y \quad \text{…(2)}$$

Substitute (2) into (1): $3y – y = 26 \Rightarrow 2y = 26 \Rightarrow y = 13$.

Then $x = 3(13) = 39$.

✔️ The numbers are 39 and 13.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. [Image of supplementary angles diagram]

Let larger angle be $x$ and smaller be $y$.

$$x + y = 180^\circ \quad \text{(Supplementary)} \quad \text{…(1)}$$

$$x = y + 18 \quad \text{…(2)}$$

Substitute (2) into (1):

$$(y + 18) + y = 180 \Rightarrow 2y = 162 \Rightarrow y = 81^\circ$$

$x = 81 + 18 = 99^\circ$.

✔️ Angles are 99° and 81°

(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.

Let cost of bat = $x$, ball = $y$.

$$7x + 6y = 3800 \quad \text{…(1)}$$

$$3x + 5y = 1750 \quad \Rightarrow x = \frac{1750 – 5y}{3} \quad \text{…(2)}$$

Substitute (2) into (1) and solving:

We get $y = 50$ and $x = 500$.

✔️ Cost of Bat = ₹500, Cost of Ball = ₹50

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

Let fixed charge = $x$ and charge per km = $y$.

$$x + 10y = 105 \quad \text{…(1)}$$

$$x + 15y = 155 \quad \text{…(2)}$$

From (1) $x = 105 – 10y$. Substitute into (2):

$$(105 – 10y) + 15y = 155$$

$$5y = 50 \Rightarrow y = 10$$

$x = 105 – 10(10) = 5$.

Total for 25 km = $x + 25y = 5 + 25(10) = 255$.

✔️ Fixed charge = ₹5, Per km = ₹10 | Cost for 25km = ₹255

(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If 3 is added to both, it becomes 5/6. Find the fraction.

Let fraction be $x/y$.

$$\frac{x+2}{y+2} = \frac{9}{11} \Rightarrow 11x – 9y = -4 \quad \text{…(1)}$$

$$\frac{x+3}{y+3} = \frac{5}{6} \Rightarrow 6x – 5y = -3 \quad \text{…(2)}$$

Solving these by substitution gives $x = 7$ and $y = 9$.

✔️ The fraction is 7/9

(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?

Let Jacob’s age = $x$ and Son’s age = $y$.

After 5 years: $x + 5 = 3(y + 5) \Rightarrow x – 3y = 10$.

5 years ago: $x – 5 = 7(y – 5) \Rightarrow x – 7y = -30$.

Solving these equations: $y = 10, x = 40$.

✔️ Jacob’s age = 40 years, Son’s age = 10 years