NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2#
Welcome to our NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2. This exercise teaches the Substitution Method to solve pairs of linear equations algebraically. The substitution method involves expressing one variable in terms of the other and substituting into the second equation to solve the system.
Substitution Method Steps:
- From one equation, express one variable in terms of the other (e.g., $x$ in terms of $y$)
- Substitute this expression into the second equation
- Solve the resulting single-variable equation
- Substitute back to find the other variable
1. Solve the following pair of linear equations by the substitution method. #
(i) $x + y = 14$; $x - y = 4$
$$x - y = 4 \quad \cdots (2)$$
From equation (1):
$$x + y = 14$$$$\Rightarrow x = 14 - y \quad \cdots (3)$$
Substituting $x = 14 - y$ in equation (2):
$$x - y = 4$$$$(14 - y) - y = 4$$
$$\Rightarrow 14 - 2y = 4$$
$$\Rightarrow 2y = 10$$
$$\Rightarrow y = 5$$
Substituting $y = 5$ in equation (3):
$$x = 14 - y$$$$x = 14 - 5$$
$$\Rightarrow x = 9$$
∴ $x = \mathbf{9}$, $y = \mathbf{5}$
(ii) $s - t = 3$; $\dfrac{s}{3} + \dfrac{t}{2} = 6$
$$\dfrac{s}{3} + \dfrac{t}{2} = 6 \quad \cdots (2)$$
From equation (1):
$$s - t = 3$$$$\Rightarrow s = t + 3 \quad \cdots (3)$$
Substituting $s = t + 3$ in equation (2):
$$\dfrac{s}{3} + \dfrac{t}{2} = 6$$$$\dfrac{t + 3}{3} + \dfrac{t}{2} = 6$$
Multiplying both sides by $6$:
$$6\left(\dfrac{t + 3}{3} + \dfrac{t}{2}\right) = 6 \times 6$$$$\Rightarrow 2(t + 3) + 3t = 36$$
$$\Rightarrow 2t + 6 + 3t = 36$$
$$\Rightarrow 5t = 30$$
$$\Rightarrow t = 6$$
Substituting $t = 6$ in equation (3):
$$s = t + 3$$$$s = 6 + 3$$
$$\Rightarrow s = 9$$
∴ $s = \mathbf{9}$, $t = \mathbf{6}$
(iii) $3x - y = 3$; $9x - 3y = 9$
$$9x - 3y = 9 \quad \cdots (2)$$
From equation (1):
$$3x - y = 3$$$$\Rightarrow y = 3x - 3 \quad \cdots (3)$$
Substituting $y = 3x - 3$ in equation (2):
$$9x - 3y = 9$$$$9x - 3(3x - 3) = 9$$
$$\Rightarrow 9x - 9x + 9 = 9$$
$$\Rightarrow 9 = 9$$
This is always true — the equation is an identity.
This means the two equations are dependent (represent the same line), so there are infinitely many solutions.
The solution set is: $\{(x, y) : y = 3x - 3\}$, i.e., all points on the line $3x - y = 3$.
(iv) $0.2x + 0.3y = 1.3$; $0.4x + 0.5y = 2.3$
$$0.4x + 0.5y = 2.3 \quad \cdots (2)$$
Multiplying both equations by $10$:
$$0.2x + 0.3y = 1.3$$$$\Rightarrow 2x + 3y = 13 \quad \cdots (3)$$
$$0.4x + 0.5y = 2.3$$
$$\Rightarrow 4x + 5y = 23 \quad \cdots (4)$$
From equation (3):
$$2x + 3y = 13$$$$\Rightarrow 2x = 13 - 3y$$
$$\Rightarrow x = \dfrac{13 - 3y}{2} \quad \cdots (5)$$
Substituting $x = \dfrac{13 - 3y}{2}$ in equation (4):
$$4x + 5y = 23$$$$4 \cdot \dfrac{13 - 3y}{2} + 5y = 23$$
$$\Rightarrow 2(13 - 3y) + 5y = 23$$
$$\Rightarrow 26 - 6y + 5y = 23$$
$$\Rightarrow -y = -3$$
$$\Rightarrow y = 3$$
Substituting $y = 3$ in equation (5):
$$x = \dfrac{13 - 3y}{2}$$$$x = \dfrac{13 - 9}{2}$$
$$\Rightarrow x = 2$$
∴ $x = \mathbf{2}$, $y = \mathbf{3}$
(v) $\sqrt{2}\,x + \sqrt{3}\,y = 0$; $\sqrt{3}\,x - \sqrt{8}\,y = 0$
$$\sqrt{3}\,x - \sqrt{8}\,y = 0 \quad \cdots (2)$$
From equation (1):
$$\sqrt{2}\,x + \sqrt{3}\,y = 0$$$$\Rightarrow \sqrt{2}\,x = -\sqrt{3}\,y$$
$$\Rightarrow x = -\dfrac{\sqrt{3}}{\sqrt{2}}\,y \quad \cdots (3)$$
Substituting $x = -\dfrac{\sqrt{3}}{\sqrt{2}}\,y$ in equation (2):
$$\sqrt{3}\,x - \sqrt{8}\,y = 0$$$$\sqrt{3}\left(-\dfrac{\sqrt{3}}{\sqrt{2}}\,y\right) - \sqrt{8}\,y = 0$$
$$\Rightarrow -\dfrac{3}{\sqrt{2}}\,y - 2\sqrt{2}\,y = 0$$
$$\Rightarrow y\left(-\dfrac{3}{\sqrt{2}} - 2\sqrt{2}\right) = 0$$
$$\Rightarrow y\left(\dfrac{-3 - 4}{\sqrt{2}}\right) = 0$$
$$\Rightarrow y \cdot \dfrac{-7}{\sqrt{2}} = 0$$
$$\Rightarrow y = 0$$
Substituting $y = 0$ in equation (3):
$$x = -\dfrac{\sqrt{3}}{\sqrt{2}}\,y$$$$x = -\dfrac{\sqrt{3}}{\sqrt{2}} \times 0$$
$$\Rightarrow x = 0$$
∴ $x = \mathbf{0}$, $y = \mathbf{0}$
(vi) $\dfrac{3x}{2} - \dfrac{5y}{3} = -2$; $\dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6}$
$$\dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6} \quad \cdots (2)$$
Multiplying both equations by $6$:
$$\dfrac{3x}{2} - \dfrac{5y}{3} = -2$$$$\Rightarrow 9x - 10y = -12 \quad \cdots (3)$$
$$\dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6}$$
$$\Rightarrow 2x + 3y = 13 \quad \cdots (4)$$
From equation (4):
$$2x + 3y = 13$$$$\Rightarrow 2x = 13 - 3y$$
$$\Rightarrow x = \dfrac{13 - 3y}{2} \quad \cdots (5)$$
Substituting $x = \dfrac{13 - 3y}{2}$ in equation (3):
$$9x - 10y = -12$$$$9 \cdot \dfrac{13 - 3y}{2} - 10y = -12$$
$$\Rightarrow \dfrac{117 - 27y}{2} - 10y = -12$$
Multiplying both sides by $2$:
$$2\left(\dfrac{117 - 27y}{2} - 10y\right) = 2 \times (-12)$$$$\Rightarrow 117 - 27y - 20y = -24$$
$$\Rightarrow -47y = -141$$
$$\Rightarrow y = 3$$
Substituting $y = 3$ in equation (5):
$$x = \dfrac{13 - 3y}{2}$$$$x = \dfrac{13 - 9}{2}$$
$$\Rightarrow x = 2$$
∴ $x = \mathbf{2}$, $y = \mathbf{3}$
2. Solve $2x + 3y = 11$ and $2x - 4y = -24$ and hence find the value of $m$ for which $y = mx + 3$. #
Answer
$$2x - 4y = -24 \quad \cdots (2)$$
From equation (1):
$$2x + 3y = 11$$$$\Rightarrow 2x = 11 - 3y$$
$$\Rightarrow x = \dfrac{11 - 3y}{2} \quad \cdots (3)$$
Substituting $x = \dfrac{11 - 3y}{2}$ in equation (2):
$$2x - 4y = -24$$$$2 \cdot \dfrac{11 - 3y}{2} - 4y = -24$$
$$\Rightarrow 11 - 3y - 4y = -24$$
$$\Rightarrow -7y = -35$$
$$\Rightarrow y = 5$$
Substituting $y = 5$ in equation (3):
$$x = \dfrac{11 - 3y}{2}$$$$x = \dfrac{11 - 15}{2}$$
$$\Rightarrow x = -2$$
So $x = -2$ and $y = 5$.
Substituting $x = -2$ and $y = 5$ in $y = mx + 3$:
$$y = mx + 3$$$$5 = m(-2) + 3$$
$$\Rightarrow -2m = 2$$
$$\Rightarrow m = -1$$
∴ $m = \mathbf{-1}$
3. 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 the larger number = $x$ and smaller number = $y$.
The difference of the two numbers is 26, so:
$$x - y = 26 \quad \cdots (1)$$One number is three times the other, so:
$$x = 3y \quad \cdots (2)$$Substituting $x = 3y$ from equation (2) in equation (1):
$$x - y = 26$$$$3y - y = 26$$
$$\Rightarrow 2y = 26$$
$$\Rightarrow y = 13$$
Substituting $y = 13$ in equation (2):
$$x = 3y$$$$x = 3 \times 13$$
$$\Rightarrow x = 39$$
∴ The two numbers are 39 and 13.
(ii) The larger of two supplementary angles exceeds the smaller by 18°. Find them.
Let the larger angle be $x$ degrees and the smaller angle be $y$ degrees.
Supplementary angles add up to 180°, so:
$$x + y = 180 \quad \cdots (1)$$The larger angle is 18° more than the smaller angle, so:
$$x - y = 18 \quad \cdots (2)$$From equation (2):
$$x - y = 18$$$$\Rightarrow x = y + 18 \quad \cdots (3)$$
Substituting $x = y + 18$ in equation (1):
$$x + y = 180$$$$(y + 18) + y = 180$$
$$\Rightarrow 2y + 18 = 180$$
$$\Rightarrow 2y = 162$$
$$\Rightarrow y = 81$$
Substituting $y = 81$ in equation (3):
$$x = y + 18$$$$x = 81 + 18$$
$$\Rightarrow x = 99$$
∴ The 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 one bat = ₹$x$ and one ball = ₹$y$.
7 bats and 6 balls cost ₹3800, so:
$$7x + 6y = 3800 \quad \cdots (1)$$3 bats and 5 balls cost ₹1750, so:
$$3x + 5y = 1750 \quad \cdots (2)$$From equation (2):
$$3x + 5y = 1750$$$$\Rightarrow 3x = 1750 - 5y$$
$$\Rightarrow x = \dfrac{1750 - 5y}{3} \quad \cdots (3)$$
Substituting $x = \dfrac{1750 - 5y}{3}$ in equation (1):
$$7x + 6y = 3800$$$$7 \cdot \dfrac{1750 - 5y}{3} + 6y = 3800$$
$$\Rightarrow \dfrac{12250 - 35y}{3} + 6y = 3800$$
Multiplying both sides by $3$:
$$3\left(\dfrac{12250 - 35y}{3} + 6y\right) = 3 \times 3800$$$$\Rightarrow 12250 - 35y + 18y = 11400$$
$$\Rightarrow 12250 - 17y = 11400$$
$$\Rightarrow -17y = -850$$
$$\Rightarrow y = 50$$
Substituting $y = 50$ in equation (3):
$$x = \dfrac{1750 - 5y}{3}$$$$x = \dfrac{1750 - 250}{3}$$
$$\Rightarrow x = 500$$
∴ Cost of one bat = ₹500 and cost of one 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 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?
Let fixed charge = ₹$x$ and charge per km = ₹$y$.
The total charge is the fixed charge plus the charge per km times the distance.
For 10 km, the charge is ₹105, so:
$$x + 10y = 105 \quad \cdots (1)$$For 15 km, the charge is ₹155, so:
$$x + 15y = 155 \quad \cdots (2)$$From equation (1):
$$x + 10y = 105$$$$\Rightarrow x = 105 - 10y \quad \cdots (3)$$
Substituting $x = 105 - 10y$ in equation (2):
$$x + 15y = 155$$$$(105 - 10y) + 15y = 155$$
$$\Rightarrow 105 + 5y = 155$$
$$\Rightarrow 5y = 50$$
$$\Rightarrow y = 10$$
Substituting $y = 10$ in equation (3):
$$x = 105 - 10y$$$$x = 105 - 100$$
$$\Rightarrow x = 5$$
For 25 km, the charge is $x + 25y$:
$$= 5 + 25 \times 10$$$$= 5 + 250$$
$$= 255$$
∴ Fixed charge = ₹5, Charge per km = ₹10, Cost for 25 km = ₹255.
(v) A fraction becomes $\dfrac{9}{11}$ if 2 is added to both numerator and denominator. If 3 is added to both, it becomes $\dfrac{5}{6}$. Find the fraction.
Let numerator = $x$ and denominator = $y$.
Adding 2 to the numerator and the denominator gives $\dfrac{9}{11}$, so:
$$\dfrac{x + 2}{y + 2} = \dfrac{9}{11}$$$$\Rightarrow 11(x + 2) = 9(y + 2)$$
$$\Rightarrow 11x + 22 = 9y + 18$$
$$\Rightarrow 11x - 9y = -4 \quad \cdots (1)$$
Adding 3 to the numerator and the denominator gives $\dfrac{5}{6}$, so:
$$\dfrac{x + 3}{y + 3} = \dfrac{5}{6}$$$$\Rightarrow 6(x + 3) = 5(y + 3)$$
$$\Rightarrow 6x + 18 = 5y + 15$$
$$\Rightarrow 6x - 5y = -3 \quad \cdots (2)$$
From equation (2):
$$6x - 5y = -3$$$$\Rightarrow 6x = 5y - 3$$
$$\Rightarrow x = \dfrac{5y - 3}{6} \quad \cdots (3)$$
Substituting $x = \dfrac{5y - 3}{6}$ in equation (1):
$$11x - 9y = -4$$$$11 \cdot \dfrac{5y - 3}{6} - 9y = -4$$
$$\Rightarrow \dfrac{55y - 33}{6} - 9y = -4$$
Multiplying both sides by $6$:
$$6\left(\dfrac{55y - 33}{6} - 9y\right) = 6 \times (-4)$$$$\Rightarrow 55y - 33 - 54y = -24$$
$$\Rightarrow y - 33 = -24$$
$$\Rightarrow y = 9$$
Substituting $y = 9$ in equation (3):
$$x = \dfrac{5y - 3}{6}$$$$x = \dfrac{45 - 3}{6}$$
$$\Rightarrow x = 7$$
∴ The fraction is $\dfrac{7}{9}$.
(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 present age = $x$ years and son’s present age = $y$ years.
In 5 years, Jacob’s age will be 3 times his son’s age, so:
$$x + 5 = 3(y + 5)$$$$\Rightarrow x + 5 = 3y + 15$$
$$\Rightarrow x - 3y = 10 \quad \cdots (1)$$
5 years ago, Jacob’s age was 7 times his son’s age, so:
$$x - 5 = 7(y - 5)$$$$\Rightarrow x - 5 = 7y - 35$$
$$\Rightarrow x - 7y = -30 \quad \cdots (2)$$
From equation (1):
$$x - 3y = 10$$$$\Rightarrow x = 3y + 10 \quad \cdots (3)$$
Substituting $x = 3y + 10$ in equation (2):
$$x - 7y = -30$$$$(3y + 10) - 7y = -30$$
$$\Rightarrow -4y + 10 = -30$$
$$\Rightarrow -4y = -40$$
$$\Rightarrow y = 10$$
Substituting $y = 10$ in equation (3):
$$x = 3y + 10$$$$x = 3 \times 10 + 10$$
$$\Rightarrow x = 40$$
∴ Jacob’s present age = 40 years and Son’s present age = 10 years.