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) $x + y = 14$; $x - y = 4$
From the first equation:
$$x = 14 - y \quad \cdots (1)$$Substituting into the second equation:
$$(14 - y) - y = 4$$$$14 - 2y = 4$$$$2y = 10 \implies y = 5$$Substituting $y = 5$ into (1):
$$x = 14 - 5 = 9$$∴ $x = \mathbf{9}$, $y = \mathbf{5}$
(ii) $s - t = 3$; $\dfrac{s}{3} + \dfrac{t}{2} = 6$
From the first equation:
$$s = t + 3 \quad \cdots (1)$$Substituting into the second equation:
$$\frac{t+3}{3} + \frac{t}{2} = 6$$Multiply through by $6$:
$$2(t+3) + 3t = 36$$$$2t + 6 + 3t = 36$$$$5t = 30 \implies t = 6$$From (1): $s = 6 + 3 = 9$
∴ $s = \mathbf{9}$, $t = \mathbf{6}$
(iii) $3x - y = 3$; $9x - 3y = 9$
From the first equation:
$$y = 3x - 3 \quad \cdots (1)$$Substituting into the second equation:
$$9x - 3(3x - 3) = 9$$$$9x - 9x + 9 = 9$$$$9 = 9$$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) $0.2x + 0.3y = 1.3$; $0.4x + 0.5y = 2.3$
Multiply both equations by 10:
$$2x + 3y = 13 \quad \cdots (1)$$$$4x + 5y = 23 \quad \cdots (2)$$From (1): $x = \dfrac{13 - 3y}{2}$
Substituting into (2):
$$4 \cdot \frac{13-3y}{2} + 5y = 23$$$$2(13 - 3y) + 5y = 23$$$$26 - 6y + 5y = 23$$$$-y = -3 \implies y = 3$$$x = \dfrac{13 - 9}{2} = 2$
∴ $x = \mathbf{2}$, $y = \mathbf{3}$
(v) $\sqrt{2}\,x + \sqrt{3}\,y = 0$; $\sqrt{3}\,x - \sqrt{8}\,y = 0$
From the first equation:
$$x = -\frac{\sqrt{3}}{\sqrt{2}}\,y \quad \cdots (1)$$Substituting into the second equation:
$$\sqrt{3}\left(-\frac{\sqrt{3}}{\sqrt{2}}\,y\right) - \sqrt{8}\,y = 0$$$$-\frac{3}{\sqrt{2}}\,y - 2\sqrt{2}\,y = 0$$$$y\left(-\frac{3}{\sqrt{2}} - 2\sqrt{2}\right) = 0$$$$y\left(\frac{-3 - 4}{\sqrt{2}}\right) = 0$$$$y \cdot \frac{-7}{\sqrt{2}} = 0 \implies y = 0$$From (1): $x = 0$
∴ $x = \mathbf{0}$, $y = \mathbf{0}$
(vi) $\dfrac{3x}{2} - \dfrac{5y}{3} = -2$; $\dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6}$
Multiply equation 1 by 6: $9x - 10y = -12 \quad \cdots (1)$
Multiply equation 2 by 6: $2x + 3y = 13 \quad \cdots (2)$
From (2): $x = \dfrac{13 - 3y}{2}$
Substituting into (1):
$$9 \cdot \frac{13-3y}{2} - 10y = -12$$$$\frac{117 - 27y}{2} - 10y = -12$$Multiply by 2:
$$117 - 27y - 20y = -24$$$$-47y = -141 \implies y = 3$$$x = \dfrac{13 - 9}{2} = 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
From (1): $x = \dfrac{11 - 3y}{2}$
Substituting into (2):
$$2 \cdot \frac{11-3y}{2} - 4y = -24$$$$11 - 3y - 4y = -24$$$$-7y = -35 \implies y = 5$$$x = \dfrac{11 - 15}{2} = -2$
So $x = -2$, $y = 5$.
Now substituting in $y = mx + 3$:
$$5 = m(-2) + 3$$$$-2m = 2 \implies m = -1$$∴ $m = \mathbf{-1}$
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 = $y$.
$$x - y = 26 \quad \cdots (1)$$$$x = 3y \quad \cdots (2)$$Substituting (2) into (1):
$$3y - y = 26 \implies 2y = 26 \implies y = 13$$$$x = 3 \times 13 = 39$$∴ 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 = $x°$ and smaller angle = $y°$.
Supplementary: $x + y = 180 \quad \cdots (1)$
Given: $x - y = 18 \quad \cdots (2)$
From (2): $x = y + 18$
Substituting into (1):
$$y + 18 + y = 180 \implies 2y = 162 \implies y = 81$$$$x = 81 + 18 = 99$$∴ 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 = $₹y$.
$$7x + 6y = 3800 \quad \cdots (1)$$$$3x + 5y = 1750 \quad \cdots (2)$$From (2): $x = \dfrac{1750 - 5y}{3}$
Substituting into (1):
$$7 \cdot \frac{1750-5y}{3} + 6y = 3800$$$$\frac{12250 - 35y}{3} + 6y = 3800$$$$12250 - 35y + 18y = 11400$$$$-17y = -850 \implies y = 50$$$x = \dfrac{1750 - 250}{3} = 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 = $₹y$.
$$x + 10y = 105 \quad \cdots (1)$$$$x + 15y = 155 \quad \cdots (2)$$From (1): $x = 105 - 10y$
Substituting into (2):
$$105 - 10y + 15y = 155$$$$5y = 50 \implies y = 10$$$$x = 105 - 100 = 5$$For 25 km: $5 + 25 \times 10 = 5 + 250 = ₹255$
∴ 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 = $y$.
$$\frac{x+2}{y+2} = \frac{9}{11} \implies 11(x+2) = 9(y+2) \implies 11x - 9y = -4 \quad \cdots (1)$$$$\frac{x+3}{y+3} = \frac{5}{6} \implies 6(x+3) = 5(y+3) \implies 6x - 5y = -3 \quad \cdots (2)$$From (2): $x = \dfrac{5y - 3}{6}$
Substituting into (1):
$$11 \cdot \frac{5y-3}{6} - 9y = -4$$$$\frac{55y - 33}{6} - 9y = -4$$$$55y - 33 - 54y = -24$$$$y = 9$$$x = \dfrac{45-3}{6} = 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.
Five years hence: $x + 5 = 3(y + 5)$
$$x - 3y = 10 \quad \cdots (1)$$Five years ago: $x - 5 = 7(y - 5)$
$$x - 7y = -30 \quad \cdots (2)$$From (1): $x = 3y + 10$
Substituting into (2):
$$3y + 10 - 7y = -30$$$$-4y = -40 \implies y = 10$$$$x = 30 + 10 = 40$$∴ Jacob’s present age = 40 years and Son’s present age = 10 years.