NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3#
Welcome to our NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3. This exercise focuses on the Elimination Method (also called the addition/subtraction method) for solving pairs of linear equations. The method involves making the coefficients of one variable equal and then adding or subtracting the equations to eliminate that variable.
Elimination Method Steps:
- Multiply one or both equations by suitable constants so that coefficients of one variable become equal
- Add or subtract the equations to eliminate that variable
- Solve the resulting equation in one variable
- Substitute back to find the other variable
1. Solve the following pair of linear equations by the elimination method and the substitution method: #
(i) $x + y = 5$ and $2x - 3y = 4$
Elimination Method:
$$x + y = 5 \quad \cdots (1)$$$$2x - 3y = 4 \quad \cdots (2)$$
Multiplying equation (1) by $2$:
$$2(x + y) = 2 \times 5$$$$\Rightarrow 2x + 2y = 10 \quad \cdots (3)$$
Subtracting equation (2) from equation (3):
$$\begin{array}{rcl} 2x + 2y &=& 10 \\ 2x - 3y &=& 4 \\ \hline 5y &=& 6 \end{array}$$$$\Rightarrow y = \dfrac{6}{5}$$
Substituting $y = \dfrac{6}{5}$ in equation (1):
$$x + y = 5$$$$x + \dfrac{6}{5} = 5$$
$$\Rightarrow x = 5 - \dfrac{6}{5}$$
$$\Rightarrow x = \dfrac{19}{5}$$
∴ $x = \dfrac{19}{5}$, $y = \dfrac{6}{5}$
(ii) $3x + 4y = 10$ and $2x - 2y = 2$
Elimination Method:
$$3x + 4y = 10 \quad \cdots (1)$$$$2x - 2y = 2 \quad \cdots (2)$$
Multiplying equation (2) by $2$:
$$2(2x - 2y) = 2 \times 2$$$$\Rightarrow 4x - 4y = 4 \quad \cdots (3)$$
Adding equation (1) and equation (3):
$$\begin{array}{rcl} 3x + 4y &=& 10 \\ 4x - 4y &=& 4 \\ \hline 7x &=& 14 \end{array}$$$$\Rightarrow x = 2$$
Substituting $x = 2$ in equation (2):
$$2x - 2y = 2$$$$2(2) - 2y = 2$$
$$\Rightarrow 4 - 2y = 2$$
$$\Rightarrow -2y = -2$$
$$\Rightarrow y = 1$$
∴ $x = \mathbf{2}$, $y = \mathbf{1}$
(iii) $3x - 5y - 4 = 0$ and $9x = 2y + 7$
Writing both equations in the form $ax + by = c$:
$$3x - 5y - 4 = 0$$$$\Rightarrow 3x - 5y = 4 \quad \cdots (1)$$
$$9x = 2y + 7$$
$$\Rightarrow 9x - 2y = 7 \quad \cdots (2)$$
Elimination Method:
Multiplying equation (1) by $3$:
$$3(3x - 5y) = 3 \times 4$$$$\Rightarrow 9x - 15y = 12 \quad \cdots (3)$$
Subtracting equation (2) from equation (3):
$$\begin{array}{rcl} 9x - 15y &=& 12 \\ 9x - 2y &=& 7 \\ \hline -13y &=& 5 \end{array}$$$$\Rightarrow y = -\dfrac{5}{13}$$
Substituting $y = -\dfrac{5}{13}$ in equation (1):
$$3x - 5y = 4$$$$3x - 5\left(-\dfrac{5}{13}\right) = 4$$
$$\Rightarrow 3x + \dfrac{25}{13} = 4$$
$$\Rightarrow 3x = 4 - \dfrac{25}{13}$$
$$\Rightarrow 3x = \dfrac{52 - 25}{13}$$
$$\Rightarrow 3x = \dfrac{27}{13}$$
$$\Rightarrow x = \dfrac{9}{13}$$
∴ $x = \dfrac{9}{13}$, $y = -\dfrac{5}{13}$
(iv) $\dfrac{x}{2} + \dfrac{2y}{3} = -1$ and $x - \dfrac{y}{3} = 3$
$$x - \dfrac{y}{3} = 3 \quad \cdots (2)$$
Multiplying equation (1) by $6$:
$$6\left(\dfrac{x}{2} + \dfrac{2y}{3}\right) = 6 \times (-1)$$$$\Rightarrow 3x + 4y = -6 \quad \cdots (3)$$
Multiplying equation (2) by $3$:
$$3\left(x - \dfrac{y}{3}\right) = 3 \times 3$$$$\Rightarrow 3x - y = 9 \quad \cdots (4)$$
Elimination Method:
Subtracting equation (4) from equation (3):
$$\begin{array}{rcl} 3x + 4y &=& -6 \\ 3x - y &=& 9 \\ \hline 5y &=& -15 \end{array}$$$$\Rightarrow y = -3$$
Substituting $y = -3$ in equation (4):
$$3x - y = 9$$$$3x - (-3) = 9$$
$$\Rightarrow 3x + 3 = 9$$
$$\Rightarrow 3x = 6$$
$$\Rightarrow x = 2$$
∴ $x = \mathbf{2}$, $y = \mathbf{-3}$
2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method: #
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $\dfrac{1}{2}$ if we only add 1 to the denominator. What is the fraction?
Let the numerator be $x$ and the denominator be $y$.
Adding 1 to the numerator and subtracting 1 from the denominator gives 1, so:
$$\dfrac{x + 1}{y - 1} = 1$$$$\Rightarrow x + 1 = y - 1$$
$$\Rightarrow x - y = -2 \quad \cdots (1)$$
Adding 1 to the denominator only gives $\dfrac{1}{2}$, so:
$$\dfrac{x}{y + 1} = \dfrac{1}{2}$$$$\Rightarrow 2x = y + 1$$
$$\Rightarrow 2x - y = 1 \quad \cdots (2)$$
Subtracting equation (1) from equation (2):
$$\begin{array}{rcl} 2x - y &=& 1 \\ x - y &=& -2 \\ \hline x &=& 3 \end{array}$$Substituting $x = 3$ in equation (1):
$$x - y = -2$$$$3 - y = -2$$
$$\Rightarrow -y = -5$$
$$\Rightarrow y = 5$$
∴ The fraction is $\dfrac{3}{5}$.
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Let Nuri’s present age be $x$ years and Sonu’s present age be $y$ years.
5 years ago, Nuri was 3 times as old as Sonu, so:
$$x - 5 = 3(y - 5)$$$$\Rightarrow x - 5 = 3y - 15$$
$$\Rightarrow x - 3y = -10 \quad \cdots (1)$$
In 10 years, Nuri will be 2 times as old as Sonu, so:
$$x + 10 = 2(y + 10)$$$$\Rightarrow x + 10 = 2y + 20$$
$$\Rightarrow x - 2y = 10 \quad \cdots (2)$$
Subtracting equation (1) from equation (2):
$$\begin{array}{rcl} x - 2y &=& 10 \\ x - 3y &=& -10 \\ \hline y &=& 20 \end{array}$$Substituting $y = 20$ in equation (2):
$$x - 2y = 10$$$$x - 2(20) = 10$$
$$\Rightarrow x - 40 = 10$$
$$\Rightarrow x = 50$$
∴ Nuri’s age = 50 years and Sonu’s age = 20 years.
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Let the tens digit be $x$ and the units digit be $y$.
The number is $10x + y$, and the number with the digits reversed is $10y + x$.
The sum of the digits is 9, so:
$$x + y = 9 \quad \cdots (1)$$Nine times the number is twice the reversed number, so:
$$9(10x + y) = 2(10y + x)$$$$\Rightarrow 90x + 9y = 20y + 2x$$
$$\Rightarrow 88x - 11y = 0$$
$$\Rightarrow 8x - y = 0 \quad \cdots (2)$$
Adding equation (1) and equation (2):
$$\begin{array}{rcl} x + y &=& 9 \\ 8x - y &=& 0 \\ \hline 9x &=& 9 \end{array}$$$$\Rightarrow x = 1$$
Substituting $x = 1$ in equation (1):
$$x + y = 9$$$$1 + y = 9$$
$$\Rightarrow y = 8$$
The number is $10x + y$:
$$10x + y = 10(1) + 8 = 18$$Verification:
$9 \times 18 = 162$
$2 \times 81 = 162$
Both are equal ✓
∴ The number is 18.
(iv) Meena went to a bank to withdraw ₹2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got 25 notes in all. Find how many notes of ₹50 and ₹100 she received.
Let the number of ₹50 notes be $x$ and the number of ₹100 notes be $y$.
Meena got 25 notes in all, so:
$$x + y = 25 \quad \cdots (1)$$The notes add up to ₹2000, so:
$$50x + 100y = 2000$$$$\Rightarrow x + 2y = 40 \quad \cdots (2)$$
Subtracting equation (1) from equation (2):
$$\begin{array}{rcl} x + 2y &=& 40 \\ x + y &=& 25 \\ \hline y &=& 15 \end{array}$$Substituting $y = 15$ in equation (1):
$$x + y = 25$$$$x + 15 = 25$$
$$\Rightarrow x = 10$$
∴ Meena received 10 notes of ₹50 and 15 notes of ₹100.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for a book kept for seven days, while Susy paid ₹21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Let the fixed charge be ₹$x$ and the charge for each extra day be ₹$y$.
Saritha kept the book for 7 days, that is 4 extra days, and paid ₹27, so:
$$x + 4y = 27 \quad \cdots (1)$$Susy kept the book for 5 days, that is 2 extra days, and paid ₹21, so:
$$x + 2y = 21 \quad \cdots (2)$$Subtracting equation (2) from equation (1):
$$\begin{array}{rcl} x + 4y &=& 27 \\ x + 2y &=& 21 \\ \hline 2y &=& 6 \end{array}$$$$\Rightarrow y = 3$$
Substituting $y = 3$ in equation (2):
$$x + 2y = 21$$$$x + 2(3) = 21$$
$$\Rightarrow x = 15$$
∴ The fixed charge is ₹15 and the charge for each extra day is ₹3.