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:
$$x + y = 5 \quad \cdots (1)$$$$2x - 3y = 4 \quad \cdots (2)$$Multiply (1) by 2: $\;2x + 2y = 10 \quad \cdots (3)$
Subtract (2) from (3):
$$(2x + 2y) - (2x - 3y) = 10 - 4$$$$5y = 6 \implies y = \frac{6}{5}$$Substitute in (1): $x = 5 - \dfrac{6}{5} = \dfrac{19}{5}$
∴ $x = \dfrac{19}{5}$, $y = \dfrac{6}{5}$
(ii) $3x + 4y = 10$ and $2x - 2y = 2$
Elimination Method:
$$3x + 4y = 10 \quad \cdots (1)$$$$2x - 2y = 2 \quad \cdots (2)$$Multiply (2) by 2: $\;4x - 4y = 4 \quad \cdots (3)$
Add (1) and (3):
$$3x + 4y + 4x - 4y = 10 + 4$$$$7x = 14 \implies x = 2$$Substitute in (2): $4 - 2y = 2 \implies y = 1$
∴ $x = \mathbf{2}$, $y = \mathbf{1}$
(iii) $3x - 5y - 4 = 0$ and $9x = 2y + 7$
Rewriting:
$$3x - 5y = 4 \quad \cdots (1)$$$$9x - 2y = 7 \quad \cdots (2)$$Elimination Method:
Multiply (1) by 3: $\;9x - 15y = 12 \quad \cdots (3)$
Subtract (2) from (3):
$$(9x - 15y) - (9x - 2y) = 12 - 7$$$$-13y = 5 \implies y = -\frac{5}{13}$$Substitute in (1): $3x - 5\left(-\dfrac{5}{13}\right) = 4$
$$3x + \frac{25}{13} = 4 \implies 3x = 4 - \frac{25}{13} = \frac{52-25}{13} = \frac{27}{13}$$$$x = \frac{9}{13}$$∴ $x = \dfrac{9}{13}$, $y = -\dfrac{5}{13}$
(iv) $\dfrac{x}{2} + \dfrac{2y}{3} = -1$ and $x - \dfrac{y}{3} = 3$
Multiply equation 1 by 6: $\;3x + 4y = -6 \quad \cdots (1)$
Multiply equation 2 by 3: $\;3x - y = 9 \quad \cdots (2)$
Elimination:
Subtract (2) from (1):
$$5y = -15 \implies y = -3$$Substitute in (2): $3x - (-3) = 9 \implies 3x = 6 \implies x = 2$
∴ $x = \mathbf{2}$, $y = \mathbf{-3}$
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 = $x$ and denominator = $y$.
Condition 1: $\dfrac{x+1}{y-1} = 1 \implies x + 1 = y - 1 \implies x - y = -2 \quad \cdots (1)$
Condition 2: $\dfrac{x}{y+1} = \dfrac{1}{2} \implies 2x = y + 1 \implies 2x - y = 1 \quad \cdots (2)$
Subtract (1) from (2):
$$(2x - y) - (x - y) = 1 - (-2)$$$$x = 3$$From (1): $3 - y = -2 \implies y = 5$
∴ 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 = $x$ years and Sonu’s present age = $y$ years.
Five years ago: $x - 5 = 3(y - 5) \implies x - 3y = -10 \quad \cdots (1)$
Ten years later: $x + 10 = 2(y + 10) \implies x - 2y = 10 \quad \cdots (2)$
Subtract (1) from (2):
$$(-2y) - (-3y) = 10 - (-10)$$$$y = 20$$From (2): $x - 40 = 10 \implies x = 50$
∴ 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 = $x$ and units digit = $y$.
The number = $10x + y$.
Condition 1: $x + y = 9 \quad \cdots (1)$
Reversed number = $10y + x$.
Condition 2: $9(10x + y) = 2(10y + x)$
$$90x + 9y = 20y + 2x$$$$88x - 11y = 0$$$$8x - y = 0 \quad \cdots (2)$$Add (1) and (2):
$$9x = 9 \implies x = 1$$From (1): $y = 9 - 1 = 8$
The number = $10(1) + 8 = \mathbf{18}$.
Verification: $9 \times 18 = 162 = 2 \times 81$ ✓