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Q1(iii):
Solve the following pair of linear equations by the elimination method and the substitution method : (iii) 3x β 5y β 4 = 0 and 9x = 2y + 7
Solution :
Given: A pair of linear equations in two variables:
(i) $3x - 5y - 4 = 0$
(ii) $9x = 2y + 7$
To Find: The values of $x$ and $y$ using both the Substitution Method and the Elimination Method.
Part 1: Substitution Method
Step 1: Standardize the equations.
Equation (i): $3x - 5y = 4$ --- (1)
Equation (ii): $9x - 2y = 7$ --- (2)
Step 2: Express one variable in terms of the other.
From equation (1), isolate $3x$:
$3x = 5y + 4$
$x = \frac{5y + 4}{3}$ --- (3)
Step 3: Substitute equation (3) into equation (2).
$9\left(\frac{5y + 4}{3}\right) - 2y = 7$
[Since $9/3 = 3$, we simplify the expression]
$3(5y + 4) - 2y = 7$
$15y + 12 - 2y = 7$ [Distributive property]
$13y + 12 = 7$ [Combining like terms]
$13y = 7 - 12$
$13y = -5$
$y = -\frac{5}{13}$
Step 4: Solve for $x$ by substituting $y$ into equation (3).
$x = \frac{5(-\frac{5}{13}) + 4}{3}$
$x = \frac{-\frac{25}{13} + 4}{3}$
$x = \frac{\frac{-25 + 52}{13}}{3}$ [Finding common denominator]
$x = \frac{27}{13 \times 3} = \frac{9}{13}$
Part 2: Elimination Method
Step 1: Align the equations.
(1) $3x - 5y = 4$
(2) $9x - 2y = 7$
Step 2: Make the coefficients of $x$ equal.
Multiply equation (1) by 3 to match the $x$-coefficient of equation (2):
$3(3x - 5y) = 3(4)$
$9x - 15y = 12$ --- (4)
Step 3: Eliminate $x$ by subtracting equation (4) from equation (2).
$(9x - 2y) - (9x - 15y) = 7 - 12$
$9x - 2y - 9x + 15y = -5$
$13y = -5$
$y = -\frac{5}{13}$
Step 4: Substitute $y$ into equation (1) to find $x$.
$3x - 5(-\frac{5}{13}) = 4$
$3x + \frac{25}{13} = 4$
$3x = 4 - \frac{25}{13}$
$3x = \frac{52 - 25}{13}$
$3x = \frac{27}{13}$
$x = \frac{27}{13 \times 3} = \frac{9}{13}$
Final Answer: $x = \frac{9}{13}, y = -\frac{5}{13}$
More Questions from Class 10 Mathematics Pair of linear equations in two variable EXERCISE 3.3
- Q1(i): Solve the following pair of linear equations by the elimination method and the substitution method : (i) x + y = 5 and 2x β 3y = 4
- Q1(ii): Solve the following pair of linear equations by the elimination method and the substitution method : (ii) 3x + 4y = 10 and 2x β 2y = 2
- Q1(iv): Solve the following pair of linear equations by the elimination method and the substitution method : (iv) $\frac{x}{2} + \frac{2y}{3} = -1$ and $x - \frac{y}{3} = 3$
- Q2(i): 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 $\frac{1}{2}$ if we only add 1 to the denominator. What is the fraction?
- Q2(ii): Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (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?
- Q2(iii): Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (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.
- Q2(iv): Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (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.
- Q2(v): Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (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.
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