Q1(iii):

Solve the following pair of linear equations by the elimination method and the substitution method : (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

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: $x = \frac{9}{13}, y = -\frac{5}{13}$