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 $x - \frac{y}{3} = 3$

Given: A pair of linear equations in two variables:

(1) $\frac{x}{2} + \frac{2y}{3} = -1$

(2) $x - \frac{y}{3} = 3$

To Find: The values of $x$ and $y$ using both the Substitution Method and the Elimination Method.

Part 1: Substitution Method

Step 1: Simplify the given equations.

For equation (1): $\frac{x}{2} + \frac{2y}{3} = -1$. Multiplying the entire equation by the Least Common Multiple (LCM) of 2 and 3, which is 6:

$6 \cdot (\frac{x}{2}) + 6 \cdot (\frac{2y}{3}) = 6 \cdot (-1)$

$3x + 4y = -6$ --- (Equation 3)

For equation (2): $x - \frac{y}{3} = 3$. Multiplying the entire equation by 3:

$3 \cdot (x) - 3 \cdot (\frac{y}{3}) = 3 \cdot (3)$

$3x - y = 9$ --- (Equation 4)

Step 2: Express one variable in terms of the other.

From Equation (4), we can isolate $y$:

$y = 3x - 9$ --- (Equation 5)

Step 3: Substitute Equation (5) into Equation (3).

$3x + 4(3x - 9) = -6$

$3x + 12x - 36 = -6$ [Distributive property]

$15x - 36 = -6$ [Combining like terms]

$15x = 30$ [Adding 36 to both sides]

$x = \frac{30}{15} = 2$

Step 4: Find the value of $y$.

Substitute $x = 2$ into Equation (5):

$y = 3(2) - 9$

$y = 6 - 9 = -3$

Part 2: Elimination Method

Step 1: Align the simplified equations.

(3) $3x + 4y = -6$

(4) $3x - y = 9$

Step 2: Eliminate one variable.

Since the coefficients of $x$ are identical ($3$), we subtract Equation (4) from Equation (3):

$(3x + 4y) - (3x - y) = -6 - 9$

$3x + 4y - 3x + y = -15$ [Distributing the negative sign]

$5y = -15$ [Combining like terms]

$y = \frac{-15}{5} = -3$

Step 3: Solve for the remaining variable.

Substitute $y = -3$ into Equation (4):

$3x - (-3) = 9$

$3x + 3 = 9$

$3x = 9 - 3$

$3x = 6$

$x = \frac{6}{3} = 2$

Final Answer: The solution to the system of equations is $x = 2$ and $y = -3$.