Q1(i):

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

Given: A pair of linear equations in two variables:

(1) $x + y = 5$

(2) $2x - 3y = 4$

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

Method 1: Substitution Method

Step 1: Express one variable in terms of the other.
From equation (1):
$x + y = 5$
$x = 5 - y$ --- (Equation 3)

Step 2: Substitute the expression into the second equation.
Substitute $x = 5 - y$ into equation (2):
$2(5 - y) - 3y = 4$ [Substituting $x$ from Eq 3 into Eq 2]

Step 3: Solve for $y$.
$10 - 2y - 3y = 4$ [Distributive property]
$10 - 5y = 4$ [Combining like terms]
$-5y = 4 - 10$ [Transposing 10 to the RHS]
$-5y = -6$
$y = \frac{-6}{-5}$
$y = \frac{6}{5}$

Step 4: Solve for $x$.
Substitute $y = \frac{6}{5}$ into equation (3):
$x = 5 - \frac{6}{5}$
$x = \frac{25 - 6}{5}$ [Finding a common denominator]
$x = \frac{19}{5}$

Method 2: Elimination Method

Step 1: Make the coefficients of one variable equal.
To eliminate $y$, multiply equation (1) by $3$ so that the coefficient of $y$ in both equations is numerically equal:
$3(x + y) = 3(5)$
$3x + 3y = 15$ --- (Equation 4)

Step 2: Add the equations to eliminate the variable.
Add equation (4) and equation (2):
$(3x + 3y) + (2x - 3y) = 15 + 4$
$3x + 2x + 3y - 3y = 19$ [Grouping like terms]
$5x = 19$
$x = \frac{19}{5}$

Step 3: Substitute the value of $x$ to find $y$.
Substitute $x = \frac{19}{5}$ into equation (1):
$\frac{19}{5} + y = 5$
$y = 5 - \frac{19}{5}$
$y = \frac{25 - 19}{5}$
$y = \frac{6}{5}$

Final Answer: The solution to the system of equations is $x = \frac{19}{5}$ and $y = \frac{6}{5}$.