Solve the following pair of linear equations by the elimination method and the substitution method:
$x+y=5$ and $2x-3y=4$

(N/A) By elimination method:
$x+y=5$ $...(1)$
$2x-3y=4$ $...(2)$
Multiplying equation $(1)$ by $2$,we obtain:
$2x+2y=10$ $...(3)$
Subtracting equation $(2)$ from equation $(3)$,we obtain:
$(2x+2y) - (2x-3y) = 10 - 4$
$5y = 6$
$y = \frac{6}{5}$ $...(4)$
Substituting the value of $y$ in equation $(1)$,we obtain:
$x + \frac{6}{5} = 5$
$x = 5 - \frac{6}{5} = \frac{25-6}{5} = \frac{19}{5}$
Therefore,$x = \frac{19}{5}, y = \frac{6}{5}$.
By substitution method:
From equation $(1)$,we obtain:
$x = 5 - y$ $...(5)$
Substituting this value in equation $(2)$,we obtain:
$2(5-y) - 3y = 4$
$10 - 2y - 3y = 4$
$10 - 5y = 4$
$-5y = 4 - 10$
$-5y = -6$
$y = \frac{6}{5}$
Substituting the value of $y$ in equation $(5)$,we obtain:
$x = 5 - \frac{6}{5} = \frac{19}{5}$
Therefore,$x = \frac{19}{5}, y = \frac{6}{5}$.