Elimination method:
Multiplying equation $(1)$ by $2,$ we get equation $(3)$
$2x +2y =10 ............. (3)$
$2x−3y =4 ........... (2)$
Subtracting equation $(2)$ from $(3),$ we get
$5y =6 ⇒ y = \frac{6}{5}$
Putting value of $y$ in $(1),$ we get
$x + \frac{6}{5}=5$
$⇒ x =5− \frac{6}{5} = \frac{{19}}{5}$
Therefore,$ x = \frac{{19}}{5}$ and $y = \frac{6}{5}$
Substitution method:
$x +y =5 .......................... (1)$
$2x−3y =4 ......................... (2)$
From equation $(1),$ we get,
$x =5−y$
Putting this in equation $(2),$ we get
$2(5−y )−3y =4$
$⇒ 10−2y−3y =4$
$⇒ 5y =6$
$⇒ y = \frac{6}{5}$
Putting value of $y$ in $(1),$ we get
$x =5−\frac{6}{5}$=$\frac{{19}}{5}$
Therefore, $x = \frac{{19}}{5}$ and $y = \frac{6}{5}$
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