Question

Solve the following system of equations by elimination method:
$3x+5y=20$
$6x-10y=-4$

Answer 1

Given system of equations
$3x+5y=\mathrm{20........}\left(i\right)$
$6x-10y=-\mathrm{4.......}\left(ii\right)$

Multiply both sides by $2$ in equation $\left(i\right)$
$6x+10y=\mathrm{40.......}\left(iii\right)$
$6x-10y=-\mathrm{4........}\left(ii\right)$

By addition

$(6x+10y)+(6x-10y)=40-4$
$\Rightarrow$ $12x=36$
$\Rightarrow$ $x=\frac{36}{12}$
$\Rightarrow$ $x=\mathrm{3.......}\left(iv\right)$
Putting the value of $x$ from $\left(iv\right)$ in equation $\left(i\right)$
$\Rightarrow$ $3\times 3+5y=20$
$\Rightarrow$ $9+5y=20$
$\Rightarrow$ $5y=20-9$
$\Rightarrow$ $5y=11$
$\Rightarrow$ $y=\frac{11}{5}$

Hence, the solution is $x=3,y=\frac{11}{5}$