Question

Solve the following system of equations by elimination method.
$3x-4y-11=0$
$5x-7y+4=0$.

Answer 1

$3x-4y-11=0...\left(i\right)$
$5x-7y+4=0...\left(ii\right)$
Multiply (i) by $7$ and (ii) by $4$, then subtract (ii) from (i)
$21x-28y-77=0$
$\underset{–}{\underset{-}{2}0x\underset{+}{-}28y\underset{-}{+}16=0}$
$x-93=0$
$x=93$
Put the value of $x$ in equation (i)
$3x-4y-11=0$
$3\times 93-4y-11=0$
$279-4y-11=0$
$268-4y=0$
$-4y=-268$
$y=\frac{-268}{-4}$
$y=67$

Hence, $x=93$ and $y=67$.