# How do you solve x + 4y = 12 and 2x - 3y = 6 using substitution?

Mar 18, 2018

$x = \frac{60}{11} \mathmr{and} y = \frac{18}{11}$

#### Explanation:

Here,$x + 4 y = 12. . . \to \left(1\right) \mathmr{and} 2 x - 3 y = 6. . . \to \left(2\right)$

From $\left(1\right) \textcolor{red}{x = 12 - 4 y} \ldots \to \left(3\right) .$

substituting value of x in $e q {u}^{n} \left(2\right)$,we get

$2 \textcolor{red}{\left(12 - 4 y\right)} - 3 y = 6 \implies 24 - 8 y - 3 y = 6$

$\implies 24 - 6 = 8 y + 3 y \implies 18 = 11 y \implies y = \frac{18}{11}$

From(3),we have

$x = 12 - 4 \left(\frac{18}{11}\right) = \frac{132 - 72}{11} = \frac{60}{11}$

from (1) $x + 4 y = 12$
$\implies L H S = \frac{60}{11} + 4 \left(\frac{18}{11}\right) = \frac{60}{11} + \frac{72}{11} = \frac{132}{11} = 12 = R H S$