# How do you solve 5x-3y= 8 and -6x+y=20 using substitution?

$x = - \frac{68}{13}$ & $y = - \frac{148}{13}$

#### Explanation:

Given equations:

$5 x - 3 y = 8 \setminus \ldots \ldots \ldots \left(1\right)$

$- 6 x + y = 20 \setminus \ldots \ldots \ldots . . \left(2\right)$

Substituting $y = 6 x + 20$ from (2) into (1), we get

$5 x - 3 \left(6 x + 20\right) = 8$

$- 13 x - 60 = 8$

$- 13 x = 68$

$x = - \frac{68}{13}$

Substituting $x = - \frac{68}{13}$ in (2), we get

$y = 6 x + 20$

$= 6 \left(- \frac{68}{13}\right) + 20$

$= - \setminus \frac{148}{13}$