# How do you solve the system of equations x + 8y = 26 and 2x + y = 5?

Aug 5, 2018

The solution is $\left(\frac{14}{15} , \frac{47}{15}\right)$.

#### Explanation:

Equation 1: $x + 8 y = 26$

Equation 2: $2 x + y = 5$

I am going to use elimination and substitution to answer this question.

Multiply Equation 1 by $- 2$.

$- 2 \left(x + 8 y\right) = 26 \times - 2$

$- 2 x - 16 y = - 52$

$- 2 x - 16 y = - 52$
$\textcolor{w h i t e}{. .}$$2 x + \textcolor{w h i t e}{. .} y = \textcolor{w h i t e}{\ldots \ldots} 5$
$- - - - - - -$
$- 15 y \textcolor{w h i t e}{\ldots \ldots \ldots} = - 47$

Divide both sides by $- 15$.

$y = \frac{- 47}{- 15}$

$y = \frac{47}{15}$

Substitute $\frac{47}{15}$ for $y$ in Equation 1 and solve for $x$.

$x + 8 \left(\frac{47}{15}\right) = 26$

$x + \frac{376}{15} = 26$

Subtract $\frac{376}{15}$ from both sides.

$x = - \frac{376}{15} + 26$

Multiply $26$ by $\frac{15}{15}$.

$x = - \frac{376}{15} + 26 \times \frac{15}{15}$

$x = - \frac{376}{15} + \frac{390}{15}$

$x = \frac{14}{15}$

Solution: $\left(\frac{14}{15} , \frac{47}{15}\right)$

graph{(x+8y-26)(2x+y-5)=0 [-10, 10, -5, 5]}