# How do you solve the system of linear equations x + 3y = 8 and 2x - y = 9?

May 18, 2017

See a solution process below:

#### Explanation:

Step 1) Solve the first equation for $x$:

$x + 3 y = 8$

$x + 3 y - \textcolor{red}{3 y} = 8 - \textcolor{red}{3 y}$

$x + 0 = 8 - 3 y$

$x = 8 - 3 y$

Step 2) Substitute $8 - 3 y$ for $x$ in the second equation and solve for $y$:

$2 x - y = 9$ becomes:

$2 \left(8 - 3 y\right) - y = 9$

$\left(2 \cdot 8\right) - \left(2 \cdot 3 y\right) - y = 9$

$16 - 6 y - y = 9$

$16 - 6 y - 1 y = 9$

$16 - 7 y = 9$

$- \textcolor{red}{16} + 16 - 7 y = - \textcolor{red}{16} + 9$

$0 - 7 y = - 7$

$- 7 y = - 7$

$\frac{- 7 y}{\textcolor{red}{- 7}} = \frac{- 7}{\textcolor{red}{- 7}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 7}}} y}{\cancel{\textcolor{red}{- 7}}} = 1$

$y = 1$

Step 3) Substitute $1$ for $y$ in the solution to the first equation at the end of Step 1 and calculate $x$:

$x = 8 - 3 y$ becomes:

$x = 8 - \left(3 \cdot 1\right)$

$x = 8 - 3$

$x = 5$

The solution is: $x = 5$ and $y = 1$ or $\left(5 , 1\right)$