# How do you solve the system x+4y=8 and 2x-5y=29 using substitution?

Mar 5, 2017

See the entire solution process below:

#### Explanation:

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

$x + 4 y - \textcolor{red}{4 y} = 8 - \textcolor{red}{4 y}$

$x + 0 = 8 - 4 y$

$x = 8 - 4 y$

Step 2) Substitute $8 - 4 y$ for $x$ in the second equation and solve for $y$:

$2 x - 5 y = 29$ becomes:

$2 \left(8 - 4 y\right) - 5 y = 29$

$\left(2 \times 8\right) - \left(2 \times 4 y\right) - 5 y = 29$

$16 - 8 y - 5 y = 29$

$16 - 13 y = 29$

$- \textcolor{red}{16} + 16 - 13 y = - \textcolor{red}{16} + 29$

$0 - 13 y = 13$

$- 13 y = 13$

$\frac{- 13 y}{\textcolor{red}{- 13}} = \frac{13}{\textcolor{red}{- 13}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 13}}} y}{\cancel{\textcolor{red}{- 13}}} = - 1$

$y = - 1$

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

$x = 8 - 4 y$ becomes:

$x = 8 - \left(4 \times - 1\right)$

$x = 8 - \left(- 4\right)$

$x = 8 + 4$

$x = 12$

The solution is $x = 12$ and $y = - 1$ or $\left(12 , - 1\right)$