# How do you solve the following linear system:  x+8y=3, 4x+5y=-2 ?

Apr 30, 2017

See the solution process below:

#### Explanation:

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

$x + 8 y = 3$

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

$x + 0 = 3 - 8 y$

$x = 3 - 8 y$

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

$4 x + 5 y = - 2$ becomes:

$4 \left(3 - 8 y\right) + 5 y = - 2$

$\left(4 \cdot 3\right) - \left(4 \cdot 8 y\right) + 5 y = - 2$

$12 - 32 y + 5 y = - 2$

$12 + \left(- 32 + 5\right) y = - 2$

$12 - 27 y = - 2$

$- \textcolor{red}{12} + 12 - 27 y = - \textcolor{red}{12} - 2$

$0 - 27 y = - 14$

$- 27 y = - 14$

$\frac{- 27 y}{\textcolor{red}{- 27}} = \frac{- 14}{\textcolor{red}{- 27}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 27}}} y}{\cancel{\textcolor{red}{- 27}}} = \frac{14}{27}$

$y = \frac{14}{27}$

Step 3) Substitute $\frac{14}{27}$ for $y$ in the solution to the first equation at the end of Step 1 and calculate $x$:

$x = 3 - 8 y$ becomes:

$x = 3 - \left(8 \cdot \frac{14}{27}\right)$

$x = 3 - \frac{112}{27}$

$x = \left(3 \cdot \frac{27}{27}\right) - \frac{112}{27}$

$x = \frac{81}{27} - \frac{112}{27}$

$x = - \frac{31}{27}$

The solution is: $x = - \frac{31}{27}$ and $y = \frac{14}{27}$ or $\left(- \frac{31}{27} , \frac{14}{27}\right)$