# How do solve the following linear system?: -4x + 4 = -y, 3x + 6y + 13 = 0 ?

May 31, 2017

$x = \frac{11}{27} \mathmr{and} y = - 2 \frac{17}{27}$

#### Explanation:

It might be better to work on the first equation a little and multiply by $- 1$ so the $y$ term is positive. Then we have an equation which is suitable for the substitution method:

$- 4 x + 4 = - y \text{ "rarr (xx-1)" } \rightarrow \textcolor{b l u e}{4 x - 4 = y}$

Substitute $\textcolor{b l u e}{\left(4 x - 4\right)}$ for $y$ in the other equation:

$3 x + 6 \textcolor{b l u e}{y} + 13 = 0$

$3 x + 6 \textcolor{b l u e}{\left(4 x - 4\right)} + 13 = 0 \text{ } \leftarrow$ only $x$ terms

$3 x + 24 x = - 13 + 24$

$27 x = 11$

$x = \frac{11}{27}$

Now use $x = \frac{11}{27}$ to find a value for $y$

$y = 4 x - 4 = 4 \left(\frac{11}{27}\right) - 4$

$y = - 2 \frac{10}{27}$

Checking into the second equation confirms these results:
$0 = 0$