# How do you solve 5x + y = 8 and –4x – 4y = – 16 using substitution?

Apr 18, 2018

$x = 1$ and $y = 3$

#### Explanation:

So always when you try to solve systems, if you find either a single y or x then use the substitution, just something I always do. Anyway so here we have

$5 x + y = 8$
$- 4 x - 4 y = - 16$

And in the first one we have a single y ( shown below)

$5 x + \textcolor{red}{y} = 8$
$- 4 x - 4 y = - 16$

Now we re-write it to get:

$\textcolor{red}{y} = 8 - 5 x$
$- 4 x - 4 y = - 16$

Put this equation :
$\textcolor{red}{y} = 8 - 5 x$

into the other to get:

$- 4 x - 4 \left(8 - 5 x\right) = - 16$

Simplify:

$- 4 x - 32 + 20 x = - 16$

Simplify more:

$16 x = 16$

Divide 16 on both sides to get:

$x = 1$

Now put that into any equation to get:

$5 + y = 8$

Simplify to get

y=3

Now check you work by putting in the numbers into the other equation to get:

$- 4 \left(1\right) - 4 \left(3\right) = - 16$
$- 4 - 12 = - 16$

So it's right with

$x = 1$
$y = 3$

Hope this helps!