# How do you solve x=y+4 and 5x+3y=-20 using substitution?

Mar 20, 2016

$y = - 5$
$x = - 1$

#### Explanation:

First we substitute in our value for $x$ (which is $y + 4$) into the other equation so that we only have one unknown, $y$.
Then you expand the bracket and simplify it to find $y$.

$5 x + 3 y = - 20$
$5 \left(y + 4\right) + 3 y = - 20$
$5 y + 20 + 3 y = - 20$
8y+20=-20)
$8 y = - 40$
$y = - \frac{40}{8}$
$y = - 5$

Now we use our value for $y$ in the original equation to find $x$.

$x = y + 4$
$x = - 5 + 4$
$x = - 1$

We can then check our answer by substituting both values into the other equation:

$\left(5 \times - 1\right) + \left(3 \times - 5\right) = - 20$
-5+ (-15) = -20)
$- 5 - 15 = - 20$
$- 20 = - 20$
So we know it's right!
It's really important to check your answers, as in a test, it could save you a whole load of marks despite only taking a few seconds:)

Hope this helps; let me know if I can do anything else:)