# How do you solve the system x+5y=4 and 3x+15y=-1 using substitution?

Mar 29, 2018

Lines are parallel so no intersection.

#### Explanation:

You have to rearrange one of the equations so that it is equal to x and y and then substitute it into the other equation

eq1 $x + 5 y = 4$ becomes $x = 4 - 5 y$

Substitute the whole equation into eq2 as $x$

$3 \left(4 - 5 y\right) + 15 y = - 1$

Solve for y

$12 - 15 y + 15 y = - 1$
$12 = - 1$

So the lines don't cross which means they are parallel