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

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Answer 1 · 2018-03-29T12:13:37

Lines are parallel so no intersection.

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

Substitute the whole equation into eq2 as $x$

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

Solve for y

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

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

How do you solve the system x+5y=4x+5y=4 and 3x+15y=-13x+15y=-1 using substitution?