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