How do you solve #2x - 6y = 6# and #-4x + 12y = -18#?

1 Answer
Jul 29, 2016

You can do it either with substitution or elimination, and you should find that there are zero solutions.

SUBSTITUTION

#2x - 6y = 6##color(white)(aaaaaaaaaaa)# (1)
#-4x + 12y = -18##color(white)(aaaaa)# (2)

Here, you would solve for one variable in terms of the other, and plug in to the unused equation. Let's look at (1).

#2x - 6y = 6#

#=> x = (6 + 6y)/2 = 3 + 3y#

So, you can plug in #x# in (2) to get:

#-4(3 + 3y) + 12y = -18#

#= -12 cancel(- 12y + 12y) = -18#

#-># invalid answer, so there's something wrong with this system already.

ELIMINATION

Just so you know how to do it, scale (1) and see what you get when you add these two equations together.

#2(2x - 6y = 6)#
#-4x + 12y = -18#
#"-------------------------"#
#0color(white)(aaaaaaaa) = -6#

Again, invalid answer. So let's show why this system has no answers.

PARALLEL LINES NEVER INTERSECT

If you compare these equations, you should recognize that if you scale both sides of (1) by #-2#, you get:

#-4x + 12y = -12##color(white)(aaaaa)# (1)
#-4x + 12y = -18##color(white)(aaaaa)# (2)

Since these equations are identical except for the constant they are equal to (#-12# vs. #-18#), these are parallel lines, which never intersect.

In other words, this has zero solutions.