How do you solve the following system: #-3x + y = -4, 4x + 2y = 0 #?

1 Answer
Mar 16, 2016

Answer:

#x=4/5# and #y=-8/5#

Explanation:

we begin with #-3x+y=-4# and #4x+2y=0#

Now, I'm going to solve for #y# in #-3x+y=-4# which looks like #y=-4+3x#.

So no we can replace the #y# in the second equation with #-4+3x#.

That gives us #4x+2(-4+3x)=0# which we can simplify to #4x-8+6x=0#. This becomes #10x=8# or #x=4/5#.

No, we can solve for #y#.

#y=-4+3(4/5)#, which is #-4+12/5# or #-8/5#.

Now we should double check our math. If we did this right then #4(4/5)+2(-8/5)# should equal #0#. Let's see if it is: #16/5-16/5# does in fact equal #0#, which means that we were right!

Nice Job!