How do you solve using elimination of #-2x+3y=-5# and #-x+6y=-1#?

1 Answer
Nov 6, 2015

Answer:

#y=1/3 and x = 3# using elimination.

Explanation:

Elimination involves multiplying one equation to make two variable equal to each other.

When I (or you) look at this equation, we can look at either variable. What I notice is that I can multiply #3y# by #-2# to make them cancel eachother out.
#-1x + 6y = -1" " => -1x + 6y = -1#
#(-2x + 3y = -5)(-2) => 4x - 6y = 10#

As you can can see above, I have lined by each #y# with each other. We can add the top equation from the bottom equation. That will leave us with #3x = 9#. We can simplify to #x = 3#.

Now we can plug #x# into either equation to solve for y. I will use #-1x + 6y = -1#.

#-1(3) + 6y = -1#
#" "-3 + 6y = -1#
#" "6y = 2#
#" "y = 1/3 and x = 3#