How do you solve the following linear system #-x + 3y = 2 , x -8y =2 #?

1 Answer
Nov 14, 2015

Answer:

Using substitution, #x = -4.4# and #y = -0.8#.

Explanation:

You can use substitution.

Our first step is to isolate a variable to one side, so that we can plug it in for variable wherever. What I notice first is that we have a positive #x# in the second equation. Therefore, I can add #8y# to each side in order to isolate the #x#. We now have #x = 2 + 8y#.

We can plug this into the first equation. #-(2 + 8y) + 3y =2#. By distributing we can get #-2 -8y + 3y =2#. By solving out this equation, we end up with #y = 0.8#.

Now we can plug #y# back into our equation that says #x = 2 + 8y#. So we can say that #x = 2 + 8(-0.8)#. That solves out to be #x = -4.4#. We can plug this back into the original equations.

#-x+3y =2 => -(-4.4) + 3(-0.8) = 2#
Then, #4.4 + -2.4 = 2#, and #2 =2 #.