#5x - 2y = -5#

#y - 5x = 3#

**Solving by Substitution**

First, you want to find the equation for a variable that you can replace in the system. #y - 5x = 3# is an equation that appears easy to re-arrange to get an equation for a variable, so we'll use it:

#y - 5x = 3#

Add #5x# to both sides to cancel out #-5x# in order to get the equation for the value of y. You should now have:

#y = 5x + 3#

Now that you have an equation for a variable, substitute these terms (#5x + 3#) in the first equation of the system. So:

#5x - 2y = -5# becomes

#5x -2(5x + 3) = -5#.

Distribute #-2# to the terms inside the parentheses. You do this by multiplying #-2# by each term, so:

#-2 * 5x = -10x#

#-2 * 3 = -6#

Re-write your equation to reflect new information:

#5x -10x - 6 = -5#

Combine like terms.

#-5x - 6 = -5#

Add #6# to both sides to cancel out #-6#. You should now have:

#-5x = 1#

Divide by #-5 to isolate for #x#. You should now have:

#x = -1/5#

Plug the value of #x# into the equation for the value of #y#:

#y = 5x + 3#

#y = 5(-1/5) + 3#

#y = -1 + 3#

#y = 2#

Plug these values back in to confirm that they're right:

#5x - 2y = -5#

#5(-1/5) - 2(2) = -5#

#-1 - 4 = -5#

#-5 = -5#

#y - 5x = 3#

#2 - 5(-1/5) = 3#

#2 - - 1 = 3#

#2 + 1 = 3#

#3 = 3#

These values are correct.