#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.
