How do you solve #–4x + 7y = –10# and #x – 5y = 9# using substitution?

Step 1) Solve the second equation for #x#:

#x - 5y = 9#

#x - 5y + color(red)(5y) = 9 + color(red)(5y)#

#x - 0 = 9 + 5y#

#x = 9 + 5y#

Step 2) Substitute #9 + 5y# for #x# in the first equation and solve for #y#:

#-4x + 7y = -10#

#-4(9 + 5y) + 7y = -10#

#-36 - 20y + 7y = -10#

#-36 - 13y = -10#

#color(red)(36) - 36 - 13y = color(red)(36) - 10#

#0 - 13y = 26#

#-13y = 26#

#(-13y)/color(red)(-13) = 26/color(red)(-13)#

#(color(red)(cancel(color(black)(-13)))y)/cancel(color(red)(-13)) = -2#

#y = -2#

Step 3) Substitute #-2# for #y# in the solution to the second equation at the end of Step 1 and calculate #x#:

#x = 9 + 5y#

#x = 9 + (5 xx -2)#

#x = 9 - 10#

#x = -1#

The solution is #x = -1# and #y = -2#

#-4x+7y=-10#
#x-5y=9#

Using the second equation, we determine a value for #x#.

#x-5y=9#

Add #5y# to each side.

#x=9+5y#

In the first equation, substitute #x# with #color(red)((9+5y))#.

#-4x+7y=-10#

#-4color(red)((9+5y))+7y=-10#

Open the brackets and simplify. The product of a negative and a positive is a negative.

#color(red)(-36-20y)+7y=-10#

#-36-13y=-10#

Add #36# to both sides.

#-13y=26#

Divide both sides by #-13#.

#y=-2#

In the second equation, substitute #y# with #color(blue)(-2)#.

#x-5color(blue)((-2))=9#

Open the brackets and simplify. The product of two negatives is a positive.

#x+10=9#

Subtract #10# from both sides.

#x=-1#