How do you solve #–4x + 7y = –10# and #x – 5y = 9# using substitution?
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See the entire solution process below:
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#
