See the entire solution process below:
Step 1) Because the first equation is already solved for #y# we can substitute #3x - 7# for #y# in the second equation and solve for #x#:
#2x + 5y = 16# becomes:
#2x + 5(3x - 7) = 16#
#2x + (5 xx 3x) - (5 xx 7) = 16#
#2x + 15x - 35 = 16#
#17x - 35 = 16#
#17x - 35 + color(red)(35) = 16 + color(red)(35)#
#17x - 0 = 51#
#17x = 51#
#(17x)/color(red)(17) = 51/color(red)(17)#
#(color(red)(cancel(color(black)(17)))x)/cancel(color(red)(17)) = 3#
#x = 3#
Step 2) Substitute #3# for #x# in the first equation and calculate #y#:
#3x - 7 = y# becomes:
#(3 xx 3) - 7 = y#
#9 - 7 = y#
#2 = y#
#y = 2#
The solution is: #x = 3# and #y = 2# or #(3, 2)#
How do you use substitution to solve the system of equations #3x - 7= y# and #2x + 5y = 16#?
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