How do you solve the system of equations #3x + 7y = 1# and #- x + y = - 7#?

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Answer 1 · 2016-12-28T22:45:47

#x = 5# and #y = -2#

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

#color(red)(x) - x + y = color(red)(x) - 7#

#0 + y = x - 7#

#y = x - 7#

Step 2) Substitute #color(red)(x - 7)# for #y# in the first equation and solve for #x#:

#3x + 7(color(red)(x - 7)) = 1#

#3x + 7x - 49 = 1#

#10x - 49 = 1#

#10x - 49 + color(red)(49) = 1 + color(red)(49)#

#10x - 0 = 50#

#10x = 50#

#(10x)/color(red)(10) = 50/color(red)(10)#

#(color(red)(cancel(color(black)(10)))x)/cancel(color(red)(10)) = 5#

#x = 5#

Step 3) Substitute #color(red)(5)# for #x# in the solution to the second equation from Step 1 and calculate #y#

#y = color(red)(5) - 7#

#y = -2#