How do you solve the following system?: #x +2y =5, -4x +y = -3#

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Answer 1 · 2018-06-29T19:53:33

See a solution process below:

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

#x + 2y = 5#

#x + 2y - color(red)(2y) = 5 - color(red)(2y)#

#x + 0 = 5 - 2y#

#x = 5 - 2y#

Step 2) Substitute #(5 - 2y)# for #x# in the second equation and solve for #y#:

#-4x + y = -3# becomes:

#-4(5 - 2y) + y = -3#

#(-4 xx 5) - (-4 xx 2y) + y = -3#

#-20 - (-8y) + y = -3#

#-20 + 8y + y = -3#

#-20 + 8y + 1y = -3#

#-20 + (8 + 1)y = -3#

#-20 + 9y = -3#

#-20 + color(red)(20) + 9y = -3 + color(red)(20)#

#0 + 9y = 17#

#9y = 17#

#(9y)/color(red)(9) = 17/color(red)(9)#

#(color(red)(cancel(color(black)(9)))y)/cancel(color(red)(9)) = 17/9#

#y = 17/9#

Step 3) Substitute #17/9# for #y# in the solution to the first equation at the end of Step 1 and calculate #x#:

#x = 5 - 2y# becomes:

#x = 5 - (2 xx 17/9)#

#x = 5 - 34/9#

#x = (9/9 xx 5) - 34/9#

#x = 45/9 - 34/9#

#x = (45 - 34)/9#

#x = 11/9#

The Solution Is:

#x = 11/9# and #y = 17/9#

Or

#(11/9, 17/9)#