How do you solve the system of equations #5x + 2y = 1# and #8x + y = 6#?

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Answer 1 · 2017-03-22T13:01:33

See the entire solution process below:

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

#8x + y = 6#

#-color(red)(8x) + 8x + y = -color(red)(8x) + 6#

#0 + y = -8x + 6#

#y = -8x + 6#

Step 2) Substitute #-8x + 6# for #y# in the first equation and solve for #x#:

#5x + 2y = 1# becomes:

#5x + 2(-8x + 6) = 1#

#5x + (2 xx -8x) + (2 xx 6) = 1#

#5x - 16x + 12 = 1#

#-11x + 12 = 1#

#-11x + 12 - color(red)(12) = 1 - color(red)(12)#

#-11x + 0 = -11#

#-11x = -11#

#(-11x)/color(red)(-11) = (-11)/color(red)(-11)#

#x = 1#

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

#y = -8x + 6# becomes:

#y = (-8 xx 1) + 6#

#y = -8 + 6#

#y = -2#

The solution is: #x = 1# and #y = -2# or #(1, -2)#