How do you solve the system of equations #-4x + 6y = 4# and #- x + 8y = 1#?

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Answer 1 · 2016-11-28T04:18:50

#{(x = -1), (y = 0) :}#

#{(-4x + 6y = 4), (-x + 8y = 1) :}#

#{(-4x + 6y = 4), ( (-x + 8y = 1) * (-4)) :}#

#{(-4x + 6y = 4), (4x - 32y = -4) :}#

Add these together to get

#-26y = 0#

Therefore

#y=0#

Plug #0# in for #y# in one of the equations like so:

#-x + (8 * 0) = 1#

#-x = 1#

#x = -1#

Check your answer by putting #0# in for #y# and #-1# in for #x# in the other equation like so:

#(-4 x -1) + (6 * 0) = 4#

#4 + 0 = 4#

#4 = 4#

:)