How do you solve the system of equations #3x + 4y = - 8# and #3x + 16y = - 32#?

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Answer 1 · 2018-05-30T17:38:44

#x=0#
#y=-2#

simply subtract each term in the second equation from the first equation:

#3x + 4y = - 8#
#3x + 16y = - 32#

#(3x-3x) + (4y-16y) = (-8 - (-32))#

#(0) + (-12y) = (-8 +32)#

#-12y = 24#

#y=-2#

now use either original equation to solve for #x#:

#3x + 4y = - 8#

#3x + 4(-2) = - 8#

#3x -8 = - 8#

#3x = 0#

#x=0#