How do you solve the system #x+y=4#, #2x+3y=0#?

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Answer 1 · 2017-06-24T06:38:08

Rearrange the equations and solve by substitution.

#x=-y+4# Let this be equation 1.

#2x+3y=0# Let this be equation 2.

Substitute equation 1 into equation 2:

#2(color(red)(-y+4))+3y=0#

#-2y+8+3y=0#

#y=-8#

Substituting #y=-8# into equation 1,

#x=-(color(red)(-8))+4#

#x=12#