How do you solve #x=y+2# and # 2x-3y= 12#?

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Answer 1 · 2018-06-05T13:22:41

#x = -6, y = -8#

[1] #x = y+2#
[2] #2x-3y=12#

Equation [1] gives the expression for #x#. This can be substituted into equation [2]:

#2(y+2) - 3y = 12#

Now we can solve for #y#:

#2y + 4 - 3y = 12#

#4 - y = 12#

#-y = 8#

[3] #y = -8#

Now we take our result in [3] and substitute back into either [1] or [2]. Let's use [1]:

#x = (-8) + 2#

#x = -6#

Hence, the solution is #x = -6# and #y = -8#.