How do you solve #x-5y=6#, #2x+3y=12# using substitution?

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Answer 1 · 2018-03-02T02:31:35

#x=6#
#y=0#

#x=5y+6#

#2(5y+6)+3y=12#

#10y+12+3y=12#

#13y+12=12#

#13y=0#

#y=0#

#x-5(0)=6#

#x=6#

By the way, here's what the graph would kind of look like:
graph{(y)^2-(x-6)^2=0 [-5, 10, -5, 5]} If you don't get it then just ignore it.