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How do you solve by substitution #4x + 6y = 62 # and #10x - 2y = 2#?

1 Answer

Jun 5, 2015

#10x-2y=2#
#10x=2+2y# ( we add #2y# on each side )
#10x-2=2y# ( we substract #2# on each side )
#5x-1=y# ( we divide by #2# on each side )

Now that we have #y#, we can use substitution in the first equation :

#4x+6*(5x-1)=62#
#4x+30x-6=62#
#34x-6=62#
#34x=62+6# ( we add #6# on each side )
#34x=68#
#x=68/34=2# ( we divide by #34# on each side )

Now that we have #x#, we can find #y# :

We know that #y=5x-1#, as we calculated it before :

#y=5*2-1=10-1=9#

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