How do you solve the system of equations #-3x - 12y = - 3# and #x + 6y = 1#?

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Answer 1 · 2016-11-22T08:56:55

#x=1# and #y=0#

From the second equation we can determine a substitution value for #x#.

#x+6y=1#

Subtract #6y# from both sides.

#x=1-6y#

In the first equation, replace #x# with #color(red)((1-6y))#.

#-3color(red)((1-6y))-12y=-3#

Open the brackets and simplify.

#-3+18y-12y=-3#

#-3+3y=-3#

Divide all terms by #3#.

#-1+y=-1#

Add #1# to both sides.

#y=0#

In the second equation, substitute #y# with #0#.

#x+6y=1#

#x+(6xx0)=1#

#x=1#