How do you solve by substitution #-x + 2y = -6# and #3x + y = 8#?

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Answer 1 · 2018-04-29T22:44:30

#(22/7,-10/7)#

In the second equation, we can easily solve for #y# by subtracting #3x# from both sides. We get:

#y=8-3x#

We've already solved for one variable in terms of the other, so we can plug into the first equation. We get

#-x+2(8-3x)=-6#

#=>-x+16-6x=-6#

#=>-7x+16=-6#

We can subtract #16# from both sides to get

#-7x=-22#

Dividing both sides by #-7#, we get

#color(blue)(x=22/7)#

We can plug into our equation for #y#, #y=8-3x#. We get

#y=8-3(22/7)#

#y=56/7-66/7#

#color(lime)(y=-10/7)#

Hope this helps!