How do you solve the following system: #2x+7y=1, x+3y=-2 #?

1 Answer
georgef
Jul 17, 2016

We use the value of #x# in one equation and substitute in the other

Explanation:

Let's consider the second equation (the first would be the same but with this one is a little easier):

#x+3y=-2#, this means that #x=-3y-2#. Now substituting #x# in the first equation we have:

#2(-3y-2)+7y=1#, and then distributing #-6y-4+7y=1#. So:

#y-4=1#, and then #y=5#. Using this value of #y# in the second equation now we know:

#x=-3y-2#, so #x=-(3*5)-2=-17#

You should always check that your answers are right by replacing the values you find in the given equations

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