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

Related questions
See all questions in Systems Using Substitution
Impact of this question
1762 views around the world
You can reuse this answer
Creative Commons License