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

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Answer 1 · 2016-07-17T21:15:19

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

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

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