How do solve the following linear system?: −2x+3y=−1,−x−7y=14?

1 Answer
Connie
Aug 29, 2016

First, solve equation #2 for x.

Explanation:

This gives you x=−7y−14 .

Now substitute this value of x into the 1st equation, like this:

−2(−7y−14)+3y=−1

Multiply the negative 2 into the parentheses:

14y+28+3y=−1

Combine like terms:

17y+28=−1

Move the 28 to the other side of the equation:

17y=−29

Divide both sides by 17 to isolate the y-term:

17y17=−2917

which gives you:

y=−2917.

Since 17 and 29 are both prime numbers, the answer cannot be reduced.

Now you have the value of y. Plug that into the either eqn:

−2x+3(−2917)=−1

and solve for x.

−2x−8717=−1
−2x=−1+8717=7017

Divide both sides by negative 2:
x=7017−2=7017⋅(−12)=−7034

Reduce:
x=−3517

Now plug x=−3517 and y=−2917 into either eqn to check the answers:

Does −2(−3517)+3(−2917)=−1 ?
7017−8717=−1 ?
−1717=−1

The fractions look odd, but it checks out.

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