How do you solve the system of equations y=-x-4y=−x−4 and y=x+2y=x+2?

2 Answers
hev1
Mar 20, 2018

x = -3x=−3 and y = -1y=−1.

Explanation:

y = -x-4y=−x−4
y = x + 2y=x+2
Substituting -x-4−x−4 for yy:
-x-4=x+2−x−4=x+2
2x = -62x=−6
x=-6/2x=−62
x=-3x=−3
Substituting -3 for x to find y:
y = 3 -4y=3−4
y=-1y=−1

Neefgi
Mar 20, 2018

x = -3x=−3
y = -1y=−1

Explanation:

Since both equations are set in terms of y (y equals), we can set both equations equal to each other:

-x-4=x+2−x−4=x+2

From here we can solve a very simple equation:

-x-x=-2x−x−x=−2x (Get like terms on both sides, x's on the left, coefficients on the right)
4+2=64+2=6
-2x=6−2x=6
x=-3x=−3

Now that we have x, we can choose one of either equations to solve for y, and plug both values in after to double check:

y=-(-3)-4y=−(−3)−4
y=3-4y=3−4
y=-1y=−1

Let's double check by using the other equation:

y=x+2y=x+2
(-1)=(-3)+2(−1)=(−3)+2
-1=-1−1=−1,
True

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