Help solving linear systems by substitution with checks?

Ok, to do subsitution you need to convert one of you equations into either

x = .... or

y = ....

form. Let's do this to the 2nd equation and shoot for y = ...

2x + y = -8 it already has the y on the correct side, we just need to move the 2x to the other side

we do this by subtractin 2x from both sides to get

y = -8 - 2x

Now we can substitute -8 - 2x in place of y. This MUST be done in the equation we didn't look at already

3x - y = -2 will now be

3x - (-8 - 2x) = -2 and we can solve to get x

3x + 8 + 2x = -2

5x + 8 = -2

5x = -2 - 8 = -10

x = -10/5 = -2

And since y = -8 - 2x, y = -8 - 2(-2) = -8 + 4 = -4

The solution is (-2,-4). That's also the point of intersection. That's actually what it means to solve

a system of two linear equations. You'll eventually run across systems that don't have solutions. This

happens when the two linear equations represent parallel lines, which never intersect.

Checking

3x - y --> 3(-2)-(-4) = -6 + 4 = -2 exactly as it should be

2x + y --> 2(-2)+-4 = -4 - 4 = -8 also as it should be.

1) Isolate one of the variables (doesn't matter which). Y has a coefficient (how many of the variables there are. The coefficient of 3x is 3) of 1, so let's get that by itself.

2) Add 2 to both sides. 3X-y+2=-2+2

3X-Y+2=0

3) Now, add Y to both sides 3X-Y+Y+2=0+Y

3X+2=Y

4) We now know what Y is (unsimplified). So, substitute the equation you got for Y back into the other equation and simplify.

2X+3X+2=-8

5X+2=-8 (combine like terms)

5X=-10 (subtract 2 from both sides)

X=-2

5) Plug the value you got for X into the other equation and solve for Y.

3(-2)-Y=-2

-6-Y=-2

-Y=4

Y=-4

6) Write the solutions as an ordered pair (-2,-4). This is the solution and the point of intersection

**Check**

Just plug your ordered pair back into either one of the original equations.

3(-2)-(-4)=-2

-6+4=-2

-2=-2 Correct!

2(-2)+(-4)=-8

-4+(-4)=-8

-8=-8 Correct!