Solving systems using substitution?

if both equations = y, therefore

3x+2 = 4x+8

3x-4x = 8-2

-x = 6

therefore

x = -6

they equal each other, because they both equal y. therefore it'll be .. :

3x+2=4x+8 now, solve for x by canceling out the two from the left side to the ride of the = sign.

3x+2-2=4x+8-2 so now its ...

3x=4x+6 because you canceled out the left side already , now cancel out the x's on the right ..

3x-4x = 4x-4x-6 so now its ...

-1x (or just -x) = -6

divide -1 from each side to get x=6

sorry if i did a really bad job explaining.

This one isn't as difficult since both equations are solved for "y." Since both of them are solved for y, you can take them and set them equal to each other.

In this case, I will substitute the second equation in for y in the first to get

4x+8 = 3x+2

Then, solve the equation for x to find your x-value.

4x - 3x + 8 = 3x - 3x +2 (Subtract 3x on both sides)

x + 8 = 2 (Simplify)

x + 8 - 8 = 2 - 8 (Subtract 8 on both sides)

x = -6

Since we found that x = -6, then we can take that value and plug it in to find the y-value. You can plug-in x into either equation. Let's plug it in to the first equation.

y = 3(-6) + 2

y = -18 + 2

y = -16

Since our x-value is -6 and our y-value is -16, the point of intersection of these two lines is (-6,-16).

Hope this helps!