Since I just transfered schools this semester, the school I was transfering from was just about to teach this, I am already expected to know this and I need some help to better understand it.

What I don't understand is, not sure how to explain, but say for instance the problem was 3x+2y=5, 2x-4y=10. I`m not sure how to rearange the problem like you would if a number wasn`t attached to the variable y. E.g. I know that 4x+y=8 would be turned to y=8-4x in the equation, but what if y has a number attatched to it? Say, instead of 4x+y=8 (in which I would change it to y=8-4x), it was 4x+3y=8. How would I rearange that to fit into the equation?
Help will be gratefully appreaciated.

EDIT: The math did not look right when the topic was posted, so I had to remove it.

I don't remember substitution, but here's how to do elimination.
You will probably need to know it, they are taught around the same time.

Using your example 3x+2y=5, 2x-4y=10
You need to get one of the variables to cancel out
Which tells you how much the other is
Then substitute the number you found out in the place of the variable
Which will give you the other variable
So:
3x+2y=5
+2x-4y=10

Multiply the top one by 2, which gives you -4y

6x-4y=10
+2x-4y=10

Then the y's cancel out, leaving you with the x's

8x=20

Divide by 8

x=2.5

Replace x with that number

3x+2y=5 ------> 3(2.5)+2y=5 -----> 7.5+2y=5

2y=-2.5

y=-1.25

(2.5,-1.25)

Hope I got it right.

try to single out one variable in one of the equations, then substitute the equivalent of the variable into the other equation, and voila

Take the two equations given:

3x+2y=5
2x-4y=10

First look for a single variable, but if there is none, just pick one. I'm going to do the 2y in the first equation.

3x+2y=5
Keep the variable you want on one side, move everything else to the other.

2y=-3x+5
Divide by coefficient (in this case, 2)

y=(-3x+5)/2
Now you have it done, you can plug it into the other equation. Since you singled out y, plug it into the y.

2x-4[(-3x+5)/2]=10
Solve it.

For any systems of equations (at least 2 equations are given), you need to single out a variable first so that you can plug that into the other equation to solve for only one variable. You can choose whichever is easier, either solve for "y" or solve for "x". Then plug what you solved for into that same variable in the other equation.

It gets harder when you get three or more variables, but the principles are the same, just solve for one variable at a time. There are several tricks to doing this (subtracting equations, etc), but since I took Algebra several years ago and Calculus involves the calculator a lot... I suppose I need to brush up on the basics. Just solve for one variable and plug what you solved for into the other equation. Then use that variable to plug into the first equation to find the answer.

Why doesn't anyone have a Calculus question I actually know that stuff haha

OK, substitution method.

Rearrange a formula to make a certain variable the subject (ie on it's own on one side) then sub the other side in for that variable in another equation. You seem to understand this OK.

Rearranging. To follow your example (making y the subject):
4x+3y=8 . . . . . . take 4x from both sides
=> 3y=8-4x . . . . divide both sides by 3 (3/3 = 1)
=> y=(8-4x)/3

So to solve the problem you suggested using substitution method -> 3x+2y=5, 2x-4y=10.
Rearrange the second equation to make y the subject:
2x-4y = 10 . . . . . . add 4y to both sides (or take away -4y, same thing)
=> 2x = 10+4y . . . take 10 from both sides
=> 2x-10 = 4y . . . . divide both sides by 4
=> (2x-10)/4 = y
=> y = (2x-10)/4

Sub into first equation and solve for x:
3x+2y = 5
=> 3x + [(2x-10)/4] = 5 . . . . times both sides by 4
=>12x + 2x - 10 = 20
=> 14x - 10 = 20 . . . . . . . . . add 10 to both sides
=> 14x = 30 . . . . . . . . . . . . divide both sides by 14
=> x = 30/14 = 15/7

Then sub x into equation for y:
y = (2x-10)/4
=> y = [(2*15/7)-10]/4
=> y = [(30-70)/7]/4
=> y = -40/28 = -10/7