Hey there!
There's two ways you can solve a system of equations (two or more equations that have a common solution). One is the elimination method and the other is the substitution method. The easier method to use would be elimination, which I'll explain down below.
Basically, you will have to get one of the terms to equal zero. You can do this by multiplying one or both of the equations to get at least one term to be the same number, except one will be positive and one will be negative. It doesn't matter which term you do.
This is easier to show than explain, so I'll solve this for you:
5x − 6y = −38
3x + 4y = 0
I would go ahead and get the y term to cancel, since you can multiply 6 by 2 to get 12 and 4 by 3 to get 12. You will multiply your entire equations by these corresponding numbers.
2(5x − 6y = −38)
3(3x + 4y = 0)
10x − 12y = −76
9x + 12y = 0
Now, just add these equations together. You'll know you did it right if one of the terms is now 0.
10x − 12y = −76
+ 9x + 12y = 0
19x = –76
You final step is to simplify as you would any other equation.
(19x)/19 = (–76)/19
x = –4
Your x value will be –4. Now, plug this x value into either of your original equations (doesn't matter which) to get your corresponding y value.
3x + 4y = 0
3(–4) + 4y = 0
(–12 + 4y) + 12 = 0 + 12
(4y)/4 = (12)/4
y = 3
Your final answer will be (–4, 3). Hope this helped you out! :-)
