For 10a), you're told y = 130. Take that value and plug it in for y in the first equation, y = 2x + 10.
Replacing y by 130 gives 130 = 2x + 10. Solve for x by first subtracting 10 from both sides:
120 = 2x.
Now divide by 2; x = 60.
Same idea for 10b) -- They tell you b = 10, so plug 10 in wherever you see b in the first equation, then solve for a.
10c) is a little trickier.
Take the second equation, x + 3y = 6, and solve it for one of the variables. You can choose either one, but this is easier to solve for x, since x isn't being multiplied by anything. This gives: x = 6 - 3y.
Now that you know what x is (even though it's not a number, but still has y in it), take that and plug it in to the first equation. Wherever you see an "x", replace it by 6-3y, since they're the same thing.
Your first equation, 3x + 5y = 12, now becomes:
3(6-3y) + 5y = 12.
Distribute on the left:
18 - 9y + 5y = 12.
Combine like terms.
18 - 4y = 12.
Now solve for y, first by subtracting 18 from both sides:
-4y = -6
And now divide by -4:
y = 3/2.
Plug that value in to EITHER equation and solve for x. I'll plug it in to the second equation, x + 3y = 6.
x + 3(3/2) = 6, so
x + 9/2 = 6.
subtract 9/2 from both sides (think of 6 as 12/2) to get
x = 3/2.
That gives your answer: x = 3/2 and y = 3/2. It's just a coincidence that they both are the same value.