(I'm reading this as a=(3c)/(2-c), but if it's a=3(c/2)-c then that would be different. In other words, I am reading this as the 3c is the whole of the numerator and the 2-c is the whole of the denominator, so please consider this a response for that scenario. See at the bottom for solving it the other way.)
a=(3c)/(2-c)
The first thing you need to do is get the (2-c) on the bottom of the fraction over to the other side. You can multiply both sides by (2-c).
a *(2-c) = (3c) /(2-c) *(2-c)
a *(2-c) = (3c) /(2-c) *(2-c)
2a -ac = 3c
Next, you want to get all the Cs on the same side of the equation. You can do this by adding ac to both sides.
2a -ac +ac = 3c +ac
2a -ac +ac = 3c +ac
2a = 3c + ac
Next, you want to isolate the Cs. You can do this by factoring out the C on the right side of the equation.
2a = 3c + ac
2a = c *(3+a)
And finally, you want to get the C alone by dividing out everything it's multiplied by from both sides, moving it down to the denominator on the other side (which is sort of the reverse of what you did in the first step).
2a /(3+a) = c *(3+a) /(3+a)
2a /(3+a) = c *(3+a) /(3+a)
2a /(3+a) = c OR
c = 2a /(3+a)
So your answer would be c = 2a / (3+a).
(Now, reading it as a=3(c/2)-c and doing it that way...)
a = 3(c/2) - c
You can multiply both sides by 2 to remove the 2 in the denominator. Keep in mind that you have to multiply all three parts of the equation by 2.
a (*2) = 3(c/2) (*2) - c (*2)
2a = 3 (c /2 *2) = 2c
The dividing by 2 and multiplying by 2 cancel out:
2a = 3 (c /2 *2) = 2c
2a = 3c - 2c
Now, you just subtract the 2c from the 3c, to get:
2a = 3c - 2c
2a = c OR
c = 2a
So your answer would be c = 2a.
