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Question 628611: Please help me solve:
C= 2a+3b/2 for a
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! To solve for something means to transform the equation so that that something is all by itself on one side of the equation. So to solve for "a" we want to transform the equation so that the "a" is all by itself on one side of the equation:
a = blah blah
or
blah blah = a
Keeping this in mind can be very helpful in helping you figure out what to do. With your equation:
we need to focus on the "a" and on what is keeping the "a" from being all by itself. It should not hard to see that if we could get rid of the 2 in front of the "a" and get rid of the 3b/2, then the "a" would be by itself on the right side of the equation. This tells what needs to be done. We need to make the 2 in front of the "a" and the 3b/2 "disappear". The only question is: How? The answer to this question is: Using procedures and properties you should have learned by now.
But before we get into things too far, I like to get rid of fractions early so that I don't have to bother with them during most of the problem. Fractions can be eliminated from an equation like this by multiplying both sides of the equation by the lowest common denominator (LCD). Since there is only one fraction, the LCD is easy to figure out. It is 2. So we multiply each side of the equation by 2. (Multiplying both sides of an equation by the same number is one of the procedures you should already know about.)
Note how I put each side of the equation in parentheses. This is an important habit because it helps us see how to do things correctly. In this case we can see that on the right side we need to use the Distributive Property (another procedure you should know) to multiply:
which simplifies to:
Now, with this fraction-less equation we can start solving for a. We want the 4 and the 3b to disappear. We can't get rid of them both in one step. It will take two steps. And we can do the steps in either order. But is is much easier one way than the other. I'll show you why later but the easy way is to get rid of the 3b first.
The 3b that we are going to eliminate is being added to the 4a. If we could somehow turn that 3b into a zero:
blah-blah = 4a + 0
then couldn't we just say
blah-blah = 4a?
(How adding a zero works is a property you should know.) So if we can turn the 3b into a zero then the 4a will end up by itself! And how do we turn the 3b into a zero? Answer: Subtract it from each side (or add it opposite to each side. (These are more procedures that should be familiar to you.) I like adding the opposite:
The 3b and the -3b add up to zero (another property you should know):
which simplifies to:
Now we want the 4 to disappear. Here the 4 is multiplied with the "a" (4a means 4 times a), not added like the 3b was. This is an important difference. Here we want tot turn the 4 into a 1, not a 0. The reason for this is that "a" and "1a" mean the same thing. So turning 4a into 1a is the same thing as turning 4a into a. So how do we turn a 4 into a 1? Answer: Divide both sides by a 4 (or multiply both sides by the reciprocal of 4). (These are more procedures that should be familiar.) I like using reciprocals when there is a fraction in front of the variable, like  . But with a whole number like 4, division is easier:
which simplifies to:
And, since the a is not all by itself, we are finished!
P.S. As we were solving for a, we reached a point where we had to get rid of a 4 or a 3b. The 4, being in front of the variable, is called a coefficient. And the 3b is a separate term. It is easier to get rid of separate terms first and get rid of any coefficients last. Here's why. Let's start from:
If want to get rid of the 4 first we would divide by 4:
Already we are in trouble. A lot of students don't know how to handle the right side properly. And they get lost and the problem is never right again. This is why getting rid of extra terms first makes things much easier.
If you're curious, the right way to continue is:
As you can see, another difficulty with this order is that the fractions come back early and we have to figure out how to deal with them. The other way the fraction did not come back until the very last step and we did not have to deal with it at all. Continuing...
which looks different but is actually equal to what we got before. The denominators are the same on the left so we can add the fractions:
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