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# C=1/2f(q+z) for q

### 2 Answers by Expert Tutors

Stephen B. | Secondary or College Math/SAT/GRE TutorSecondary or College Math/SAT/GRE Tutor
4.8 4.8 (114 lesson ratings) (114)
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It is a little unclear when you state the problem what exactly is going on with that fraction. This is a common issue when you try to type up equations. On the right side, you have a fraction bar and then a bunch of stuff after it. We need to clear up what is supposed to be in the numerator and what is supposed to stay in the denominator. This is important not just for typing up questions on the internet to get help but in using your calculator as well. If what the equation means is that both "f" and "q+z" are in the denominator, but then on your calculator you type something 1 / 2 * f * ( q + z ), most calculators (e.g. TI 83/84 and almost any basic or scientific model) will interpret that  as meaning only the 2 is on the bottom and the rest is on top.

If you meant the f and the q+z are on top, then Steve's answer is good.

If you meant the f and the q+z are supposed to be on the bottom, then we have a little bit different process.
First, write down the operations that are happening to q, and in what order they happen from inside out, that is, according to the regular PEMDAS rule:
2. 2f is multiplied
3. The result is divided into 1 (as opposed to being divided by 1).

So, to solve for q, we "invert" the order of operations and do the inverse operation of our 3 steps, and also in reverse order.

1. To get the big thing out of the denominator (Step 3), we will multiply both sides by 2f(q+z). Now we have:
C*2f(q+z) = 1, or as we normally write it,
2Cf(q+z) = 1
2. Now we have to adjust our steps just a little bit because we have the *C part too. So we will change Step 2 above to "2Cf is multiplied." We invert that by dividing, so we have:
q+z = 1/[2Cf]. Notice how I put brackets around the entire denominator so it's clear!
3. Finally to invert Step 1 we subtract z:
q = 1/[2Cf] - z

Hope that helps!
Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
5.0 5.0 (3 lesson ratings) (3)
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C=(1/2) f(q+z)

Kendra, is there a function f() involved?

Assuming there isn't and f is just a variable:

Multiply both sides by 2/f:

2C/f = q + z

Subtract z from both sides:

2C/f - z = q