I have plugged in numbers and seen that the rearranged right side is equivalent to the left side, but have been unable to even begin to know how to go about rearranging the left side of the equation into the right side equivalent. Can you show me how?

## A - AB/(B+C)= A/[1+B/C], how do you rearrange the left side to get the right side?

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# 2 Answers

Hello Todd,

Let me walk you through this step by step.

- A - AB/(B+C) = A/[1+(B/C)] is what you started out with. Since we're only wanting to rearrange the left side, let's focus on it.
- You should multiple the "A" by "(B+C)/(B+C)". This works because you're multiplying the "top and bottom" of the fraction "A/1" by the same term, so it's still equivalent. This leaves you with A(B+C)/(B+C) - AB/(B+C).
- Next, distribute the "A(B+C)" to get (AB+AC)/(B+C) - AB/(B+C)
- Now you can "combine" the fractions as (AB + AC - AB)/(B+C)
- Subtract the AB to get AC/(B+C)
- Now is where the problem gets a bit tricky. Divide both the top and the bottom of the fraction by "C". This will give you (AC/C)/[(B+C)/C]
- The top of the fraction (AC/C) simply becomes "A", while the bottom becomes [B/C + C/C], so you have A/[B/C + C/C]. The C/C = 1, so you have A/[1+B/C].

It's a lot easier to write the steps out on paper to follow along with how to do it. Let me know if that explanation makes sense!

A - [AB/(B+C)]= A/[1+B/C]

First, let's replace the denominator on the right hand side with its equivalent [(C + B)/C)

A - [AB/(B + C)]= A/[C + B)/C]

Now, remember that dividing by [(C + B)/C] is the same as multiplying by [C/(C + B)] so that

A - [AB/(B + C)] = [AC/(C + B)]

Now, multiply both sides by (B + C) to get rid of the denominator on the left hand side:

A(B + C) - AB = [AC/(C + B)]*(B + C)

AB + AC - AB = AC

and we are left with

AC = AC which is true. We have, therefore, proven that the two expressions are equivalent.

Also, if you really want to impress your instructor, write the letters QED after your proof. These stand for "Quod erat demonstrandum" (which is what we were trying to prove).

# Comments

William,

Very succinct and clear! Many thanks. I am trying to learn electronic analysis and diagnostic procedures. Your answer to my inquiry was helpful in understanding numerous transformation equations methods including matrix theory calculation rationale.

My Sincere thanks,

Todd R. Parody, Ph.D; Genetics and Biochemistry.

## Comments

This would give you a complete review:

http://www.khanacademy.org/math/arithmetic.

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