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### Solving Equations for Variables

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Date: 02/03/2002 at 19:13:27
From: Anonymous
Subject: Solving formulas for variables

Hello!

How do you solve the following equation for (M) and then (m)?

k = M-m/1+Mm

I tried to solve this, but I all I got was k+kMm+m = M. The real
answer is M = m+k/1-km; m = M-k/1+km.  Please show me how you get
there, step by step.

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Date: 02/04/2002 at 02:07:28
From: Doctor Jeremiah
Subject: Re: Solving formulas for variables

Hi there!

First we need to change the question. Here is what the question really
is:

k = (M-m)/(1+Mm)

Notice the extra parentheses. Without them the question would be
determined as:

k = M - m/1 + Mm

Which is:

k = M - m + Mm

And that isn't what you are expecting.

So with the parentheses where they belong:

k = (M-m)/(1+Mm)

The first thing to do is to get rid of the parentheses, but the
division sign is in the way, so first let's get rid of the division.

Move the denominator over to the left-hand side of the equation. Since
the right side is divided by the denominator, we can move it to the
other side by multiplying both sides by the denominator:

k = (M-m)/(1+Mm)
(1+Mm)k = (1+Mm)(M-m)/(1+Mm)

This works because now the right-hand side has (1+Mm) on both the top
and bottom of the division. And (1+Mm) divided by (1+Mm) is 1 so:

k = (M-m)/(1+Mm)
(1+Mm)k = (1+Mm)(M-m)/(1+Mm)
(1+Mm)k = (M-m) x (1+Mm)/(1+Mm)
(1+Mm)k = (M-m) x 1
(1+Mm)k = (M-m)
(1+Mm)k = M - m

Okay, so we got rid of one set of parentheses. Now we need to get rid
of the other set. There is a k multiplied by the parenthesis so we
need to do something about that. If we "distribute" that k into the
parenthesis, then we can get rid of the parenthesis.

To "distribute" the k we need to multiply all the terms inside the
parentheses by k:

(1+Mm)k = M - m
(1k + Mmk) = M - m
1k + Mmk = M - m
k + Mmk = M - m

Now we don't have any parenthesis. Now we have to decide what to solve
for. I will choose M. That means we must get all the Ms on one side of
the equals sign (by themselves) with everything else on the other
side.  That means moving the term that has the M on the left side of
the equals sign over to the right side of the equals sign.

Since that term is added to the left side, one way to move it to the
right side is to subtract it from both sides of the equation:

k + Mmk = M - m
k + Mmk - Mmk = M - m - Mmk

This works because Mmk - Mmk is 0 and anything plus 0 is unchanged:

k + Mmk = M - m
k + Mmk - Mmk = M - m - Mmk
k + 0 = M - m - Mmk
k = M - m - Mmk

Now we have all the Ms on one side of the equals sign, but they need
to be by themselves, so we need to get rid of the m term that is on
the right side.

One way to get it off the right side would be to move it to the left
side.  How?  Since it is subtracted from the right side we can move it
by adding m to both sides:

k = M - m - Mmk
k + m = M - m - Mmk + m

Now just rearrange a little:

k = M - m - Mmk
k + m = M - m - Mmk + m
k + m = M + m - m - Mmk

Now m - m is 0 and anything added to 0 is unchanged:

k = M - m - Mmk
k + m = M - m - Mmk + m
k + m = M + m - m - Mmk
k + m = M + 0 - Mmk
k + m = M - Mmk

Now all the Ms are on the right side, by themselves. That's good. How
do we combine terms to get just one M? Well, remember when we
distributed earlier. What happens if we un-distribute?  M(1-mk) is the
same as M - Mmk so:

k + m = M - Mmk
k + m = M(1 - mk)

We know it is the same because if we distribute the M into the
parenthesis we get M - Mmk right back.

Now we need the M by itself. That means moving the (1 - mk) term.
Since it's multiplied onto the right side, we can move it to the left
side by dividing both sides by (1 - mk):

k + m = M(1 - mk)
(k + m)/(1 - mk) = M(1 - mk)/(1 - mk)

This works because (1 - mk) divided by (1 - mk) is 1, and anything
times 1 is unchanged:

k + m = M(1 - mk)
(k + m)/(1 - mk) = M(1 - mk)/(1 - mk)
(k + m)/(1 - mk) = M x 1
(k + m)/(1 - mk) = M

If we start back where I said we had a decision to make and choose m
instead, we can do the same sort of thing and end up with:

m = (M - k)/(Mk + 1)

Try it, and if you get stuck please write back.

- Doctor Jeremiah, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Basic Algebra
Middle School Algebra

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