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### Solving Equations with Negative Coefficients

```Date: 12/13/2005 at 00:24:10
From: Alejandra
Subject: negative coefficient

When you solve an equation and the x or the variable is negative, like
8 - x = 3 do I multiply it by -1 so that it would be positive?

```

```
Date: 12/13/2005 at 09:13:18
From: Doctor Peterson
Subject: Re: negative coefficient

Hi, Alejandra.

There are two ways you can go here.

One is, as you suggest, to multiply both sides by -1:

8 - x = 3
Subtract 8 from both sides
-x = -5
Multiply both sides by -1
x = 5

(Actually, this is just the same as the usual last step, dividing both
sides by the coefficient, which is -1.)

The other approach (which helps even more with inequalities) is to
avoid negative coefficients by moving the x to the other side:

8 - x = 3
8 = x + 3
Subtract 3 from both sides
5 = x

Here I added x to both sides so that it would end up with a positive
coefficient; then I isolated it on the right side.  The result is the
same both ways.

You can do the same thing in more complicated cases like this:

8 - 5x = 2 - x

Here, since the -5 on the left is less than the -1 coefficient on the
right, I choose to get rid of the SMALLER multiple of x by adding 5x
to both sides:

8 = 2 + 4x

Now I have a positive coefficient, and the rest is easy:

6 = 4x

6/4 = x

x = 3/2

If I kept x on the left, I would have to divide by a negative
coefficient--which works fine, but gives me one more place where I can
make a mistake:

8 - 5x = 2 - x

8 - 4x = 2

-4x = -6

x = -6/-4 = 3/2

It's the step of dividing by a negative, and the resulting sign change
(in inequalities), that can be error-prone.

If you have any further questions, feel free to write back.

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

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