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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
                 Add x to both sides
      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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