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# Rewritten question! Solve Linear Equation

Yes, the original problem is similar to the one you mention on the bottom of the reply.  Except I made an error in the problem as well.  It should read:

x      -  x+1  =    -4

------     ------     -------

(x^2 - 16) (x^2 - 4x) (x^2 + 4x)

Thank you and my apologies for my math grammar!

### 1 Answer by Expert Tutors

Gene G. | You can do it! I'll show you how.You can do it! I'll show you how.
5.0 5.0 (254 lesson ratings) (254)
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Original equation: x\(x^2-4x) -x+1/x^2-4x=-4/x^2+4x
(I suspect that there may be some missing parentheses. See my note below.)
Assuming the "\" should be "/", and yes, the caret is the correct symbol for an exponent, here's the equation you wrote in a more readable form:
x                  1              -4
------------ - x + ----- - 4x = ----- + 4x
(x^2 - 4x)          x^2           x^2
This will wind up being a higher order polynomial.
First, let’s move everything to the left side of the equation:
x                 1              4
------------ - x + ----- - 4x + ----- - 4x = 0
(x^2 - 4x)         x^2            x^2
Now we can simplify the first fraction, combine the “x” terms, and combine the two fractions with “x^2” in the denominator:
1              5
------- - 9x + ----- = 0
(x - 4)          x^2
Multiply all terms by x^2:
x^2
-------- - 9x^3 + 5 = (0 * x^2) = 0
(x – 4)
Multiply all terms by (x-4):
x^2 - (9x^3) (x – 4) + 5*(x-4) = 0
x^2 – (9x^4 - 36x^3) + 5x – 20 = 0
x^2 – 9x^4 + 36x^3 + 5x - 20 = 0
Put the highest order terms first:
- 9x^4 + 36x^3 + x^2 + 5x - 20 = 0
If this really was the correct equation, you can now work on solving a 4th order polynomial. If that is the case, we can go further.

NOTE:
If your equation should actually look like this: x / (x^2 - 4x) - (x+1)/(x^2 - 4x) = -4 / (x^2 + 4x)
x (x+1) -4
------------ - ------------- = -------------
(x^2 - 4x) (x^2 - 4x) (x^2 + 4x)
it would be a much different problem. Check your problem statement and see if we need to rewrite the original equation!

Gene G

That looks more like it.  Now it really IS a linear equation!  I thought it looked too difficult.

x            x+1               -4
------------ - ----------- = -------------
(x^2 - 16)   (x^2 - 4x)     (x^2 + 4x)
No problem about the grammar. We all had to learn it somewhere along the way.

First, look at the denominators: Each one can be reduced to something a little simpler:

(x^2 - 16) can be factored into (x-4)(x+4)

(x^2 - 4x) is (x)(x-4)

(x^2 + 4x) is (x)(x+4)

That is the trick to get you started. Next, you need to multiply both sides of the equation by all of the factors that you see in denominators. It’s a good idea to multiply by one factor at a time to keep it simple. There are only 3 different factors: (x), (x-4) and (x+4).
x               x+1              -4
------------ - ----------- = -----------
(x-4)(x+4)    (x)(x - 4)     (x)(x + 4)

Multiply by (x-4)
x       x+1     -4(x-4)
------ - ----- = ------------
(x+4)     x       (x)(x + 4)

Multiply by (x+4)
(x+1)(x+4)     -4(x-4)
x - ------------- = --------
x                x

Multiply by (x)
x^2 - (x+1)(x+4) = -4(x-4)
Expand the multiplied terms
x^2 – (x^2 + 5x + 4) = -4x + 16
Get rid of the parentheses (apply the “-“ sign)
x^2 – x^2 - 5x - 4 = -4x + 16
Move x-terms to the left, constants to the right (and x^2 – x^2 = 0)
- 5x + 4x = 4 + 16
- x = 20
x = -20
This should be your answer if I didn’t get a sign wrong somewhere. It looks believable, but check it carefully.  Unless you're really comfortable with what you did, go back and work through it again until you can do it without looking.

Thank you, that answer is correct and reminded me to "apply the negative" in similar exercises! =)