SOLUTION: Solve the equation. {{{ 1/(x+4)=2/(x^2+3x-4)-1/(1-x) }}}

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Question 1054591: Solve the equation.

++1%2F%28x%2B4%29=2%2F%28x%5E2%2B3x-4%29-1%2F%281-x%29+

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

You didn't put enough parentheses when you typed in your equation:
It should be:

1/(x+4)=2/(x^2+3x-4)-1/(1-x) 

not this 1/x+4=2/x^2+3x-4-1/1-x

You want it to mean the same as this:

1%5E%22%22%2F%28x%5E%22%22%2B4%29%22%22=%22%222%5E%22%22%2F%28x%5E2%2B3x-4%29-1%5E%22%22%2F%281-x%5E%22%22%29 

Not the same as this cross%281%5E%22%22%2Fx%5E%22%22%2B4=2%5E%22%22%2Fx%5E2%2B3x-4-1%5E%22%22%2F1-x%5E%22%22%29

--------------------------------------------------

1%5E%22%22%2F%28x%5E%22%22%2B4%29%22%22=%22%222%5E%22%22%2F%28x%5E2%2B3x-4%29-1%5E%22%22%2F%281-x%5E%22%22%29

Before we get a common denominator we must simplify.
We factor the denominator x²+3x-4 as (x+4)(x-1) and change 
the denominator 1-x to descending order -x+1, and then
factor out a negative sign as -(x-1) by changing the 
sign inside the parentheses.  So now we have

1%2F%28x%2B4%29%22%22=%22%222%2F%28%28x%2B4%29%28x-1%29%29-1%2F%28-%28x-1%29%29
 
or

1%2F%28x%2B4%29%22%22=%22%222%2F%28%28x%2B4%29%28x-1%29%29%2B1%2F%28x-1%29  <--eq. #1

Now the common denominator is (x+4)(x-1).  Since it is
an equation and not just an expression without an equal
sign, we may multiply both sides by the LCD over 1, which
I'll write as red%28%28%28x%2B4%29%28x-1%29%29%2F1%29

red%28%28%28x%2B4%29%28x-1%29%29%2F1%29expr%281%2F%28x%2B4%29%29%22%22=%22%22

Notice that we have multiplied that LCD by every term of 
eq. #1 above.

Now we can cancel and eventually we'll end up with no fractions:

red%28%28%28cross%28x%2B4%29%29%28x-1%29%29%2F1%29expr%281%2F%28cross%28x%2B4%29%29%29%22%22=%22%22

And if we ignore the ones in the denominators, we have just:

red%28%28x-1%29%291%22%22=%22%222%2Bred%28%28x%2B4%29%291

or 

x-1%22%22=%22%222%2Bx%2B4

x-1%22%22=%22%226%2Bx

Subtract x from both sides:

-1%22%22=%22%226

This is false, so there is no solution
to the given problem, but we could not
have known that it had no solution until 
we got to a false equation containing no 
variables.  Other very similar equations
to this one do have solutions because the
variables x's don't all cancel out as they 
did in this equation.

Edwin