Solve \[\frac{2x+4}{x^2+4x-5}=\frac{2-x}{x-1}\]for $x$.
We notice that the denominator on the left factors, giving us \[\frac{2x+4}{(x-1)(x+5)}=\frac{2-x}{x-1}.\]As long as $x\neq1$ we are allowed to cancel $x-1$ from the denominators, giving \[\frac{2x+4}{x+5}=2-x.\]Now we can cross-multiply to find \[2x+4=(2-x)(x+5)=-x^2-3x+10.\]We simplify this to  \[x^2+5x-6=0\]and then factor to  \[(x-1)(x+6)=0.\]Notice that since $x-1$ is in the denominator of the original equation, $x=1$ is an extraneous solution.  However $x=\boxed{-6}$ does solve the original equation.
-6
Level 5
Algebra
Problem: Solve \[\frac{2x+4}{x^2+4x-5}=\frac{2-x}{x-1}\]for $x$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the equation, first, simplify both sides. The left side can be factored. The denominator \(x^2 + 4x - 5\) factors to \((x + 5)(x - 1)\). Thus, the equation becomes:

\[\frac{2x + 4}{(x + 5)(x - 1)} = \frac{2 - x}{x - 1}.\]

Next, eliminate the common denominator by multiplying both sides by \((x + 5)(x - 1)\), assuming \(x \neq 1\) to avoid division by zero. This results in:

\[2x + 4 = (2 - x)(x + 5).\]

Now, expand the right side:

\[2x + 4 = 2x + 10 - x^2 - 5x.\]

Rearranging gives:

\[2x + 4 = -x^2 - 3x + 10.\]

Next, move all terms to one side to set the equation to zero:

\[x^2 + 5x - 6 = 0.\]

This quadratic can be factored as:

\[(x + 6)(x - 1) = 0.\]

Setting each factor to zero gives the potential solutions \(x = -6\) and \(x = 1\). However, since \(x = 1\) was excluded earlier due to the denominator, the only valid solution is:

\[x = -6.\]

Thus, the final answer is \(-6\).<|END_OF_LATENT|> The final answer is $-6$.