Answer:
C. \(\frac{w^2-7w+12}{w^2+w-20}\)
Step-by-step explanation:
We have been given a fraction \(\frac{w-3}{w+5}\) and we are asked to write the equivalent fraction of our given fraction with the denominator \(w^2+w-10\).
First of all we will divide \(w^2+w-10\) by \(w+5\) to find the expression that we need to multiply with our given fraction to get the denominator \(w^2+w-10\)
Let us factor out \(w^2+w-10\) by splitting the middle term
\(w^2+w-10\)
\(w^2+5w-4w-20\)
\(w(w+5)-4(w+5)\)
\((w+5)(w-4)\)
\(\frac{w^2+w-20}{w+5}=\frac{(w+5)(w-4)}{(w+5)}=(w-4)\)
Now we will multiply the numerator and denominator of \(\frac{w-3}{w+5}\) by \((w-4)\) to find the equivalent fraction.
\(\frac{w-3}{w+5}*\frac{(w-4)}{(w-4)}\)
\(\frac{(w-3)*(w-4)}{(w+5)*(w-4)}\)
\(\frac{w^2-4w-3w+12}{w^2-4w+5w-20}\)
\(\frac{w^2-7w+12}{w^2+w-20}\)
Therefore, option C is the correct choice.
