Answer
$$\frac{x^2-4x+3}{x^2+4x-5}=\frac{x-3}{x+5}$$
Explanation
| $$ \begin{aligned}\frac{x^2-4x+3}{x^2+4x-5}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x-3}{x+5}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{x^2-4x+3}{x^2+4x-5} $ to $ \dfrac{x-3}{x+5} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x-1}$. $$ \begin{aligned} \frac{x^2-4x+3}{x^2+4x-5} & =\frac{ \left( x-3 \right) \cdot \color{blue}{ \left( x-1 \right) }}{ \left( x+5 \right) \cdot \color{blue}{ \left( x-1 \right) }} = \\[1ex] &= \frac{x-3}{x+5} \end{aligned} $$ |
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Simplify Rational Expressions