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Consider the expression below.
[tex]\[
\frac{3x+6}{x^2-x-6} + \frac{2x}{x^2+x-12}
\][/tex]

Place the steps required to determine the sum of the two expressions in the correct order.
[tex]\[
\begin{array}{l}
\frac{3(x+2)}{(x+2)(x-3)} + \frac{2x}{(x-3)(x+4)} \\
\frac{3(x+4)}{(x-3)(x+4)} + \frac{2x}{(x-3)(x+4)} \\
\frac{(3x+12)+2x}{(x-3)(x+4)} \\
\frac{3}{(x-3)} + \frac{2x}{(x-3)(x+4)} \\
\frac{(3x+12)+2x}{(x-9)/(x-1)} \\
\frac{5x+6}{x^2-x+12} \\
\end{array}
\][/tex]

Sagot :

To determine the sum of the two given expressions, we need to follow these steps in the correct order:

1. Factor the denominators of each fraction:
[tex]\[ \frac{3(x+2)}{(x+2)(x-3)}+\frac{2 x}{(x-3)(x+4)} \][/tex]
2. Simplify the first fraction by canceling the common factor in the numerator and the denominator:
[tex]\[ \frac{3}{(x-3)}+\frac{2 x}{(x-3)(x+4)} \][/tex]
3. Rewrite the first fraction with the common denominator (x - 3)(x + 4):
[tex]\[ \frac{3(x+4)}{(x-3)(x+4)}+\frac{2 x}{(x-3)(x+4)} \][/tex]
4. Combine the numerators over the common denominator:
[tex]\[ \frac{(3(x+4) + 2 x)}{(x-3)(x+4)} \][/tex]

The correct order of steps is:

- [tex]\(\frac{3(x+2)}{(x+2)(x-3)}+\frac{2 x}{(x-3)(x+4)} \)[/tex]
- [tex]\(\frac{3}{(x-3)}+\frac{2 x}{(x-3)(x+4)}\)[/tex]
- [tex]\(\frac{3(x+4)}{(x-3)(x+4)}+\frac{2 x}{(x-3)(x+4)}\)[/tex]
- [tex]\(\frac{(3(x+4)+2x)}{(x-3)(x+4)}\)[/tex]

These steps lead us to the required result.
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