Question

Please help me asap!

Answer 1

Answer:

$\frac{7x + 7}{(x-3)(x+4)}$

Excluded values are 3 and -4

Step-by-step explanation:

To simplify the expression multiply to create a common denominator.

Multiply the first fraction by (x-3) and the second fraction by (x+4).

$\frac{3}{x+4} + \frac{4}{x-3}\\\\\frac{3(x-3)}{(x+4)(x-3)} + \frac{4(x+4)}{(x-3)(x+4)}$

$\frac{3(x-3)}{(x+4)(x-3)} + \frac{4(x+4)}{(x-3)(x+4)} \\\\\frac{3(x-3) + 4(x+4)}{(x-3)(x+4)}\\\\\frac{3x - 9 + 4x + 16}{(x-3)(x+4)}\\\\\frac{7x + 7}{(x-3)(x+4)}$

Excluded values are values which make the denominator 0. This means (x-3)(x+4) cannot be 0. This means x cannot be 3 or -4. These are excluded values.