Question

The subject is operations on rational expressions.

Answer 1

Answer:

$\frac{2(x-1)(2x+9)}{(x-3)(x-2)}$

Step-by-step explanation:

$\frac{4x}{(x-3)}+\frac{6}{(x+2)}$

= $\frac{4x(x+2)}{(x-3)(x+2)}+\frac{6(x-3)}{(x+2)(x-3)}$

Now we have done the denominators of each term of the expression equal.

Further we add the terms,

$\frac{4x(x+2)}{(x-3)(x+2)}+\frac{6(x-3)}{(x+2)(x-3)}$

= $\frac{4x(x+2)+6(x-3)}{(x-3)(x+2)}$

= $\frac{4x^{2}+8x+6x-18}{(x-3)(x+2)}$

= $\frac{4x^{2}+14x-18}{(x-3)(x-2)}$

Now factorize the numerator of the fraction.

4x² + 14x - 18 = 2(2x² + 7x - 9)

                     = 2(2x² + 9x - 2x - 9)

                     = 2[x(2x + 9) - 1(2x + 9)]

                     = 2(x - 1)(2x + 9)

Therefore, $\frac{2(x-1)(2x+9)}{(x-3)(x-2)}$ will be the answer.