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stellarik [79]
4 years ago
6

The subject is operations on rational expressions.

Mathematics
1 answer:
Pie4 years ago
6 0

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.

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Answer:

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Step-by-step explanation:

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