Answer to Question #343472 in Algebra for Bethsheba Kiap

Question

In the following problems, perform the operations and simplify as much as possible.

a. (x2 + 2x)/(3x2 – 18x + 24) ÷ (x2 – x – 6)/(x

2 – 4x + 4)

b. (x2 + 6x + 9)/x/(x + 3)

c. 1/(3x – 1) + x/(x2 – 9)

Answer 1

a. $\frac{x^2+2x}{3x^2-18x+24}*\frac{x^2-4x+4}{x^2-x-6}=\\$

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

b. $\frac{(x^2+6x+9)}{x(x+3)}=\frac{(x+3)^2}{x(x+3)}=\frac{x+3}{x}$

c. $\frac{1}{3x-1}+\frac{x}{x^2-9}=\frac{x^2-9+3x^2-x}{(3x-1)(x^2-9)}=\frac{4x^2-x-9}{(3x-1)(x-3)(x+3)}$