The book says the answer is supposed to be x+2 for this question with the restrictions of x does not equal -2, $\displaystyle \frac12$
I couldn't get that
The question is:
$\displaystyle \frac{2x^2-8}{6x+3}\div\frac{6x-12}{18x+9}$
The book says the answer is supposed to be x+2 for this question with the restrictions of x does not equal -2, $\displaystyle \frac12$
I couldn't get that
The question is:
$\displaystyle \frac{2x^2-8}{6x+3}\div\frac{6x-12}{18x+9}$
$\displaystyle \frac{2x^2-8}{6x+3}\div\frac{6x-12}{18x+9} = \frac{2x^2-8}{6x+3} \times \frac{18x+9}{6x-12}$
$\displaystyle = \frac{2x^2-8}{6x+3} \times \frac{3(6x+3)}{6x-12}$
$\displaystyle = \frac{3(2x^2-8)}{6x - 12} $
$\displaystyle = \frac{6(x^2-4)}{6(x - 2)} $
$\displaystyle = \frac{6(x - 2)(x + 2)}{6(x - 2)} $
$\displaystyle = (x + 2)$
Hello, t-dot!
Factor all the expressions:$\displaystyle \frac{2x^2-8}{6x+3} \div \frac{6x-12}{18x+9}\;\; {\color{blue}=\;\;\frac{2x^2-8}{6x+3} \times \frac{18x+9}{6x-12}}$
Book's answer: .$\displaystyle x+2$
. . $\displaystyle 2x^2-8\:=\:2(x^2-4)\:=\:2(x-2)(x+2)$
. . $\displaystyle 6x+3\:=\:3(2x+1)$
. . $\displaystyle 6x-12 \:=\:6(x-2)$
. . $\displaystyle 18x + 9 \:=\:9(2x+1)$
The problem becomes: .$\displaystyle \frac{\not{2}(x-2)(x+2)}{\not3(2x+1)} \times \frac{\not9(2x+1)}{\not6(x-2)} \;\;=\;\;x + 2$