How do you simplify and find the restrictions for $\frac{3 x - 2}{x + 3} + \frac{7}{x^{2} - x - 12}$ $(3x-2)/(x+3)+7/(x^2-x-12)$ ?

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How do you simplify and find the restrictions for $\frac{3 x - 2}{x + 3} + \frac{7}{x^{2} - x - 12}$ $(3x-2)/(x+3)+7/(x^2-x-12)$ ?

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Answer 1 · 2016-08-08T20:23:51

$= \frac{(x - 3) (3 x - 5)}{(x - 4) (x + 3)}$ $=((x-3)(3x-5))/((x-4)(x+3)$
Restrictions are
$x \neq 4$ $x!=4$
$x \neq - 3$ $x!=-3$

Explanation:

$\frac{3 x - 2}{x + 3} + \frac{7}{x^{2} - x - 12}$ $(3x-2)/(x+3)+7/(x^2-x-12)$
$= \frac{3 x - 2}{x + 3} + \frac{7}{x^{2} - 4 x + 3 x - 12}$ $=(3x-2)/(x+3)+7/(x^2-4x+3x-12)$
$= \frac{3 x - 2}{x + 3} + \frac{7}{x (x - 4) + 3 (x - 4)}$ $=(3x-2)/(x+3)+7/(x(x-4)+3(x-4)$
$= \frac{3 x - 2}{x + 3} + \frac{7}{(x - 4) (x + 3)}$ $=(3x-2)/(x+3)+7/((x-4)(x+3))$
$= \frac{(3 x - 2) (x - 4) + 7}{(x - 4) (x + 3)}$ $=((3x-2)(x-4)+7)/((x-4)(x+3))$
$= \frac{3 x^{2} - 12 x - 2 x + 8 + 7}{(x - 4) (x + 3)}$ $=(3x^2-12x-2x+8+7)/((x-4)(x+3))$
$= \frac{3 x^{2} - 14 x + 15}{(x - 4) (x + 3)}$ $=(3x^2-14x+15)/((x-4)(x+3))$
$= \frac{3 x^{2} - 9 x - 5 x + 15}{(x - 4) (x + 3)}$ $=(3x^2-9x-5x+15)/((x-4)(x+3))$
$= \frac{3 x (x - 3) - 5 (x - 3)}{(x - 4) (x + 3)}$ $=(3x(x-3)-5(x-3))/((x-4)(x+3))$
$= \frac{(x - 3) (3 x - 5)}{(x - 4) (x + 3)}$ $=((x-3)(3x-5))/((x-4)(x+3)$
Restrictions are
$x - 4 \neq 0$ $x-4!=0$
or
$x \neq 4$ $x!=4$
$x + 3 \neq 0$ $x+3!=0$
or
$x \neq - 3$ $x!=-3$

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How do you simplify and find the restrictions for $\frac{3 x - 2}{x + 3} + \frac{7}{x^{2} - x - 12}$ $(3x-2)/(x+3)+7/(x^2-x-12)$ ?

1 Answer

Explanation:

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How do you simplify and find the restrictions for 3x−2x+3+7x2−x−12(3x-2)/(x+3)+7/(x^2-x-12)?

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How do you simplify and find the restrictions for $\frac{3 x - 2}{x + 3} + \frac{7}{x^{2} - x - 12}$ $(3x-2)/(x+3)+7/(x^2-x-12)$ ?