Question

Solve for x: 4 over x plus 4 over quantity x squared minus 9 equals 3 over quantity x minus 3. (2 points) Select one: a. x = -4 and x = -9 b. x = 4 and x = -9 c. x = -4 and x = 9 d. x = 4 and x = 9

Answer 1

Answer:

c. x = -4 or x = 9

Step-by-step explanation:

$\dfrac{4}{x}+\dfrac{4}{x^2-9}=\dfrac{3}{x-3}$

Domain:

$x\neq0\ \wedge\ x^2-9\neq0\ \wedge\ x-3\neq0\\\\x\neq0\ \wedge\ x\neq\pm3$

solution:

$\dfrac{4}{x}+\dfrac{4}{x^2-3^2}=\dfrac{3}{x-3}$

use (a - b)(a + b) = a² - b²

$\dfrac{4}{x}+\dfrac{4}{(x-3)(x+3)}=\dfrac{3}{x-3}$

multiply both sides by (x - 3) ≠ 0

$\dfrac{4(x-3)}{x}+\dfrac{4(x-3)}{(x-3)(x+3)}=\dfrac{3(x-3)}{x-3}$

cancel (x - 3)

$\dfrac{4(x-3)}{x}+\dfrac{4}{x+3}=3$

subtract $\frac{4(x-3)}{x}$ from both sides

$\dfrac{4}{x+3}=3-\dfrac{4(x-3)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x}{x}-\dfrac{(4)(x)+(4)(-3)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x-\bigg(4x-12\bigg)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x-4x-(-12)}{x}\\\\\dfrac{4}{x+3}=\dfrac{-x+12}{x}$

cross multiply

$(4)(x)=(x+3)(-x+12)$

use FOIL

$4x=(x)(-x)+(x)(12)+(3)(-x)+(3)(12)\\\\4x=-x^2+12x-3x+36$

subtract 4x from both sides

$0=-x^2+12x-3x+36-4x$

combine like terms

$0=-x^2+(12x-3x-4x)+36\\\\0=-x^2+5x+36$

change the signs

$x^2-5x-36=0\\\\x^2-9x+4x-36=0\\\\x(x-9)+4(x-9)=0\\\\(x-9)(x+4)=0$

The product is 0 if one of the factors is 0. Therefore:

$x-9=0\ \vee\ x+4=0$

$x-9=0$            add 9 to both sides

$x=9\in D$

$x+4=0$          subtract 4 from both sides

$x=-4\in D$