Answer:
Step-by-step explanation:
[tex]\dfrac{4}{x}+\dfrac{4}{x^2-9}=\dfrac{3}{x-3}[/tex]
Domain:
[tex]x\neq0\ \wedge\ x^2-9\neq0\ \wedge\ x-3\neq0\\\\x\neq0\ \wedge\ x\neq\pm3[/tex]
solution:
[tex]\dfrac{4}{x}+\dfrac{4}{x^2-3^2}=\dfrac{3}{x-3}[/tex]
use (a - b)(a + b) = a² - b²
[tex]\dfrac{4}{x}+\dfrac{4}{(x-3)(x+3)}=\dfrac{3}{x-3}[/tex]
multiply both sides by (x - 3) ≠ 0
[tex]\dfrac{4(x-3)}{x}+\dfrac{4(x-3)}{(x-3)(x+3)}=\dfrac{3(x-3)}{x-3}[/tex]
cancel (x - 3)
[tex]\dfrac{4(x-3)}{x}+\dfrac{4}{x+3}=3[/tex]
subtract [tex]\frac{4(x-3)}{x}[/tex] from both sides
[tex]\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}[/tex]
cross multiply
[tex](4)(x)=(x+3)(-x+12)[/tex]
use FOIL
[tex]4x=(x)(-x)+(x)(12)+(3)(-x)+(3)(12)\\\\4x=-x^2+12x-3x+36[/tex]
subtract 4x from both sides
[tex]0=-x^2+12x-3x+36-4x[/tex]
combine like terms
[tex]0=-x^2+(12x-3x-4x)+36\\\\0=-x^2+5x+36[/tex]
change the signs
[tex]x^2-5x-36=0\\\\x^2-9x+4x-36=0\\\\x(x-9)+4(x-9)=0\\\\(x-9)(x+4)=0[/tex]
The product is 0 if one of the factors is 0. Therefore:
[tex]x-9=0\ \vee\ x+4=0[/tex]
[tex]x-9=0[/tex] add 9 to both sides
[tex]x=9\in D[/tex]
[tex]x+4=0[/tex] subtract 4 from both sides
[tex]x=-4\in D[/tex]
