[tex]Domain:\\\\x^2-9\neq0\ \wedge\ x-3\neq0\ \wedge\ 4x-12\neq0\\\\x^2\neq9\ \wedge\ x\neq3\ \wedge\ 4x\neq12\\\\x\neq-3\ \wedge\ x\neq3\ \wedge\ x\neq3\ \wedge\ x\neq3\\\\x\in\mathbb{R}-\{-3,\ 3\}\\-----------------------[/tex]
[tex]\dfrac{x}{x^2-9}+\dfrac{1}{x-3}=\dfrac{1}{4x-12}\\\\\dfrac{x}{x^2-3^2}+\dfrac{1}{x-3}=\dfrac{1}{4(x-3)}\qquad\text{multiply both sides by 4}\\\\\dfrac{4x}{(x-3)(x+3)}+\dfrac{4}{x-3}=\dfrac{1}{x-3}\qquad\text{subtract}\ \dfrac{4}{x-3}\ \text{from both sides}\\\\\dfrac{4x}{(x-3)(x+3)}=\dfrac{-3}{x-3}\\\\\dfrac{4x}{(x-3)(x+3)}=\dfrac{-3(x+3)}{(x-3)(x+3)}\iff4x=-3(x+3)\\\\4x=-3x-9\qquad\text{add 3x to both sides}\\\\7x=-9\qquad\text{divide both sides by 7}\\\\\boxed{x=-\dfrac{9}{7}}\to\boxed{A.}[/tex]
Answer:
[tex]x=-\frac{9}{7}[/tex]
Step-by-step explanation:
[tex]\frac{x}{x^2-9} +\frac{1}{x-3} =\frac{1}{4x-12}[/tex]
To solve for x , factor all the denominator to find LCD
[tex]x^2-9= x^2-3^3= (x+3)(x-3)[/tex]
[tex]4x-12= 4(x-3)[/tex]
LCD is 4(x+3)(x-3)
multiply LCD with each fraction
[tex]x \cdot 4 +1 \cdot 4(x+3) = 1 \cdot (x+3)[/tex]
[tex]4x +4x+12=x+3[/tex]
Combine like terms
[tex]8x+12=x+3[/tex]
Subtract x from both sides
[tex]7x+12=3[/tex]
Subtract 12 from both sides
[tex]7x=-9[/tex]
Divide both sides by 7
[tex]x=-\frac{9}{7}[/tex]
