Question

What is the solution to the rational equation x/x^2-9 + 1/x-3= 1/4x-12?

A. x=-9/7
B. x=0
C. x=15/7
D. x=9

Answer 1

`Domain:\\\\x^2-9\\eq0\ \wedge\ x-3\\eq0\ \wedge\ 4x-12\\eq0\\\\x^2\\eq9\ \wedge\ x\\eq3\ \wedge\ 4x\\eq12\\\\x\\eq-3\ \wedge\ x\\eq3\ \wedge\ x\\eq3\ \wedge\ x\\eq3\\\\x\in\mathbb{R}-\{-3,\ 3\}\\-----------------------`

`(x)/(x^2-9)+(1)/(x-3)=(1)/(4x-12)\\\\(x)/(x^2-3^2)+(1)/(x-3)=(1)/(4(x-3))\qquad\text{multiply both sides by 4}\\\\(4x)/((x-3)(x+3))+(4)/(x-3)=(1)/(x-3)\qquad\text{subtract}\ (4)/(x-3)\ \text{from both sides}\\\\(4x)/((x-3)(x+3))=(-3)/(x-3)\\\\(4x)/((x-3)(x+3))=(-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=-(9)/(7)}\to\boxed{A.}`

Answer 2

`x=-(9)/(7)`

Explanation:

`(x)/(x^2-9) +(1)/(x-3) =(1)/(4x-12)`

To solve for x , factor all the denominator to find LCD

`x^2-9= x^2-3^3= (x+3)(x-3)`

`4x-12= 4(x-3)`

LCD is 4(x+3)(x-3)

multiply LCD with each fraction

`x \cdot 4 +1 \cdot 4(x+3) = 1 \cdot (x+3)`

`4x +4x+12=x+3`

Combine like terms

`8x+12=x+3`

Subtract x from both sides

`7x+12=3`

Subtract 12 from both sides

`7x=-9`

Divide both sides by 7

`x=-(9)/(7)`