Question

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

Answer 1

\[\frac{x}{x^2 - 9} - \frac{1}{x + 3} = \frac{1}{4x - 12}\]

Make sure denominators are factored:
\[\frac{x}{(x+3)(x-3)} - \frac{1}{x + 3} = \frac{1}{4(x-3)}\]

Multiply both sides by 4:
\[\frac{4x}{(x+3)(x-3)} - \frac{4}{x + 3} = \frac{1}{(x-3)}\]

Add 4/(x+3) to both sides; Subtract 1/(x -3) from both sides:
\[\frac{4x}{(x+3)(x-3)} - \frac{1}{(x-3)} = \frac{4}{x + 3}\]

Rewrite as:
\[\frac{4x}{(x+3)} \dot\ \frac{1}{(x-3)} - \frac{1}{(x-3)} = \frac{4}{x + 3}\]

Observe that 1/(x-3) is a common term; Factor it out:
\[\left(\frac{4x}{x+3} - 1\right)\frac{1}{x-3}= \frac{4}{x + 3}\]

Change 1 to (x + 3)/(x+3) then combine fractions inside parentheses:
\[\left(\frac{4x}{x+3} - \frac{x+3}{x+3}\right)\frac{1}{x-3}= \frac{4}{x + 3}\]

\[\left(\frac{4x -(x+3)}{x+3} \right)\frac{1}{x-3}= \frac{4}{x + 3}\]
\[\left(\frac{4x -x -3 }{x+3} \right)\frac{1}{x-3}= \frac{4}{x + 3}\]
\[\left(\frac{3x -3 }{x+3} \right)\frac{1}{x-3}= \frac{4}{x + 3}\]

Rewrite the fraction on the left as:
\[\frac{3x -3 }{(x+3)(x-3)} = \frac{4}{x + 3}\]
Multiply both sides by (x + 3):

\[\frac{3x - 3}{x-3} = 4\]

Multiiply both sides by (x - 3)

\[3x - 3 = 4(x-3)\]

Distribute:

\[3x - 3 = 4x - 12\]

Finish solving for x