Question

solve for all values of x.
3x - \frac{3x + 7}{x + 9} = 2x

Explanation:

Step1: Eliminate the fraction

Multiply both sides by \(x + 9\) (note \(x
eq - 9\)) and \(2x\) (note \(x
eq0\)) to get rid of denominators:
\(2x(3x - 3)=(3x + 7)\times2x\)? Wait, no, the original equation is \(3x-\frac{3x + 7}{x + 9}=2x\). Let's correct that. First, move \(2x\) to the left and the fraction to the right: \(3x-2x=\frac{3x + 7}{x + 9}\), so \(x=\frac{3x + 7}{x + 9}\). Now multiply both sides by \(x + 9\) (\(x
eq - 9\)): \(x(x + 9)=3x + 7\)

Step2: Expand and simplify

Expand the left side: \(x^{2}+9x = 3x + 7\)
Subtract \(3x\) and \(7\) from both sides: \(x^{2}+9x-3x - 7=0\)
Simplify: \(x^{2}+6x - 7 = 0\)

Step3: Factor the quadratic

Factor \(x^{2}+6x - 7\): we need two numbers that multiply to \(-7\) and add to \(6\). Those numbers are \(7\) and \(-1\). So \((x + 7)(x - 1)=0\)

Step4: Solve for x

Set each factor equal to zero:
\(x+7 = 0\) gives \(x=-7\)
\(x - 1=0\) gives \(x = 1\)

We need to check for excluded values ( \(x
eq0\) and \(x
eq - 9\) ). Both \(-7\) and \(1\) are not excluded, so they are valid.

Answer:

\(x=-7\) or \(x = 1\)