Question

Solution(s) of the equation 9 over x plus 9 over x-2=12

Answer 1

To solve the equation

\[
\frac{9}{x} + \frac{9}{x - 2} = 12,
\]

we can start by finding a common denominator for the left-hand side. The common denominator will be \(x(x - 2)\). Rewriting the equation with the common denominator, we have:

\[
\frac{9(x - 2) + 9x}{x(x - 2)} = 12.
\]

Now simplifying the numerator:

\[
9(x - 2) + 9x = 9x - 18 + 9x = 18x - 18.
\]

So the equation becomes:

\[
\frac{18x - 18}{x(x - 2)} = 12.
\]

Next, we can eliminate the fraction by multiplying both sides by \(x(x - 2)\):

\[
18x - 18 = 12x(x - 2).
\]

Expanding the right-hand side gives:

\[
18x - 18 = 12x^2 - 24x.
\]

Now, we can rearrange all terms to one side of the equation:

\[
0 = 12x^2 - 24x - 18x + 18,
\]
which simplifies to:

\[
0 = 12x^2 - 42x + 18.
\]

Dividing the entire equation by 6 to simplify yields:

\[
0 = 2x^2 - 7x + 3.
\]

Next, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \(a = 2\), \(b = -7\), and \(c = 3\).

Calculating the discriminant:

\[
b^2 - 4ac = (-7)^2 - 4 \cdot 2 \cdot 3 = 49 - 24 = 25.
\]

Now we can find the solutions:

\[
x = \frac{-(-7) \pm \sqrt{25}}{2 \cdot 2} = \frac{7 \pm 5}{4}.
\]

Calculating the two possible values for \(x\):

1. \( x = \frac{7 + 5}{4} = \frac{12}{4} = 3 \)
2. \( x = \frac{7 - 5}{4} = \frac{2}{4} = \frac{1}{2} \)

Thus, the solutions to the equation are:

\[
x = 3 \quad \text{and} \quad x = \frac{1}{2}.
\]

Finally, we should check if either \(x\) values make the original equation undefined (division by zero):

- For \(x = 3\), \( x - 2 \) is \(1\) (valid).
- For \(x = \frac{1}{2}\), \( x - 2 \) is \(-\frac{3}{2}\) (valid).

Both solutions are valid, thus:

The solutions are \(x = 3\) and \(x = \frac{1}{2}\).