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

1 answer

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 \(x = \frac{1}{2}\).