Solve for x
`(1)/( 2x -3) + (1)/(x -5) =1," " x ne 3/2,5`

Answer

To solve the equation

\[ \frac{1}{2x - 3} + \frac{1}{x - 5} = 1 \]

where \( x \neq \frac{3}{2} \) and \( x \neq 5 \), we will follow these steps:

### Step 1: Find a common denominator
The common denominator for the fractions on the left side is \((2x - 3)(x - 5)\).

### Step 2: Rewrite the equation
Multiply both sides of the equation by the common denominator:

\[ (2x - 3)(x - 5) \left( \frac{1}{2x - 3} + \frac{1}{x - 5} \right) = (2x - 3)(x - 5) \cdot 1 \]

This simplifies to:

\[ (x - 5) + (2x - 3) = (2x - 3)(x - 5) \]

### Step 3: Simplify the left side
Combine like terms on the left side:

\[ x - 5 + 2x - 3 = 3x - 8 \]

### Step 4: Expand the right side
Now expand the right side:

\[ (2x - 3)(x - 5) = 2x^2 - 10x - 3x + 15 = 2x^2 - 13x + 15 \]

### Step 5: Set the equation to zero
Now we have:

\[ 3x - 8 = 2x^2 - 13x + 15 \]

Rearranging gives:

\[ 0 = 2x^2 - 13x - 3x + 15 + 8 \]

This simplifies to:

\[ 0 = 2x^2 - 16x + 23 \]

### Step 6: Write the quadratic equation
Rearranging gives us the quadratic equation:

\[ 2x^2 - 16x + 23 = 0 \]

### Step 7: Use the quadratic formula
To find the roots, we will use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In our case, \( a = 2 \), \( b = -16 \), and \( c = 23 \).

### Step 8: Calculate the discriminant
Calculate \( b^2 - 4ac \):

\[ b^2 - 4ac = (-16)^2 - 4 \cdot 2 \cdot 23 = 256 - 184 = 72 \]

### Step 9: Substitute into the quadratic formula
Now substitute back into the formula:

\[ x = \frac{16 \pm \sqrt{72}}{4} \]

### Step 10: Simplify \(\sqrt{72}\)
We can simplify \(\sqrt{72}\):

\[ \sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2} \]

### Step 11: Final expression for \(x\)
Substituting back gives:

\[ x = \frac{16 \pm 6\sqrt{2}}{4} = \frac{16}{4} \pm \frac{6\sqrt{2}}{4} = 4 \pm \frac{3\sqrt{2}}{2} \]

Thus, the solutions are:

\[ x = 4 + \frac{3\sqrt{2}}{2} \quad \text{and} \quad x = 4 - \frac{3\sqrt{2}}{2} \]