Solve for x :
`(x+ 3)/(x + 2) = (3x -7)/(2x - 3) , x =2 , (3)/(2)`

Answer

To solve the equation \(\frac{x + 3}{x + 2} = \frac{3x - 7}{2x - 3}\), we will use cross multiplication. Here are the steps:

### Step 1: Cross Multiply
Cross multiplying gives us:
\[ (x + 3)(2x - 3) = (x + 2)(3x - 7) \]

### Step 2: Expand Both Sides
Now, we will expand both sides of the equation.
- Left side:
\[ x(2x) + x(-3) + 3(2x) + 3(-3) = 2x^2 - 3x + 6x - 9 = 2x^2 + 3x - 9 \]
- Right side:
\[ x(3x) + x(-7) + 2(3x) + 2(-7) = 3x^2 - 7x + 6x - 14 = 3x^2 - x - 14 \]

### Step 3: Set the Equation to Zero
Now, we set the equation to zero by moving all terms to one side:
\[ 2x^2 + 3x - 9 - (3x^2 - x - 14) = 0 \]
This simplifies to:
\[ 2x^2 + 3x - 9 - 3x^2 + x + 14 = 0 \]
Combining like terms gives:
\[ - x^2 + 4x + 5 = 0 \]
or
\[ x^2 - 4x - 5 = 0 \quad \text{(Multiplying through by -1)} \]

### Step 4: Factor the Quadratic Equation
Next, we factor the quadratic equation:
\[ (x - 5)(x + 1) = 0 \]

### Step 5: Solve for x
Setting each factor to zero gives us the solutions:
\[ x - 5 = 0 \quad \Rightarrow \quad x = 5 \]
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]

### Step 6: Check for Extraneous Solutions
We need to check if our solutions are valid given the restrictions \(x \neq 2\) and \(x \neq \frac{3}{2}\):
- \(x = 5\) is valid.
- \(x = -1\) is also valid.

### Final Answer
The solutions to the equation are:
\[ x = 5 \quad \text{and} \quad x = -1 \]

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