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

Options

A. 5, 1
B. 5 , -1
C. 2, -3
D. -2 , 3

Explanation

To solve the equation \(\frac{x+3}{x+2} = \frac{3x-7}{2x-3}\), we will follow these steps:

### Step 1: Cross Multiply
We start by cross multiplying to eliminate the fractions. This gives us:
\[ (x + 3)(2x - 3) = (3x - 7)(x + 2) \]

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

### 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:
\[ - x^2 + 4x + 5 = 0 \]
To make it easier, we can multiply the entire equation by -1:
\[ x^2 - 4x - 5 = 0 \]

### Step 4: Factor the Quadratic Equation
Now we will factor the quadratic equation \(x^2 - 4x - 5 = 0\). We need two numbers that multiply to \(-5\) and add to \(-4\). The numbers \(-5\) and \(1\) work:
\[ (x - 5)(x + 1) = 0 \]

### Step 5: Solve for x
Now we set each factor equal to zero:
1. \(x - 5 = 0 \Rightarrow x = 5\)
2. \(x + 1 = 0 \Rightarrow x = -1\)

Thus, the solutions to the equation are:
\[ x = 5 \quad \text{and} \quad x = -1 \]