If `(ax)/((x+2)(2x-3)) =(2)/(x+2) +(3)/(2x-3)`, then `a=`

Options

A. `4`
B. `5`
C. `7`
D. None of these

Answer

To solve the equation

\[ \frac{ax}{(x+2)(2x-3)} = \frac{2}{x+2} + \frac{3}{2x-3}, \]

we will follow these steps:

### Step 1: Combine the right-hand side into a single fraction.

To combine the fractions on the right-hand side, we need a common denominator, which is \((x+2)(2x-3)\). Thus, we rewrite the right-hand side:

\[ \frac{2}{x+2} = \frac{2(2x-3)}{(x+2)(2x-3)} = \frac{4x - 6}{(x+2)(2x-3)}, \]

\[ \frac{3}{2x-3} = \frac{3(x+2)}{(x+2)(2x-3)} = \frac{3x + 6}{(x+2)(2x-3)}. \]

Now, we can combine these:

\[ \frac{4x - 6 + 3x + 6}{(x+2)(2x-3)} = \frac{(4x + 3x - 6 + 6)}{(x+2)(2x-3)} = \frac{7x}{(x+2)(2x-3)}. \]

### Step 2: Set the left-hand side equal to the combined right-hand side.

Now we have:

\[ \frac{ax}{(x+2)(2x-3)} = \frac{7x}{(x+2)(2x-3)}. \]

### Step 3: Compare the numerators.

Since the denominators are the same, we can equate the numerators:

\[ ax = 7x. \]

### Step 4: Solve for \(a\).

To find \(a\), we can divide both sides by \(x\) (assuming \(x \neq 0\)):

\[ a = 7. \]

Thus, the value of \(a\) is

\[ \boxed{7}. \]