Question

solve: \\(\frac{2}{x - 2} - \frac{3}{x + 3} = \frac{2x + 3}{x^2 + x - 6}\\)\
\\(\bigcirc\\) x=2\
\\(\bigcirc\\) x=-1\
\\(\bigcirc\\) x=-3\
\\(\bigcirc\\) x=3

Explanation:

Step1: Factor the denominator

First, factor the quadratic denominator \(x^2 + x - 6\). We need two numbers that multiply to \(-6\) and add to \(1\). Those numbers are \(3\) and \(-2\), so \(x^2 + x - 6=(x + 3)(x - 2)\).

The equation becomes:

$$ \frac{2}{x - 2}-\frac{3}{x + 3}=\frac{2x + 3}{(x + 3)(x - 2)} $$

Step2: Multiply through by the LCD

The least common denominator (LCD) of the fractions is \((x + 3)(x - 2)\). Multiply each term by the LCD to eliminate the fractions:

$$ (x + 3)(x - 2)\cdot\frac{2}{x - 2}-(x + 3)(x - 2)\cdot\frac{3}{x + 3}=(x + 3)(x - 2)\cdot\frac{2x + 3}{(x + 3)(x - 2)} $$

Simplify each term:

• For the first term, \((x - 2)\) cancels out: \(2(x + 3)\)
• For the second term, \((x + 3)\) cancels out: \(-3(x - 2)\)
• For the third term, \((x + 3)(x - 2)\) cancels out: \(2x + 3\)

So we have the equation:

$$ 2(x + 3)-3(x - 2)=2x + 3 $$

Step3: Expand and simplify

Expand the left - hand side:

$$ 2x+6-3x + 6=2x + 3 $$

Combine like terms on the left - hand side:

$$ (2x-3x)+(6 + 6)=2x + 3 $$

$$ -x+12=2x + 3 $$

Step4: Solve for \(x\)

Add \(x\) to both sides:

$$ 12=3x + 3 $$

Subtract \(3\) from both sides:

$$ 9 = 3x $$

Divide both sides by \(3\):

$$ x = 3 $$

We need to check for extraneous solutions. The original equation has restrictions \(x
eq - 3\) and \(x
eq2\) (because these values make the denominators zero). When \(x = 3\), the denominators \(x-2=3 - 2 = 1\), \(x + 3=3+3 = 6\), and \((x + 3)(x - 2)=(3 + 3)(3 - 2)=6\times1 = 6\) are all non - zero.

Answer:

x = 3 (or in the option format: D. x = 3, assuming the options are labeled A: x = 2, B: x=-1, C: x=-3, D: x = 3)