Question

Solve for $x$: $\frac{4}{x} - \frac{5}{2x + 3} = 3.$

Answer 1

Step 1: Find the common denominator The denominator is $x(2x + 3)$. Rewrite: $\frac{4(2x + 3) - 5x}{x(2x + 3)} = 3.$ Step 2: Simplify the numerator $\frac{8x + 12 - 5x}{x(2x + 3)} = 3 \implies \frac{3x + 12}{x(2x + 3)} = 3.$ Step 3: Cross-multiply $3x + 12 = 3x(2x + 3) \implies 3x + 12 = 6x^2 + 9x.$ Rearrange: $6x^2 + 6x - 12 = 0.$ Simplify: $x^2 + x - 2 = 0.$ Step 4: Solve the quadratic equation Factorize: $x^2 + x - 2 = (x + 2)(x - 1) = 0.$ $x = -2 \quad \text{or} \quad x = 1.$ Step 5: Check solutions For $x = -2$, $2x + 3 = -1$, so the denominator is undefined. Therefore, $x = -2$ is not valid. Correct Answer: $x = 1$.

Answer 2

Step 1: Find the common denominator The denominator is $x(2x + 3)$. Rewrite: $\frac{4(2x + 3) - 5x}{x(2x + 3)} = 3.$ Step 2: Simplify the numerator $\frac{8x + 12 - 5x}{x(2x + 3)} = 3 \implies \frac{3x + 12}{x(2x + 3)} = 3.$ Step 3: Cross-multiply $3x + 12 = 3x(2x + 3) \implies 3x + 12 = 6x^2 + 9x.$ Rearrange: $6x^2 + 6x - 12 = 0.$ Simplify: $x^2 + x - 2 = 0.$ Step 4: Solve the quadratic equation Factorize: $x^2 + x - 2 = (x + 2)(x - 1) = 0.$ $x = -2 \quad \text{or} \quad x = 1.$ Step 5: Check solutions For $x = -2$, $2x + 3 = -1$, so the denominator is undefined. Therefore, $x = -2$ is not valid. Correct Answer: $x = 1$.