Question

Solve the following equation: \[ 2\left( \frac{2x - 1}{x + 3} \right) - 3\left( \frac{x + 3}{2x - 1} \right) = 5, \quad x \neq 3, \frac{1}{2}. \]

Hint:

When solving rational equations, eliminate the fractions by multiplying both sides by the least common denominator and then simplify the resulting quadratic equation.

Answer 1

We are given the equation: \[ 2\left( \frac{2x - 1}{x + 3} \right) - 3\left( \frac{x + 3}{2x - 1} \right) = 5. \] Step 1: Simplify the equation. First, expand both terms on the left-hand side: \[ \frac{2(2x - 1)}{x + 3} - \frac{3(x + 3)}{2x - 1} = 5. \] This simplifies to: \[ \frac{4x - 2}{x + 3} - \frac{3x + 9}{2x - 1} = 5. \] Step 2: Solve the equation. To simplify this further, we multiply both sides of the equation by \( (x + 3)(2x - 1) \) to eliminate the denominators: \[ (4x - 2)(2x - 1) - (3x + 9)(x + 3) = 5(x + 3)(2x - 1). \] Expand each term: \[ (4x - 2)(2x - 1) = 8x^2 - 4x - 4x + 2 = 8x^2 - 8x + 2, \] \[ (3x + 9)(x + 3) = 3x^2 + 9x + 9x + 27 = 3x^2 + 18x + 27, \] \[ 5(x + 3)(2x - 1) = 5(2x^2 + 6x - x - 3) = 5(2x^2 + 5x - 3) = 10x^2 + 25x - 15. \] Now, substitute these into the equation: \[ 8x^2 - 8x + 2 - (3x^2 + 18x + 27) = 10x^2 + 25x - 15. \] Simplifying: \[ 8x^2 - 8x + 2 - 3x^2 - 18x - 27 = 10x^2 + 25x - 15, \] \[ 5x^2 - 26x - 25 = 10x^2 + 25x - 15. \] Step 3: Move all terms to one side. Bring all terms to one side of the equation: \[ 5x^2 - 26x - 25 - 10x^2 - 25x + 15 = 0. \] Simplifying: \[ -5x^2 - 51x - 10 = 0. \] Step 4: Solve the quadratic equation. Multiply through by \(-1\) to simplify: \[ 5x^2 + 51x + 10 = 0. \] Now, solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] where \( a = 5 \), \( b = 51 \), and \( c = 10 \). Substituting these values: \[ x = \frac{-51 \pm \sqrt{51^2 - 4(5)(10)}}{2(5)} = \frac{-51 \pm \sqrt{2601 - 200}}{10} = \frac{-51 \pm \sqrt{2401}}{10}. \] \[ x = \frac{-51 \pm 49}{10}. \] Thus, we have two solutions for \( x \): \[ x = \frac{-51 + 49}{10} = \frac{-2}{10} = -0.2 \quad \text{or} \quad x = \frac{-51 - 49}{10} = \frac{-100}{10} = -10. \]
Conclusion:
The solutions are \( x = -0.2 \) and \( x = -10 \).