16. \( \frac{2 x-3}{x^{2}-4}-\frac{x-1}{2 x^{2}-5 x+2}=\frac{1}{2 x^{2}+3 x-2} \)

To solve the equation \[ \frac{2x - 3}{x^2 - 4} - \frac{x - 1}{2x^2 - 5x + 2} = \frac{1}{2x^2 + 3x - 2}, \] we first factor the denominators where possible. The term \( x^2 - 4 \) factors as \( (x - 2)(x + 2) \). Next, let's factor \( 2x^2 - 5x + 2 \). To do this, we look for two numbers that multiply to \( 2 \cdot 2 = 4 \) and add to \( -5 \). These numbers are \( -4 \) and \( -1 \), allowing us to break down the middle term: \[ 2x^2 - 4x - x + 2 = 2x(x - 2) - 1(x - 2) = (2x - 1)(x - 2). \] Now, we also factor \( 2x^2 + 3x - 2 \). This polynomial factors into two binomials based on finding numbers that multiply to \( -4 \) (2 times -2) and adds to \( 3 \), which are \( 4 \) and \( -1 \): \[ 2x^2 + 4x - x - 2 = 2x(x + 2) - 1(x + 2) = (2x - 1)(x + 2). \] So we can rewrite the equation as: \[ \frac{2x - 3}{(x - 2)(x + 2)} - \frac{x - 1}{(2x - 1)(x - 2)} = \frac{1}{(2x - 1)(x + 2)}. \] To eliminate the fractions, we multiply the entire equation by the common denominator, which is \((x - 2)(x + 2)(2x - 1)\): \[ (2x - 3)(2x - 1) - (x - 1)(x + 2) = (x - 2). \] Now, we expand both sides: 1. Left-hand side: - First term: \( (2x - 3)(2x - 1) = 4x^2 - 2x - 6x + 3 = 4x^2 - 8x + 3 \), - Second term: \( -(x - 1)(x + 2) = -(x^2 + 2x - x - 2) = -x^2 - x + 2 \). Combine these: \[ (4x^2 - 8x + 3) + (-x^2 - x + 2) = 4x^2 - x^2 - 8x - x + 3 + 2 = 3x^2 - 9x + 5. \] 2. Right-hand side remains \( x - 2 \). So we equate the expanded form: \[ 3x^2 - 9x + 5 = x - 2. \] Move everything to one side: \[ 3x^2 - 9x - x + 5 + 2 = 0 \Rightarrow 3x^2 - 10x + 7 = 0. \] Now we can solve the quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 7}}{2 \cdot 3} = \frac{10 \pm \sqrt{100 - 84}}{6} = \frac{10 \pm \sqrt{16}}{6} = \frac{10 \pm 4}{6}. \] This gives us: \[ x = \frac{14}{6} = \frac{7}{3} \quad \text{and} \quad x = \frac{6}{6} = 1. \] Thus, the solutions are \( x = \frac{7}{3} \) and \( x = 1 \). Finally, we check these solutions against the original equation to ensure they don’t cause any denominator to go to zero: - For \( x = 1 \): The denominators check out (not zero). - For \( x = \frac{7}{3} \): Check the denominators, they also do