Question Progress \( 2-\frac{x+1}{x-2}-\frac{x-4}{x+2} \) can be written as a single fraction in the form \( \frac{a x+b}{x^{2}-4} \) where \( a \) and \( b \) are integers. Work out the value of \( a \) and the value of \( b \).

To combine the expression \( 2-\frac{x+1}{x-2}-\frac{x-4}{x+2} \) into a single fraction, we first rewrite \( 2 \) in terms of the common denominator \( (x-2)(x+2) \), which is \( x^2 - 4 \). We have: \[ 2 = \frac{2(x^2 - 4)}{x^2 - 4} = \frac{2x^2 - 8}{x^2 - 4} \] Now we rewrite the other fractions with the same denominator: \[ -\frac{x+1}{x-2} = -\frac{(x+1)(x+2)}{(x-2)(x+2)} = -\frac{x^2 + 3x + 2}{x^2 - 4} \] \[ -\frac{x-4}{x+2} = -\frac{(x-4)(x-2)}{(x+2)(x-2)} = -\frac{x^2 - 2x + 8}{x^2 - 4} \] Now we combine them: \[ \frac{2x^2 - 8 - (x^2 + 3x + 2) - (x^2 - 2x + 8)}{x^2 - 4} \] Now simplify the numerator: \[ 2x^2 - 8 - x^2 - 3x - 2 - x^2 + 2x - 8 \] Combine like terms: \[ (2x^2 - x^2 - x^2) + (-3x + 2x) + (-8 - 2 - 8) = 0x^2 - 1x - 18 = -x - 18 \] So we have: \[ \frac{-x - 18}{x^2 - 4} \] From this, we identify \( a = -1 \) and \( b = -18 \). Therefore, the values are: - \( a = -1 \) - \( b = -18 \)