Solve for the unknown variable i a) \( \frac{3 a-1}{a^{2}-2 a-3}+\frac{3 a}{a+1}+\frac{2}{3-a}=0 \)

To solve the equation \( \frac{3 a-1}{a^{2}-2 a-3}+\frac{3 a}{a+1}+\frac{2}{3-a}=0 \), let's simplify and solve step by step. First, we factor the quadratic in the first term's denominator: \[ a^{2}-2a-3 = (a-3)(a+1) \] Substituting this back into the equation, we have: \[ \frac{3a-1}{(a-3)(a+1)} + \frac{3a}{a+1} + \frac{2}{3-a} = 0 \] Next, notice that \( \frac{2}{3-a} \) can be rewritten as \( -\frac{2}{a-3} \): \[ \frac{3a-1}{(a-3)(a+1)} + \frac{3a}{a+1} - \frac{2}{a-3} = 0 \] Now to combine the fractions. The common denominator will be \( (a-3)(a+1) \): \[ \frac{(3a-1) + 3a(a-3) - 2(a+1)}{(a-3)(a+1)} = 0 \] Now we simplify the numerator: 1. Expanding \( 3a(a-3) \) gives \( 3a^2 - 9a \). 2. Expanding \( -2(a+1) \) gives \( -2a - 2 \). Adding these gives: \[ (3a - 1) + (3a^2 - 9a) - (2a + 2) = 3a^2 - 8a - 3 \] Setting the numerator equal to zero: \[ 3a^2 - 8a - 3 = 0 \] We apply the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 3, b = -8, c = -3 \): \[ a = \frac{8 \pm \sqrt{(-8)^2 - 4(3)(-3)}}{2(3)} \] \[ a = \frac{8 \pm \sqrt{64 + 36}}{6} \] \[ a = \frac{8 \pm \sqrt{100}}{6} \] \[ a = \frac{8 \pm 10}{6} \] This results in two possible solutions: \[ a = \frac{18}{6} = 3 \quad \text{or} \quad a = \frac{-2}{6} = -\frac{1}{3} \] Thus, the unknown variable \( a \) can take values \( 3 \) or \( -\frac{1}{3} \).