Solve \( \frac{7}{x-3}+\frac{10}{x+1}=3 \)

To solve the equation \( \frac{7}{x-3}+\frac{10}{x+1}=3 \), we first find a common denominator for the fractions, which is \((x-3)(x+1)\). Rewriting the equation with the common denominator: \[ \frac{7(x+1) + 10(x-3)}{(x-3)(x+1)} = 3 \] Now, expand the numerator: \[ 7(x+1) + 10(x-3) = 7x + 7 + 10x - 30 = 17x - 23 \] This turns our equation into: \[ \frac{17x - 23}{(x-3)(x+1)} = 3 \] Next, we cross-multiply to eliminate the fraction: \[ 17x - 23 = 3(x-3)(x+1) \] Now let’s expand the right side: \[ 3(x-3)(x+1) = 3(x^2 + x - 3x - 3) = 3(x^2 - 2x - 3) = 3x^2 - 6x - 9 \] Setting the entire equation: \[ 17x - 23 = 3x^2 - 6x - 9 \] Rearranging all terms to one side gives: \[ 0 = 3x^2 - 6x - 9 - 17x + 23 \] Combining like terms results in: \[ 0 = 3x^2 - 23x + 14 \] Now we need to factor or use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 3 \), \( b = -23 \), and \( c = 14 \): Calculating the discriminant: \[ b^2 - 4ac = (-23)^2 - 4(3)(14) = 529 - 168 = 361 \] Since 361 is a perfect square, we can continue with the quadratic formula: \[ x = \frac{23 \pm \sqrt{361}}{2 \cdot 3} = \frac{23 \pm 19}{6} \] Then, the two potential solutions are: 1. \( x = \frac{42}{6} = 7 \) 2. \( x = \frac{4}{6} = \frac{2}{3} \) Thus, the solutions to the original equation are \( x = 7 \) and \( x = \frac{2}{3} \).