Solve the Following Quadratic Equation by Factorization
Question:
\[ \frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3}, \qquad x\ne 3,5 \]Solution
Given:
\[ \frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3} \]Multiplying both sides by \(3(x-3)(x-5)\):
\[ 3(x-2)(x-5)+3(x-4)(x-3) =10(x-3)(x-5) \] \[ 3(x^2-7x+10)+3(x^2-7x+12) =10(x^2-8x+15) \] \[ 6x^2-42x+66 =10x^2-80x+150 \] \[ 4x^2-38x+84=0 \] \[ 2x^2-19x+42=0 \]Factorizing:
\[ 2x^2-12x-7x+42=0 \] \[ 2x(x-6)-7(x-6)=0 \] \[ (x-6)(2x-7)=0 \]Therefore,
\[ x-6=0 \quad \text{or} \quad 2x-7=0 \] \[ x=6 \quad \text{or} \quad x=\frac{7}{2} \]Both values satisfy the condition \(x\ne3,5\).