Solve the Following Quadratic Equation by Factorization
Question:
\[ \frac{x-4}{x-5}+\frac{x-6}{x-7}=\frac{10}{3}, \qquad x\ne 5,7 \]Solution
Given:
\[ \frac{x-4}{x-5}+\frac{x-6}{x-7}=\frac{10}{3} \]Cross-multiplying by \(3(x-5)(x-7)\):
\[ 3(x-4)(x-7)+3(x-6)(x-5)=10(x-5)(x-7) \] \[ 3(x^2-11x+28)+3(x^2-11x+30) =10(x^2-12x+35) \] \[ 6x^2-66x+174 =10x^2-120x+350 \] \[ 4x^2-54x+176=0 \] \[ 2x^2-27x+88=0 \]Factorizing:
\[ 2x^2-16x-11x+88=0 \] \[ 2x(x-8)-11(x-8)=0 \] \[ (x-8)(2x-11)=0 \]Therefore,
\[ x-8=0 \quad \text{or} \quad 2x-11=0 \] \[ x=8 \quad \text{or} \quad x=\frac{11}{2} \]Both values satisfy the condition \(x\ne5,7\).