Solve the Following Quadratic Equation by Factorization
Question:
\[ \frac{2x}{x-4}+\frac{2x-5}{x-3}=\frac{25}{3}, \qquad x\ne 3,4 \]Solution
Given:
\[ \frac{2x}{x-4}+\frac{2x-5}{x-3}=\frac{25}{3} \]Multiplying both sides by \(3(x-4)(x-3)\), we get:
\[ 6x(x-3)+3(2x-5)(x-4)=25(x-4)(x-3) \] \[ 6x^2-18x+6x^2-39x+60=25x^2-175x+300 \] \[ 12x^2-57x+60=25x^2-175x+300 \] \[ 13x^2-118x+240=0 \]Factorizing:
\[ 13x^2-78x-40x+240=0 \] \[ 13x(x-6)-40(x-6)=0 \] \[ (13x-40)(x-6)=0 \]Therefore,
\[ 13x-40=0 \quad \text{or} \quad x-6=0 \] \[ x=\frac{40}{13} \quad \text{or} \quad x=6 \]Both values satisfy the condition \(x\ne 3,4\).