Solve the Following Quadratic Equation by Factorization
Question:
\[ \frac{x-1}{x-2}+\frac{x-3}{x-4}=\frac{10}{3}, \qquad x\ne2,4 \]Solution
Given,
\[ \frac{x-1}{x-2}+\frac{x-3}{x-4}=\frac{10}{3} \]Multiplying both sides by \(3(x-2)(x-4)\), we get
\[ 3(x-1)(x-4)+3(x-3)(x-2) = 10(x-2)(x-4) \] \[ 3(x^2-5x+4)+3(x^2-5x+6) = 10(x^2-6x+8) \] \[ 6x^2-30x+30 = 10x^2-60x+80 \] \[ 4x^2-30x+50=0 \] \[ 2x^2-15x+25=0 \]Factorizing:
\[ 2x^2-10x-5x+25=0 \] \[ 2x(x-5)-5(x-5)=0 \] \[ (x-5)(2x-5)=0 \]Therefore,
\[ x-5=0 \quad \text{or} \quad 2x-5=0 \] \[ x=5 \] or \[ x=\frac{5}{2} \]Both values satisfy the condition \(x\ne2,4\).