Question

​19. Solve the following quadratic equations for x:
​(ii) 𝑥−1/𝑥−2+𝑥−3/𝑥−4=10/3,𝑥≠2,𝑥≠4

Answer 1

$$\begin{gathered} \frac{x-1}{x-2}+\frac{x-3}{x-4}=\frac{10}{3} \\ \Rightarrow \frac{{\displaystyle (x-1)(x-4)+(x-3)(x-2)}}{(x-2)(x-4)}=\frac{10}{3} \\ \Rightarrow \frac{{\displaystyle x^2-5x+4+x^2-5x+6}}{(x-2)(x-4)}=\frac{10}{3} \\ \Rightarrow \frac{{\displaystyle 2x^2-10x+10}}{(x-2)(x-4)}=\frac{10}{3} \\ \Rightarrow \frac{{\displaystyle x^2-5x+5}}{(x-2)(x-4)}=\frac{5}{3} \\ \Rightarrow 3\left(x^2-5x+5\right)=5(x-2)(x-4) \\ \Rightarrow 3x^2-15x+15=5x^2-30x+40 \\ \Rightarrow 2x^2-15x+25=0 \\ \Rightarrow 2x^2-10x-5x+25=0 \\ \Rightarrow 2x(x-5)-5(x-5)=0 \\ x=5,\frac{5}{2} \end{gathered}$$