Question

Solve the given equation:

$\frac{x-4}{x-5}+\frac{x-6}{x-7}=\frac{10}{3}$

Answer 1

We have,

$\frac{x-4}{x-5}+\frac{x-6}{x-7}=\frac{10}{3}$

On combining and solving we get

$$\begin{gathered} \frac{x-4}{x-5}+\frac{x-6}{x-7}=\frac{10}{3} \\ \Rightarrow \frac{(x-7)(x-4)+(x-5)(x-6)}{(x-5)(x-7)}=\frac{10}{3} \\ \Rightarrow \frac{x^2-4x-7x+28+x^2-6x-5x+30}{x^2-7x-5x+35}=\frac{10}{3} \\ \Rightarrow \frac{2x^2-22x+58}{x^2-12x+35}=\frac{10}{3} \\ \Rightarrow 3(2x^2-22x+58)=10(x^2-12x+35) \\ \Rightarrow 6x^2-66x+174=10x^2-120x+350 \\ \Rightarrow 10x^2-120x+350-6x^2+66x-174=0 \\ \Rightarrow 4x^2-54x+176=0 \\ \Rightarrow 2x^2-27x+88=0 \\ \Rightarrow 2x^2-16x-11x+88=0 \\ \Rightarrow 2x(x-8)-11(x-8)=0 \\ \Rightarrow (x-8)(2x-11)=0 \end{gathered}$$

Either,

$$\begin{gathered} x-8=0\text{or}2x-11=0 \\ \Rightarrow x=8\text{or}2x=11 \\ \Rightarrow x=8\text{or}x=\frac{11}{2} \end{gathered}$$

Hence, $\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{8}\mathbf{}\mathbf{\text{or}}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{11}}{\mathbf{2}}$are the solution of given equation.