Question

Please solve [Quadratic equation]​

Answer 1

Let me take $\displaystyle \frac{2x-3}{x-1}$ as $k$.

Thus, $\displaystyle \frac{x-1}{2x-3}$ becomes $\displaystyle \frac{1}{k}$.

So,

$\displaystyle \begin{aligned}&\frac{2x-3}{x-1}-4\left(\frac{x-1}{2x-3}\right)=3\\ \\ \Longrightarrow\ \ &k-4\left(\frac{1}{k}\right)=3\\ \\ \Longrightarrow\ \ &k-\frac{4}{k}=3\\ \\ \Longrightarrow\ \ &\frac{k^2-4}{k}=3\\ \\ \Longrightarrow\ \ &k^2-4=3k\end{aligned}$

$\displaystyle \begin{aligned}\\ \\ \Longrightarrow\ \ &k^2-3k-4=0\\ \\ \Longrightarrow\ \ &k^2+k-4k-4=0\\ \\ \Longrightarrow\ \ &k(k+1)-4(k+1)=0\\ \\ \Longrightarrow\ \ &(k+1)(k-4)=0\\ \\ \\ \therefore\ \ &k=-1\ \ \ ; \ \ \ k=4\\ \\ \Longrightarrow\ \ &\frac{2x-3}{x-1}=-1\ \ \ ; \ \ \ \frac{2x-3}{x-1}=4\end{aligned}$

$\displaystyle \begin{aligned}&\text{Taking \displaystyle \frac{2x-3}{x-1}=-1 ,}\\ \\ \Longrightarrow\ \ &2x-3=-1(x-1)\\ \\ \Longrightarrow\ \ &2x-3=1-x\\ \\ \Longrightarrow\ \ &2x+x=1+3\\ \\ \Longrightarrow\ \ &3x=4\\ \\ \Longrightarrow\ \ &x=\large \text{ \bold{\frac{4}{3}} }\end{aligned}$

$\displaystyle \begin{aligned}&\text{Taking \displaystyle \frac{2x-3}{x-1}=4 ,}\\ \\ \Longrightarrow\ \ &2x-3=4(x-1)\\ \\ \Longrightarrow\ \ &2x-3=4x-4\\ \\ \Longrightarrow\ \ &4-3=4x-2x\\ \\ \Longrightarrow\ \ &1=2x\\ \\ \Longrightarrow\ \ &x=\large \text{ \bold{\frac{1}{2}} }\end{aligned}$

Hence the values of x are 4/3 and 1/2.