Chapter 12.5, Problem 21WE

To determine

Solve the quadratic equation.

Answer to Problem 21WE

  $x\in \left\{3,\frac{2}{3}\right\}$

Explanation of Solution

Given:

The equation is :

  $\frac{4x-3}{2x-3}=\frac{5x-3}{x+1}$

Calculation:

The given equation $\frac{4x-3}{2x-3}=\frac{5x-3}{x+1}$ .

So, Cross multiply :

  $\begin{array}{l}(4x-3)(x+1)=\left(2x-3\right)\left(5x-3\right)\\ \Rightarrow 4x^2+4x-3x-3=10x^2-6x-15x+9\\ \Rightarrow 4x^2+x-3=10x^2-21x+9\\ \Rightarrow 6x^2-22x+12=0\\ \Rightarrow 3x^2-11x+6=\mathrm{0...............................}[\text{divide each side by 2]}\\ \Rightarrow 3x^2-9x-2x+6=\mathrm{0........................}[\text{factor the middle term]}\\ \Rightarrow 3x(x-3)-2(x-3)=0\\ \Rightarrow (3x-2)(x-3)=0\\ \Rightarrow 3x-2=0\text{ or }x-3=\mathrm{0..................}[\text{zero product property]}\\ \Rightarrow x=\frac{2}{3}\text{ or }x=3\end{array}$

So, $x\in \left\{3,\frac{2}{3}\right\}$