Question

Solve for $x$:

$2\left(\frac{2x-1}{x+3}\right)-3\left(\frac{x+3}{2x-1}\right)=5$ given that $x\ne -3,x\ne \frac{1}{2}$

Answer 1

Step 1:

Reduce the given equation in the quadratic form:

The given equation can also be written as:

$2\left(\frac{2x-1}{x+3}\right)-3\left(\frac{1}{{\displaystyle \frac{2x-1}{x+3}}}\right)=5$

Substitute $\frac{2x-1}{x+3}=y$ in the above equation.

$$\begin{gathered} 2\left(\frac{2x-1}{x+3}\right)-3\left(\frac{1}{{\displaystyle \frac{2x-1}{x+3}}}\right)=5 \\ \begin{array}{rcl}& \Rightarrow & 2y-\frac{3}{y}=5\\ & \Rightarrow & \frac{2y^2-3}{y}=5\\ & \Rightarrow & 2y^2-3=5y\\ & \Rightarrow & 2y^2-5y-3=0\end{array} \end{gathered}$$

Step 2:

Calculate the value of $x$:

Use the factorization method to solve the obtained quadratic equation.

$$\begin{gathered} 2y^2-5y-3=0 \\ \begin{array}{rcl}& \Rightarrow & 2y^2-6y+1y-3=0\\ & \Rightarrow & 2y\left(y-3\right)+1\left(y-3\right)=0\\ & \Rightarrow & \left(2y+1\right)\left(y-3\right)=0\\ & \Rightarrow & 2y+1=0\mathrm{or}y-3=0\\ & \Rightarrow & y=-\frac{1}{2}\mathrm{or}y=3\end{array} \end{gathered}$$

When $y=-\frac{1}{2}$,

$\begin{array}{rcl}\frac{2x-1}{x+3}& =& -\frac{1}{2}\\ & \Rightarrow & 2\left(2x-1\right)=-1\left(x+3\right)\left(\mathrm{On}\mathrm{cross}\mathrm{multiplication}\right)\\ & \Rightarrow & 4x-2=-x-3\\ & \Rightarrow & 4x+x=-3+2\\ & \Rightarrow & 5x=-1\\ & \Rightarrow & x=-\frac{1}{5}\end{array}$

When $y=3$,

$\begin{array}{rcl}\frac{2x-1}{x+3}& =& 3\\ & \Rightarrow & 2x-1=3\left(x+3\right)\left(\mathrm{On}\mathrm{cross}\mathrm{multiplication}\right)\\ & \Rightarrow & 2x-1=3x+9\\ & \Rightarrow & 2x-3x=9+1\\ & \Rightarrow & -x=10\\ & \Rightarrow & x=-10\end{array}$

Therefore, $x=-\frac{1}{5}$ and $x=-10$ are the required solution.