Question

Solve the following equations for $x$:

$\left(\frac{2x+3}{x+1}\right)+6\left(\frac{x+1}{2x+3}\right)=7$

Answer 1

Step 1:

Express the given equation into the standard quadratic form:

The given equation can also be written as,

$\begin{array}{rcl}\left(\frac{2x+3}{x+1}\right)+6\left(\frac{x+1}{2x+3}\right)& =& 7\\ & \Rightarrow & \left(\frac{2x+3}{x+1}\right)+6\left(\frac{1}{{\displaystyle \frac{2x+3}{x+1}}}\right)=7\end{array}$

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

$\begin{array}{rcl}a+\frac{6}{a}& =& 7\\ & \Rightarrow & a+\frac{6}{a}-7=0\\ & \Rightarrow & a^2-7a+6=0\end{array}$

Step 2:

Solve the given equation:

Use the factorization method to solve the obtained quadratic equation.

$$\begin{gathered} a^2-7a+6=0 \\ \begin{array}{rcl}& \Rightarrow & a^2-6a-1a+6=0\\ & \Rightarrow & a\left(a-6\right)-1\left(a-6\right)=0\\ & \Rightarrow & \left(a-1\right)\left(a-6\right)=0\\ & \Rightarrow & a-1=0\mathrm{or}a-6=0\\ & \Rightarrow & a=1\mathrm{or}a=6\end{array} \end{gathered}$$

Consider the case when $a=1$.

$\begin{array}{rcl}\frac{2x+3}{x+1}& =& 1\\ & \Rightarrow & 2x+3=x+1\\ & \Rightarrow & 2x-x=1-3\\ \therefore x& =& -2\end{array}$

Consider the case when $a=6$.

$\begin{array}{rcl}\frac{2x+3}{x+1}& =& 6\\ & \Rightarrow & 2x+3=6x+6\\ & \Rightarrow & 2x-6x=6-3\\ & \Rightarrow & -4x=3\\ & \Rightarrow & x=\frac{3}{-4}\\ \therefore x& =& -\frac{3}{4}\end{array}$

Hence, the required solution is $x=-2\mathrm{and}x=-\frac{3}{4}$.