Question

Find the real solutions of the equation $\frac{x}{2x+7}-\frac{x+1}{x+3}=1$.

Short answer

The given equation has two real solutions $x=-\frac{7}{3},-4$.

Step-by-step solution

Step 1. By method of quadratic formula.

The roots of the quadratic equation of the form $a{x}^2+bx+c=0$where $a\ne 0$can be evaluated by using the quadratic formula shown below.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Step 2. Find roots of quadratic equation by quadratic formula.

The given equation is$\frac{x}{2x+7}-\frac{x+1}{x+3}=1$which can be written as,

localid="1643835978901" $$\begin{gathered} \frac{x}{2x+7}-\frac{x+1}{x+3}=1....\left(\text{Given}\right) \\ \text{or,\hspace{0.17em}}x\left(x+3\right)-\left(x+1\right)\left(2x+7\right)=\left(x+3\right)\left(2x+7\right)\mathrm{....}\left(\text{Simplify}\right) \\ -x^2-6x-7=2x^2+13x+21 \\ \text{or,}3x^2+19x+28=0....\left(1\right) \end{gathered}$$

Evaluate roots of (1) by quadratic formula as shown below.

$$\begin{gathered} x=\frac{-\left(19\right)\pm \sqrt{{\left(19\right)}^2-4\left(3\right)\left(28\right)}}{2\left(3\right)} \\ =\frac{-19\pm \sqrt{25}}{6} \\ =\frac{-19\pm 5}{6} \\ =-\frac{7}{3},-4 \end{gathered}$$

Step 3. DESCRIPTION OF STEPS.

The given equation has two distinct real solutions $x=-\frac{7}{3},-4$.