College Algebra, Chapter 1, 1.5, Section 1.5, Problem 24

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


$
\begin{equation}
\begin{aligned}

\frac{x}{2x + 7} - \frac{x + 1}{x + 3} =& 1
&& \text{Given}
\\
\\
x(x + 3) - (x + 1)(2x + 7) =& (2x + 7)(x + 3)
&& \text{Multiply the LCD } (2x + 7)(x + 3)
\\
\\
x^2 + 3x - 2x^2 - 7x - x - 7 =& 2x^2 + 6x + 7x + 21
&& \text{Expand using FOIL method and Distributive Property}
\\
\\
3x^2 + 18x + 30 =& 0
&& \text{Combine like terms}
\\
\\
x^2 + 6x + 10 =& 0
&& \text{Divide both sides by } 3
\\
\\
x^2 + 6x =& -10
&& \text{Subtract } 10
\\
\\
x^2 + 6x + 9 =& -10 + 9
&& \text{Complete the square: add } \left( \frac{6}{2} \right)^2 = 9
\\
\\
(x + 3)^2 =& -1
&& \text{Perfect Square}
\\
\\
x + 3 =& \pm \sqrt{-1}
&& \text{Take the square root}
\\
\\
x =& -3 \pm \sqrt{-1}
&& \text{Subtract } 3
\\
\\
x =& -3 \pm \sqrt{i^2}
&& \text{Recall that } i^2 = -1
\\
\\
x =& -3 \pm i

\end{aligned}
\end{equation}
$



There are no real solutions.