College Algebra, Chapter 1, 1.5, Section 1.5, Problem 20

Find all real solutions of the equation $\displaystyle \frac{10}{x} - \frac{12}{x - 3} + 4 = 0$


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\begin{equation}
\begin{aligned}

\frac{10}{x} - \frac{12}{x - 3} + 4 =& 0
&& \text{Given}
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\frac{10}{x} - \frac{12}{x - 3} =& -4
&& \text{Subtract}4
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10(x - 3) - 12(x) =& -4x(x - 3)
&& \text{Multiply the LCD } x(x - 3)
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10x - 30 - 12x =& -4x^2 + 12x
&& \text{Expand}
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4x^2 - 14x - 30 =& 0
&& \text{Combine like terms}
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x^2 - \frac{7}{2} x - \frac{15}{2} =& 0
&& \text{Divide both sides of the equation by } 4
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x^2 - \frac{7}{2} x =& \frac{15}{2}
&& \text{Add } \frac{15}{2}
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x^2 - \frac{7}{2} x + \frac{49}{16} =& \frac{15}{2} + \frac{49}{16}
&& \text{Complete the square: add } \left( \frac{\displaystyle \frac{-7}{2}}{2} \right)^2 = \frac{49}{16}
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\left(x - \frac{7}{4} \right)^2 =& \frac{169}{16}
&& \text{Perfect Square}
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x - \frac{7}{4} =& \pm \sqrt{\frac{169}{16}}
&& \text{Take the square root}
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x =& \frac{7}{4} \pm \frac{13}{4}
&& \text{Add } \frac{7}{4} \text{ and simplify}
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x =& 5 \text{ and } x = \frac{-3}{2}
&& \text{Solve for } x

\end{aligned}
\end{equation}
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