Skip to main content
Contents Dark Mode Prev Up Next \(\newcommand{\N}{\mathbb N}
\newcommand{\Z}{\mathbb Z}
\newcommand{\Q}{\mathbb Q}
\newcommand{\R}{\mathbb R}
\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\renewcommand{\neg}{\sim}
\newcommand{\brac}[1]{\left( #1 \right)}
\newcommand{\rad}[1]{#1 \ \text{rad}}
\newcommand{\eval}[1]{\left. #1 \right|}
\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}
\newcommand{\ceil}[1]{\left\lceil #1 \right\rceil}
\newcommand{\ang}[1]{#1^\circ}
\newcommand{\crossmethod}[4]{
\begin{tikzpicture}[baseline=(M.base)]
\node (M) at (0,0) {$#1$};
\node (P) at (0,-1) {$#2$};
\node (N) at (1.5,0) {$#3$};
\node (Q) at (1.5,-1) {$#4$};
\draw (M) -- (Q);
\draw (P) -- (N);
\end{tikzpicture}
}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\definecolor{fillinmathshade}{gray}{0.9}
\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
\)
Section 7.6 Solving Rational Equations
Subsection 7.6.1 Examples
Exercise Group 7.6.1 . Equations With \(x\) in the Denominator.
(a)
\(\dfrac{2}{x}+\dfrac{5}{3}=\dfrac{7}{x}\)
(b) (c)
\(\dfrac{x-2}{x}=\dfrac{2-x}{x+1}\)
(d)
\(\dfrac{6}{x}+\dfrac{x}{2}=4\)
(e)
\(\dfrac{2x}{5}=\dfrac{x^2-5x}{5x}\)
Answer .
\(x=-5\) (reject
\(x = 0\) )
(f) (g)
\(\dfrac{1}{2x} - \dfrac{3}{8} = 1 - \dfrac{2}{x}\)
(h)
\(2x=\dfrac{3}{x}-\dfrac{5}{2}\)
(i)
\(\dfrac{5}{3x}-\dfrac{1}{9}=\dfrac{4}{x}\)
(j)
\(\dfrac{1}{2x}+\dfrac{4}{x}=\dfrac{9}{2x}\)
Answer .
all real
\(x\) with
\(x\neq 0\)
(k)
\(\dfrac{x-1}{x}-\dfrac{x+2}{x^2}=\dfrac{x+1}{3x}\)
Answer .
\(x=\dfrac{7 \pm \sqrt{97}}{4}\)
(l)
\(\dfrac{x}{4}-\dfrac{7}{x}=3\)
(m)
\(\dfrac{3-x}{3x}+\dfrac{1}{4}=\dfrac{1}{2x}\)
Exercise Group 7.6.2 . Equations With Linear Expressions in the Denominator.
(a) (b)
\(\dfrac{2x}{x-3}=1-\dfrac{6}{x-3}\)
(c)
\(\dfrac{2}{2x+1}=\dfrac{5}{4-x}\)
(d)
\(\dfrac{10}{x+3}+\dfrac{10}{3}=6\)
(e) (f)
\(\dfrac{2}{x+1}+\dfrac{1}{x+1}=3\)
(g)
\(\dfrac{5}{x-2}=\dfrac{4}{x+3}\)
(h) (i)
\(\dfrac{3}{x}+\dfrac{4}{x+1}=2\)
(j)
\(\dfrac{3}{3x+2}=\dfrac{6}{5x}\)
(k)
\(1-\dfrac{x}{x+3}=\dfrac{3}{2x-5}\)
(l)
\(\dfrac{2}{x+2}=\dfrac{3}{x+6}\)
(m)
\(\dfrac{x+1}{x-2}=\dfrac{x+3}{x-4}\)
(n)
\(\dfrac{x}{x+3}=\dfrac{8}{x+6}\)
(o)
\(\dfrac{2}{x}=\dfrac{x+3}{5}\)
(p)
\(\dfrac{7}{x+3}=\dfrac{5}{x-9}\)
(q)
\(\dfrac{3}{x-1}+\dfrac{1}{2x-2}=\dfrac{7}{4}\)
Exercise Group 7.6.3 . Equations With Linear Denominators 2.
(a)
\(\dfrac{x+3}{x-1}=2x+1\)
(b)
\(\dfrac{6}{x+1}+\dfrac{5}{x-1}=1\)
(c)
\(\dfrac{2x+28}{x+4}+3x = 9\)
(d) (e)
\(\dfrac{x+3}{x-4}=\dfrac{x-1}{x+2}\)
(f)
\(\dfrac{2x}{2x+1}=\dfrac{5}{4-x}\)
(g)
\(\dfrac{x+2}{2x+1}-\dfrac{x}{3}=\dfrac{3}{4x+2}\)
Answer .
\(x=\dfrac{3}{2}\) (reject
\(x = -\dfrac{1}{2}\) )
(h)
\(\dfrac{x}{3x+18} = \dfrac{1}{x}\)
(i)
\(\dfrac{x}{x+2} = \dfrac{9}{x+9}\)
(j)
\(\dfrac{x}{3x-9} = \dfrac{1}{x-3}\)
Answer .
no solution (reject
\(x = 3\) )
(k)
\(\dfrac{1}{x+2}+\dfrac{24}{x+3}=13\)
Answer .
\(x=-\dfrac{27}{13}, -1\)
(l)
\(\dfrac{-2}{x-1}=\dfrac{x-8}{x+1}\)
(m)
\(\dfrac{1}{x-3}=\dfrac{x+2}{7x+14}\)
(n)
\(\dfrac{3}{x}-\dfrac{2}{x-1}=-\dfrac{1}{x}\)
(o)
\(\dfrac{4}{2x-1}=\dfrac{2}{x+3}\)
(p) (q)
\(\dfrac{2x+3}{x+5}+\dfrac{1}{2}=\dfrac{7}{2x+10}\)
(r)
\(\dfrac{x}{2x+2}+\dfrac{2x}{4x+4}=\dfrac{2x-3}{x+1}\)
(s)
\(\dfrac{2}{x}-\dfrac{x}{5x-12}=0\)
Exercise Group 7.6.4 . Equations With Linear Denominators 3.
(a) (b) (c)
\(\dfrac{x}{x-2}=\dfrac{2}{x-2}+2\)
Answer .
no solution (reject
\(x = 2\) )
(d)
\(\dfrac{3x}{x-3}+2 = \dfrac{3x-1}{x+3}\)
Answer .
\(x = -\dfrac{21}{2}, 1\)
(e)
\(\dfrac{3x}{4x+1}-1 = \dfrac{x}{2x-1}\)
Answer .
\(x = \dfrac{-1 \pm \sqrt{7}}{6} \approx 0.27, -0.61\)
(f)
\(\dfrac{2}{x-2} = \dfrac{x + 2}{x}\)
Answer .
\(x = 1 \pm \sqrt{5} \approx -1.24,3.24\)
(g)
\(\dfrac{x^2+3}{x-1}=\dfrac{4}{x-1}\)
Answer .
\(x=-1\) (reject
\(x=1\) )
(h)
\(\dfrac{4x+1}{x+4}+\dfrac{3x-1}{x+1} = 2\)
Answer .
\(x = 1, -\dfrac{11}{5}\)
(i)
\(\dfrac{2x-5}{x+10}=\dfrac{1}{x-6}\)
Answer .
\(x=\dfrac{9 \pm \sqrt{41}}{2}\)
(j)
\(\dfrac{2x}{x-4}=\dfrac{8}{x-4}+1\)
Answer .
no solution (reject
\(x = 4\) )
(k)
\(\dfrac{1}{x}-\dfrac{1}{45}=\dfrac{1}{2x-3}\)
Answer .
\(x=\dfrac{24 \pm 3\sqrt{34}}{2}\)
(l)
\(\dfrac{x}{x+3}-2=\dfrac{-3}{x+3}\)
Answer .
no solution (reject
\(x = -3\) )
(m)
\(\dfrac{2x+3}{3x-1}=\dfrac{x+2}{4}\)
Answer .
\(x = \dfrac{3 \pm \sqrt{177}}{6}\)
(n)
\(\dfrac{1}{x}=\dfrac{2}{x}+1+\dfrac{1}{1-x}\)
Answer .
\(x=\dfrac{1 \pm \sqrt{5}}{2}\)
(o)
\(\dfrac{26}{x+5}=1+\dfrac{3}{x-2}\)
Answer .
\(x=10\pm\sqrt{43} \approx 3.44,16.56\)
(p)
\(\dfrac{x}{2x+2}+\dfrac{2x-16}{4x+4}=\dfrac{2x-3}{x+1}\)
Answer .
no solution (reject
\(x=-1\) )
(q)
\(\dfrac{5}{4x-2}-\dfrac{1}{1-2x}=\dfrac{7}{3x+6}\)
Most of the time, the quadratic will factor into the other linear factors already in the equation. This isn’t true every single time, but you can use it to make educated guesses.
Exercise Group 7.6.5 . Equations With Quadratic Denominators 1.
(a)
\(\dfrac{2x}{x^2-1}=\dfrac{2}{x+1}+\dfrac{1}{1-x}\)
(b)
\(\dfrac{6}{x-3}=\dfrac{x+3}{x^2-9}-5\)
Answer .
\(x=2\) (reject
\(x=-3\) )
(c)
\(\dfrac{x}{x+4}-\dfrac{2-x}{x^2+3x-4}=\dfrac{1}{x-1}\)
(d)
\(\dfrac{x^2+x+2}{x+1}-x=\dfrac{x^2-5}{x^2-1}\)
Answer .
\(x=3\) (reject
\(x=-1\) )
(e)
\(\dfrac{9}{x^2+x-2} + \dfrac{1}{x^2-x} = \dfrac{4}{x-1}\)
Answer .
\(x = -\dfrac{1}{2}\) (reject
\(x = 1\) )
(f)
\(\dfrac{x}{x-3}-\dfrac{6}{x+1}=\dfrac{12}{x^2-2x-3}\)
(g)
\(\dfrac{x}{x-2}=\dfrac{8}{x^2-4}\)
Answer .
\(x=-4\) (reject
\(x=2\) )
(h)
\(\dfrac{2x+3}{x-1}-\dfrac{2}{x+3}=\dfrac{5-6x}{x^2+2x-3}\)
(i)
\(\dfrac{x+2}{x+3}+\dfrac{3}{x^2+3x}=\dfrac{1}{x}\)
Answer .
\(x=-1\) (reject
\(x=0\) )
(j)
\(\dfrac{x+1}{x+3}+\dfrac{x-3}{x-2}=\dfrac{x^2-11x}{x^2+x-6}\)
(k)
\(\dfrac{3}{x+2} + \dfrac{5}{x-3} = \dfrac{3x}{x^2-x-6}\)
(l)
\(\dfrac{x}{x+5} - \dfrac{2}{x-9} = \dfrac{-11x+15}{x^2 - 4x - 45}\)
Answer .
\(x = 5\) (reject
\(x = -5\) )
(m)
\(\dfrac{x}{x-2} + \dfrac{x}{x^2 - 4} = \dfrac{x+3}{x+2}\)
(n)
\(\dfrac{4x-1}{x+2}-\dfrac{x+1}{x-2}=\dfrac{x^2-4x+24}{x^2-4}\)
Answer .
\(x=6\) (reject
\(x=-2\) )
(o)
\(4+\dfrac{2x-1}{x^2-x}=\dfrac{x}{x-1}\)
Answer .
\(x=-\dfrac{1}{3}\) (reject
\(x=1\) )
(p)
\(\dfrac{9}{x-3}-\dfrac{4}{x-6}=\dfrac{18}{x^2-9x+18}\)
Exercise Group 7.6.6 . Equations With Quadratic Denominators 2.
(a)
\(\dfrac{x}{x^2+x-2}+\dfrac{x}{x^2-1}=\dfrac{x}{x^2+3x+2}\)
(b)
\(\dfrac{3-2x}{x+1}-\dfrac{10}{x^2-1}=\dfrac{2x+3}{1-x}\)
Answer .
no solution (reject
\(x=1\) )
(c)
\(\dfrac{x}{x+2}-3=\dfrac{-6}{x^2-4}\)
Answer .
\(x=\dfrac{-1\pm\sqrt{37}}{2} \approx 2.54,-3.54\)
(d)
\(\dfrac{3}{x-3}+\dfrac{5}{x+2}=\dfrac{5x}{x^2-x-6}\)
Answer .
no solution (reject
\(x=3\) )
(e)
\(\dfrac{x+5}{2x+4}=\dfrac{x}{x-3}-\dfrac{2x+9}{x^2-x-6}\)
Answer .
\(x=-1\) (reject
\(x=3\) )
(f)
\(\dfrac{x-5}{2x+10}+\dfrac{8}{25-x^2}=\dfrac{x}{x-5}\)
Answer .
\(x = -10 \pm \sqrt{109} \approx -20.44,0.44\)
(g)
\(\dfrac{1}{(x-1)^2}-3=\dfrac{2}{1-x}\)
(h)
\(\dfrac{4x}{x^2-9} - \dfrac{5}{x+3}=2\)
Answer .
\(x = \dfrac{-1 \pm \sqrt{265}}{4}\)
(i)
\(\dfrac{2}{x^2-4}+\dfrac{10}{6x+12}=\dfrac{1}{x-2}\)
(j)
\(\dfrac{3x}{x+2}-\dfrac{5}{x-3}=\dfrac{-25}{x^2-x-6}\)
Answer .
\(x=\dfrac{5}{3}\) (reject
\(x=3\) )
(k)
\(\dfrac{1}{x-4}-\dfrac{1}{x-2}=\dfrac{2x}{x^2-6x+8}\)
(l)
\(\dfrac{3x}{x+2}-\dfrac{9}{x-3}=-\dfrac{25}{x^2-x-6}\)
Answer .
\(x=\dfrac{9 \pm 2\sqrt{15}}{3}\)
(m)
\(\dfrac{x}{x-5}-\dfrac{3}{x+1}=\dfrac{30}{x^2-4x-5}\)
Answer .
\(x=-3\) (reject
\(x=5\) )
(n)
\(\dfrac{5}{x-7}-\dfrac{1}{2x}=\dfrac{9x+7}{2x^2-14x}\)
Answer .
all real numbers
\(x\text{,}\) with
\(x \neq 0,7\)
(o)
\(\dfrac{2}{x+5}+\dfrac{20}{x^2-25}=\dfrac{-3}{5-x}\)
Answer .
no solution (reject
\(x=-5\) )
(p)
\(\dfrac{x^2}{x^2-x-2}=\dfrac{2x}{x^2+x-6}\)
(q)
\(\dfrac{3}{x^2+x-2}-\dfrac{1}{x^2-1}=\dfrac{7}{2(x^2+3x+2)}\)
Exercise Group 7.6.7 . Advanced Examples.
(a)
\(\dfrac{x}{x-3}+\dfrac{1}{x-2}-\dfrac{1}{x+2}=\dfrac{x-12}{x^3-3x^2-4x+12}\)
(b)
\(\dfrac{3x-7}{x^2-5x+6}+\dfrac{2x+8}{9-x^2}-\dfrac{x+2}{x^2+x-6}=0\)
