Solving Equations

Solve the equations:

(a) \(\dfrac{3}{x-2}=\dfrac{1}{x-1}+\dfrac{7}{( x-1) ( x-2) }\)

(b) \(x^{3}-x^{2}-4x+4=0\)

(c) \(\sqrt{x-1}=x-7\)

(d) \(\vert 1-x\vert =2\)

A-4

The set of real numbers that a variable can assume is called the domain of the variable.

Solution (a) First, notice that the domain of the variable is \( \{ x~|~x\neq 1,\,x\neq 2\} \). Now clear the equation of rational expressions by multiplying both sides by \(( x-1) ( x-2) \). \[ \begin{array}{@{\hspace*{-7.1pc}}rcl@{\qquad}ll} \dfrac{3}{x-2}&=&\dfrac{1}{x-1}+\dfrac{7}{( x-1) ( x-2) }& \\ ( x-1) ( x-2) \dfrac{3}{x-2} &=&( x-1) ( x-2) \left[ \dfrac{1}{x-1}+\dfrac{7}{( x-1) ( x-2) }\right] & {\color{#0066A7}{\hbox{Multiply both sides by} }}\\ &&&{\color{#0066A7}{\hbox{\(( {x-1}) ( { x-2}).\)}}} \\ 3x-3&=&( x-1) ( x-2) \dfrac{1}{x-1} + ( x-1) ( x-2) \dfrac{7}{( x-1) ( x-2) } & {\color{#0066A7}{\hbox{Distribute on both sides.}}} \\ 3x-3&=&( x-2) +7 & {\color{#0066A7}{\hbox{Simplify.}}} \\ 3x-3&=&x+5 & \\ 2x&=&8\hbox{ } & \\ x&=&4 & \end{array} \] Since 4 is in the domain of the variable, the solution is \(4\).

(b) We group the terms of \(x^{3}-x^{2}-4x+4=0\), and factor by grouping. \[ \begin{array}{rclll} x^{3}-x^{2}-4x+4&=&0 & \\ [4pt] ( x^{3}-x^{2}) -( 4x-4) &=&0 & {\color{#0066A7}{\hbox{Group the terms.}}} \\[4pt] x^{2}( x-1) -4( x-1) &=&0 & {\color{#0066A7}{\hbox{Factor out the common factor}}} \\[-2pt] &&&{\color{#0066A7}{\hbox{from each group.}}}\\[4pt] ( x^{2}-4) ( x-1) &=&0 & {\color{#0066A7}{\hbox{Factor out the common factor \((x-1).\)}}} \\[4pt] ( x-2) ( x+2) ( x-1) &=&0 & {\color{#0066A7}{\hbox{\({x}^{2}-4=( {x-2}) ( {x+2})\)}}} \\[4pt] x-2=0\hbox{ or }x+2&=&0\hbox{ or }x-1=0 & {\color{#0066A7}{\hbox{Set each factor equal to 0.}}} \\[3pt] x\hbox{ }\,=\,\hbox{ }2\hbox{ }x&=&-2\hbox{ }x\hbox{ }\,=\,\hbox{ }1 & {\color{#0066A7}{\hbox{Solve.}}} \end{array} \] The solutions are \(-2\), 1, and 2.

Squaring both sides of an equation may lead to extraneous solutions. Check all apparent solutions.

(c) We square both sides of the equation since the index of a square root is \(2\). \[ \begin{array}{@{{-8pt}}rclll} \hbox{ }\sqrt{x-1}&=&x-7 & \\[4.5pt] ( \sqrt{x-1}) ^{2}&=& ( x-7) ^{2} & {\color{#0066A7}{\hbox{Square both sides.}}} \\[4.5pt] \hbox{}x-1&=&x^{2}-14x+49\hbox{} & \\[4.5pt] x^{2}-15x+50&=&0 & {\color{#0066A7}{\hbox{Put in standard form.}}} \\[4.5pt] ( x-10) ( x-5) &=&0 & {\color{#0066A7}{\hbox{Factor. }}} \\[4.5pt] \hbox{ }x=10\hbox{ or }x&=&5 & {\color{#0066A7}{\hbox{Set each factor equal to 0 and solve.}}} \end{array} \]

Check: \(x=10\): \(\sqrt{x-1}=\sqrt{10-1}=\sqrt{9}=3\) and \( x-7=10-7=3\)
\(x=5\): \(\sqrt{x-1}=\sqrt{5-1}=\sqrt{4}=2\) and \( x-7=5-7=-2\)

The apparent solution \(5\) is extraneous; the only solution of the equation is \(10\).

\(\vert a\vert =a\) if \(a\geq 0\); \(\vert a\vert =-a\) if \(a<0\). If\( \vert x \vert =b\), \(b\geq 0\), then \(x=b\) or \(x=-b\).

(d) \begin{array}[t]{@{{-.8pc}}rclrcl@{\quad}l} \vert 1-x \vert &=&2 & & & \\[5pt] \hbox{ }1-x&=&2 & \hbox{or}\quad 1-x&=&-2 & {\color{#0066A7}{\hbox{The expression inside the absolute }}} \\[-2pt] &&&&&&{\color{#0066A7}{\hbox{value bars equals 2 or \(-2\).}}} \\[5pt] -x&=&1 & -x&=&-3 & {\color{#0066A7}{\hbox{Simplify.}}} \\[5pt] x&=&-1 & x&=&3 & {\color{#0066A7}{\hbox{Simplify.}}} \end{array} The solutions are \(-1\) and \(3\).