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

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.

(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 $10$.

(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 $3$.