‏إظهار الرسائل ذات التسميات chapter-6. إظهار كافة الرسائل
‏إظهار الرسائل ذات التسميات chapter-6. إظهار كافة الرسائل

الثلاثاء، 13 يوليو 2021

Exercise (6.4) - Absolute Value Inequality

  1. Find the solution sets of the following inequalities. Illustrate each of the inequalities on the number line.

    (a) $|x-1|<3$


    $\begin{array}{l}\text{(a)}\ \ \ \ \ \ \ |x-1|\ <3\\\\\ \ \ \ \ \ \therefore \ \ -3<x-1<3\\\\\ \ \ \ \ \ \therefore \ \ -3+1<x-1+1<3+1\\\\\ \ \ \ \ \ \therefore \ \ -2<x<4\\\\\ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \{x\ |\ -2<x<4\}\end{array}$

    (b) $|x+5|>2$


    $\begin{array}{l}\text{(b)}\ \ \ \ \ \ \ \begin{array}{*{20}{l}} {|x+5|\ >2} \end{array}\\\\\ \ \ \ \ \ \therefore \ \ x+5<-2\ \ (\text{or})\ \ x+5>2\\\\\ \ \ \ \ \ \therefore \ \ x+5-5<-2-5\ \ (\text{or})\ \ x+5-5>2-5\\\\\ \ \ \ \ \ \therefore \ \ x<-7\ \ (\text{or})\ \ x>-3\\\\\ \ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \{x\ |\ x<-7\ \ (\text{or})\ \ x>-3\}\end{array}$

    (c) $|x-2| \leq 4$


    $\begin{array}{l}\text{(c)}\ \ \ \ \ \ \ |x-2|\ \le \ 4\\\\\ \ \ \ \ \ \therefore \ \ -4\le x-2\le 4\\\\\ \ \ \ \ \ \therefore \ \ -4+2\le x-2+2\le 4+2\\\\\ \ \ \ \ \ \therefore \ \ -2\le x\le 6\\\\\ \ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \{x\ |\ -2\le x\le 6\}\end{array}$

    (d) $|x+1| \geq 2$


    $\begin{array}{l}\text{(d)}\ \ \ \ \ \ \ |x+1|\ \ge 2\\\\\ \ \ \ \ \ \therefore \ \ x+1\le -2\ \ (\text{or})\ \ x+1\ge 2\\\\\ \ \ \ \ \ \therefore \ \ x+1-15\le -2-1\ \ (\text{or})\ \ x+1-1\ge 2-1\\\\\ \ \ \ \ \ \therefore \ \ x\le -3\ \ (\text{or})\ \ x\ge 1\\\\\ \ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \{x\ |\ x\le -3\ \ (\text{or})\ \ x\ge 1\}\end{array}$

    (e) $|x-3|>0$


    $\begin{array}{l}\text{(e)}\ \ \ \ \ \ \ |x-3|\ >0\\\\\ \ \ \ \ \ \therefore \ \ x-3<0\ \ (\text{or})\ \ x-3>0\\\\\ \ \ \ \ \ \therefore \ \ x-3+3<0+3\ \ (\text{or})\ \ x-3+3>0+3\\\\\ \ \ \ \ \ \therefore \ \ x<3\ \ (\text{or})\ \ x>3\\\\\ \ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \mathbb{R}\backslash \{3\}\end{array}$

    (f) $|x+3| \leq 0$


    $\begin{array}{l}\text{(f)}\ \ \ \ \ \ \ |x+3|\ \le 0\\\\\ \ \ \ \ \ \ \ \ \text{Since}\ |x+3|\ \text{cannot be negative,}\ \\\\\ \ \ \ \ \ \ \ \ x+3=0\\\\\ \ \ \ \ \ \therefore \ \ x=-3\\\\\ \ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \{-3\}\end{array}$


  2. Find the solution sets of the following inequalities.

    (a) $|2 x-1|<4$


    $\begin{array}{l}\text{(a)}\ \ \ \ \ \ |2x-1|<4\\\\\ \ \ \ \ \ \therefore \ \ -4<2x-1<4\\\\\ \ \ \ \ \ \therefore \ \ -4+1<2x-1+1<4+1\\\\\ \ \ \ \ \ \therefore \ \ -3<2x<5\\\\\ \ \ \ \ \ \therefore \ \ -\displaystyle\frac{3}{2}<\displaystyle\frac{{2x}}{2}<\displaystyle\frac{5}{2}\\\\\ \ \ \ \ \ \therefore \ \ -\displaystyle\frac{3}{2}<x<\displaystyle\frac{5}{2}\\\\\ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \left\{ {x\ |\ -\displaystyle\frac{3}{2}<x<\displaystyle\frac{5}{2}} \right\}\end{array}$

    (b) $|3 x+5|>6$


    $\begin{array}{l}\text{(b)}\ \ \ \ \ \ \ |3x+5|>6\\\\\ \ \ \ \ \ \therefore \ \ 3x+5<-6\ \ (\text{or})\ \ 3x+5>6\\\\\ \ \ \ \ \ \therefore \ \ 3x+5-5<-6-5\ \ (\text{or})\ \ 3x+5-5>6-5\\\\\ \ \ \ \ \ \therefore \ \ 3x<-11\ \ (\text{or})\ \ 3x>1\\\\\ \ \ \ \ \ \therefore \ \ \displaystyle\frac{{3x}}{3}<-\displaystyle\frac{{11}}{3}\ \ (\text{or})\ \ \displaystyle\frac{{3x}}{3}>\displaystyle\frac{1}{3}\\\\\ \ \ \ \ \ \therefore \ \ x<-\displaystyle\frac{{11}}{3}\ \ (\text{or})\ \ x>\displaystyle\frac{1}{3}\\\\\ \ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \left\{ {x\ |\ x<-\displaystyle\frac{{11}}{3}\ \ (\text{or})\ \ x>\displaystyle\frac{1}{3}} \right\}\end{array}$

    (c) $|4 x-2| \leq 4$


    $\begin{array}{l}\text{(c)}\ \ \ \ \ \ |4x-2|\ \le 4\\\\\ \ \ \ \ \ \therefore \ \ -4\le 4x-2\le 4\\\\\ \ \ \ \ \ \therefore \ \ -4+2\le 4x-2+2\le 4+2\\\\\ \ \ \ \ \ \therefore \ \ -2\le 4x\le 6\\\\\ \ \ \ \ \ \therefore \ \ -\displaystyle\frac{2}{4}\le \displaystyle\frac{{4x}}{4}\le \displaystyle\frac{6}{4}\\\\\ \ \ \ \ \ \therefore \ \ -\displaystyle\frac{1}{2}\le x\le \displaystyle\frac{3}{2}\\\\\ \ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \left\{ {x\ |\ -\displaystyle\frac{1}{2}\le x\le \displaystyle\frac{3}{2}} \right\}\end{array}$

    (d) $|2 x+1| \geq 2$


    $\begin{array}{l}\text{(d)}\ \ \ \ \ \ |2x+1|\ \ge 2\\\\\ \ \ \ \ \ \therefore \ \ 2x+1\le -2\ \ (\text{or})\ \ 2x+1\ge 2\\\\\ \ \ \ \ \ \therefore \ \ 2x+1-1\le -2-1\ \ (\text{or})\ \ 2x+1-1\ge 2-1\\\\\ \ \ \ \ \ \therefore \ \ 2x\le -3\ \ (\text{or})\ \ 2x\ge 1\\\\\ \ \ \ \ \ \therefore \ \ \displaystyle\frac{{2x}}{2}\le -\displaystyle\frac{3}{2}\ \ (\text{or})\ \ \displaystyle\frac{{2x}}{2}\ge \displaystyle\frac{1}{2}\\\\\ \ \ \ \ \ \therefore \ \ x\le -\displaystyle\frac{3}{2}\ \ (\text{or})\ \ x\ge \displaystyle\frac{1}{2}\\\\\ \ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \left\{ {x\ |\ x\le -\displaystyle\frac{3}{2}\ \ (\text{or})\ \ x\ge \displaystyle\frac{1}{2}} \right\}\end{array}$

    (e) $|2 x-3|>0$


    $ \begin{array}{l}\text{(e)}\ \ \ \ \ \ \ |2x-3|\ >0\\\\\ \ \ \ \ \ \therefore \ \ 2x-3<0\ \ (\text{or})\ \ 2x-3>0\\\\\ \ \ \ \ \ \therefore \ \ 2x-3+3<0+3\ \ (\text{or})\ \ 2x-3+3>0+3\\\\\ \ \ \ \ \ \therefore \ \ 2x<3\ \ (\text{or})\ \ 2x>3\\\\\ \ \ \ \ \ \therefore \ \ \displaystyle\frac{{2x}}{2}<\displaystyle\frac{3}{2}\ \ (\text{or})\ \ \displaystyle\frac{{2x}}{2}>\displaystyle\frac{3}{2}\\\\\ \ \ \ \ \ \therefore \ \ x<\displaystyle\frac{3}{2}\ \ (\text{or})\ \ x>\displaystyle\frac{3}{2}\\\\\ \ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \mathbb{R}\backslash \left\{ {\displaystyle\frac{3}{2}} \right\}\end{array}$

    (f) $|5 x+3| \leq 0$


    $\begin{array}{l}\text{(f)}\ \ \ \ \ \ \ |5x+3|\ \le 0\\\\\ \ \ \ \ \ \ \ \ \text{Since}\ |5x+3|\ \text{cannot be negative,}\ \\\\\ \ \ \ \ \ \ \ \ 5x+3=0\\\\\ \ \ \ \ \ \therefore \ \ x=-\displaystyle\frac{3}{5}\\\\\ \ \ \ \ \ \ \ \ \ \text{Solution Set}\ =\ \left\{ {-\displaystyle\frac{3}{5}} \right\}\end{array}$

Exercise (6.3) - Solutions, Modulus Equations

Keys to determine solutions of the equation $|x-p|=q$


  • When $q<0,|x-p|=q$ has no solution.

  • When $q=0,|x-p|=0$ has only one solution $p$.

  • When $q>0$, the equation $|x-p|=q$ can be seen $x-p=q$ or $x-p=-q \Rightarrow x=p+q$ or $x=p-q$

Exercise (6.3)

  1. Find the solutions of the following equations. Illustrate each of the equations on the number line.

    (a) $|x-5|=3$

    (b) $|x+3|=2$

    (c) $|x-4|=1$


    $\begin{array}{l}(\text{a})\ |x-5|=3\\\\ \ \ \ \ \ \ x-5=-3\ \ \text{or}\ \ x-5=3\\\\ \ \ \ \ \ \ x=2\ \ \text{or}\ \ x=8\end{array}$

    $\begin{array}{l}(\text{b})\ \ |x+3|=2\\\\\ \ \ \ \ \ x+3=-2\ \ \text{or}\ \ x+3=2\\\\\ \ \ \ \ \ x=-5\ \ \text{or}\ \ x=-1\end{array}$

    $\begin{array}{l}(\text{c})\ \ |x-4|=1\\\\\ \ \ \ \ \ x-4=-1\ \ \text{or}\ \ x-4=1\\\\\ \ \ \ \ \ x=3\ \ \text{or}\ \ x=5\end{array}$


  2. Find the solutions of the following equations.

    (a) $|2 x-5|=4$

    (b) $|-2 x-4|=3$

    (c) $|5 x+10|=2$


    $\begin{array}{l}(\text{a})\ \ |2x-5|=4\\\\\ \ \ \ \ \ 2x-5=-4\ \ \text{or}\ \ 2x-5=4\\\\\ \ \ \ \ \ 2x=1\ \ \text{or}\ \ 2x=9\\\\\ \ \ \ \ \ x=\displaystyle\frac{1}{2}\ \ \text{or}\ \ x=\displaystyle\frac{9}{2}\\\\(\text{b})\ \ |-2x-4|=3\\\\\ \ \ \ \ \ -2x-4=-3\ \ \text{or}\ \ -2x-4=3\\\\\ \ \ \ \ \ -2x=1\ \ \text{or}\ \ -2x=7\\\\\ \ \ \ \ \ x=-\displaystyle\frac{1}{2}\ \ \text{or}\ \ x=-\displaystyle\frac{7}{2}\\\\(\text{c})\ \ |5x+10|=2\\\\\ \ \ \ \ \ 5x+10=-2\ \ \text{or}\ \ 5x+10=2\\\\\ \ \ \ \ \ 5x=-12\ \ \text{or}\ \ 5x=8\\\\\ \ \ \ \ \ x=-\displaystyle\frac{{12}}{5}\ \ \text{or}\ \ x=\displaystyle\frac{8}{5}\end{array}$

  3. Solve the following equations.

    (a) $3|x−4|-4=8$

    (b) $2|x−5|+3=9$

    (c) $ \left| {\displaystyle\frac{2}{3}x-4} \right|+11=3$


    $\begin{array}{l}(\text{a})\ \ \ \ 3\left| {x-4} \right|-4=8\\\\\ \ \ \ \ \ \ \ 3\left| {x-4} \right|=12\\\\\ \ \ \ \ \ \ \ \left| {x-4} \right|=4\\\\\ \ \ \ \ \ \ \ x-4=-4\quad \text{or }\quad x-4=4\\\\\ \ \ \ \ \ \ \ x=0\quad \text{or }\quad x=8\\\\(\text{b})\ \ \ \ 2\left| {x-5} \right|+3=9\\\\\ \ \ \ \ \ \ \ 2\left| {x-5} \right|=6\\\\\ \ \ \ \ \ \ \ \left| {x-5} \right|=3\\\\\ \ \ \ \ \ \ \ x-5=-3\quad \text{ or }\quad x-5=3\\\\\ \ \ \ \ \ \ \ x=2\quad \text{or }\quad x=8\\\\(\text{c})\ \ \ \ \left| {\displaystyle\frac{2}{3}x-4} \right|+11=3\\\\\ \ \ \ \ \ \ \ \left| {\displaystyle\frac{2}{3}x-4} \right|=-8<0\\\\\ \ \ \ \ \ \ \ \therefore \ \text{There is no solution.}\end{array}$

  4. Solve the following equations.

    (a) $|5x−1|=|2x+3|$

    (b) $|7x−3|=|3x+7|$

    (c) $|6x−5|=|3x+4|$


    $\begin{array}{l}(\text{a})\ \ \ \ \left| {5x-1} \right|=\left| {2x+3} \right|\\\\\ \ \ \ \ \ \ \ 5x-1=-\ (2x+3)\text{ or }\quad 5x-1=2x+3\\\\\ \ \ \ \ \ \ \ 5x-1=-2x-3\quad \text{ or }\quad 5x-1=2x+3\\\\\ \ \ \ \ \ \ \ 7x=-2\quad \text{or }\quad 3x=4\\\\\ \ \ \ \ \ \ \ x=-\displaystyle\frac{2}{7}\quad \text{or }\quad x=\displaystyle\frac{4}{3}\\\\(\text{b})\ \ \ \ \left| {7x-3} \right|=\left| {3x+7} \right|\\\\\ \ \ \ \ \ \ \ 7x-3=-\ (3x+7)\text{ or }\quad 7x-3=3x+7\\\\\ \ \ \ \ \ \ \ 7x-3=-3x-7\quad \text{ or }\quad 7x-3=3x+7\\\\\ \ \ \ \ \ \ \ 10x=-4\quad \text{or }\quad 4x=10\\\\\ \ \ \ \ \ \ \ x=-\displaystyle\frac{2}{5}\quad \text{or }\quad x=\displaystyle\frac{5}{2}\\\\(\text{c})\ \ \ \ \left| {6x-5} \right|=\left| {3x+4} \right|\\\\\ \ \ \ \ \ \ \ 6x-5=-\ (3x+4)\text{ or }\quad 6x-5=3x+4\\\\\ \ \ \ \ \ \ \ 6x-5=-3x-4\quad \text{ or }\quad 6x-5=3x+4\\\\\ \ \ \ \ \ \ \ 9x=1\quad \text{or }\quad 3x=9\\\\\ \ \ \ \ \ \ \ x=\displaystyle\frac{1}{9}\quad \text{or }\quad x=3\end{array}$

No (3) နှင့် No (4) သည် ပြဌာန်းချက်တွင် မပါဝင်ပါ။ ထပ်တိုးလေ့ကျင့်နိုင်ရန် ပေါင်းထည့်ပေးခြင်း ဖြစ်ပါသည်။

الجمعة، 10 يوليو 2020

Graph of $y=|x-h|+k$ and $y=-|x-h|+k$ : Exercise (6.1) - Solutions


Graph of the Function $y = |x − h| + k$


The graph of the absolute value function$y = |x − h| + k$ can be seen as the translation of $h$-units horizontally and $k$-units vertically of the graph $y = |x|$.

Graph of the Function $y = -|x − h| + k$


The graph of the absolute value function$y = -|x − h| + k$ can be seen as the translation of $h$-units horizontally and $k$-units vertically of the graph $y = -|x|$.


1.           Compare the graphs of the following functions to the graph of $y=|x|$.

             (a)    $y=|x-3|-2$

             (b)    $y=|x+1|+3$

             (c)    $y=|x-2|+3$

Show/Hide Solution



(a) The graph of $y=|x-3|-2$ is the translation of positive 3 units horizontally and negative 2 units vertically of the graph $y=|x| .$


(b) The graph of $y=|x+1|+3$ is the translation of negative 1 unit horizontally and positive 3 units vertically of the graph $y=|x| .$


(c) The graph of $y=|x-2|+3$ is the translation of positive 2 units horizontally and positive 3 units vertically of the graph $y=|x| .$


2.           Compare the graphs of the following functions to the graph of $y=-|x|$.

             (a)    $y=-|x+3|+2$

             (b)    $y=-|x-4|+1$

             (c)    $y=-|x+4|-1$

Show/Hide Solution



(a) The graph of $y=-|x+3|+2$ is the translation of negative 3 units horizontally and positive 2 units vertically of the graph $y=-|x| .$



(b) The graph of $y=-|x-4|+1$ is the translation of positive 4 units horizontally and positive 1 unit vertically of the graph $y=-|x| .$



(c) The graph of $y=y=-|x+4|-1$ is the translation of negative 4 units horizontally and negative 1 unit vertically of the graph $y=-|x| .$