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။

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