Solve absolute value inequality :

$\left|3(x-1)\right|+2\le 20$

tripiverded9
2021-12-10
Answered

Solve absolute value inequality :

$\left|3(x-1)\right|+2\le 20$

You can still ask an expert for help

lalilulelo2k3eq

Answered 2021-12-11
Author has **38** answers

Step 1

Given inequality is

$\left|3(x-1)\right|+2\le 20$

Step 2

$\left|3(x-1)\right|+2\le 20$

Subtract 2 from both sides,

$\left|3(x-1)\right|+2-2\le 20-2$

$\left|3(x-1)\right|\le 18$

We know, if$\left|x\right|\le a$ , then $-a\le x\le a\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}x\le a\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}x\ge -a$

Hence, given inequality becomes,

$-18\le 3(x-1)\le 18$

Divide all sides by 3,

$-\frac{18}{3}\le \frac{3(x-1)}{3}\le \frac{18}{3}$

$-6\le x-1\le 6$

Add 1 on all sides,

$-6+1\le x-1+1\le 6+1$

$-5\le x\le 7$

Hence, the solution is$x\le 7\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}x\ge -5$ .

Given inequality is

Step 2

Subtract 2 from both sides,

We know, if

Hence, given inequality becomes,

Divide all sides by 3,

Add 1 on all sides,

Hence, the solution is

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$y+x{y}^{2}-{x}^{3}+2x{z}^{4}=0\phantom{\rule{0ex}{0ex}}-x-{y}^{3}-3{x}^{2}y+3y{z}^{4}=0\phantom{\rule{0ex}{0ex}}-\frac{5}{2}{y}^{2}{z}^{3}-2{x}^{2}{z}^{3}-\frac{{z}^{7}}{2}=0$

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$y+x{y}^{2}-{x}^{3}+2x{z}^{4}=0\phantom{\rule{0ex}{0ex}}-x-{y}^{3}-3{x}^{2}y+3y{z}^{4}=0\phantom{\rule{0ex}{0ex}}-\frac{5}{2}{y}^{2}{z}^{3}-2{x}^{2}{z}^{3}-\frac{{z}^{7}}{2}=0$

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