trefoniu1

2022-01-24

Solve the following equations and inequalities:
1) $-{x}^{2}+2x>1$
2) $|1-3x|=5$
3) $|4x-3|\le 2$

terorimaox

Expert

a)$-{x}^{2}+2x>1$
$-{x}^{2}+2x-1>0$
$-\left({x}^{2}-2x+1\right)>0$
${x}^{2}-2x+1<0$
$\left(x-1{\right)}^{2}<0$
Square of any number cannot be negative.Thus, there is no solution.

Aiden Cooper

Expert

Consider the part (2):
$|1-3x|=5$
$±\left(1-3x\right)=5$
$1-3x=5$
$-3x=4$
$x=-\frac{4}{3}$
$-\left(1-3x\right)=5$
$-1+3=5$
$3x=6$
$x=2$
Hence, the solution is $x=-\frac{4}{3},2$
Consider the part (3):
$|4-3x|\le 2$
$±\left(4-3x\right)\le 2$
$4-3x\le 2$
$-3x\le -2$
$3x\ge 2$
$x\ge \frac{2}{3}$
$-\left(4-3x\right)\le 2$
$-4+3x\le 2$
$3x\le 6$
$x\le 2$
Thus, the solution is $x=\left[\frac{2}{3},2\right]$

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