Blericker74

2022-07-01

Solve, please. Quick if you can
1) $-{x}^{2}+2x>1$
2) $|1-3x|=5$
3) $|4x-3|\le 2$

behk0

Expert

1) $-{x}^{2}+2x>1$
$-{x}^{2}+2x>1$
$-{x}^{2}+2x-1>0$
$-\left({x}^{2}-2x+1\right)>0$
$\left({x}^{2}-2x+1\right)<0$
$\left(x-1{\right)}^{2}<0$

Square of a term is always positive, so there is no $x\in R$, such that (x−1)2<0.
Hence, there is no solution

2) $|1-3x|=5$
If $|x|=a$ and $a>0$ , then $x=a$ or $x=-a$.
So, $|1-3x|=5$ implies $1-3x=5$ or $1-3x=-5$
Then, we have:
$1-3x=5$
$3x=1-5$
$3x=-4$
$x=\frac{-4}{3}$
or $1-3x=-5$
$3x=1+5$
$3x=6$
$x=2$
Therefore, the solution is $x=-\frac{4}{3}$ or $x=2$

ttyme411gl

Expert

(3) Given: $|4x-3|\le 2$

If $|x|\le a$ and $a>0$, then $-a\le x\le a$
$|4x-3|\le 2$ implies $-2\le 4x-3\le 2$

And you get
$-2\le 4x-3\le 2$
$-2\le 4x-3$ and $4x-3\le 2$
$4x\ge -2+3$ and $4x\le 2+3$
$4x\ge 1$ and $4x\le 5$
$x\ge \frac{1}{4}$ and $x\le \frac{5}{4}$

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