How do you solve $∣ ∣ ∣ \frac{5}{2 x - 1} ∣ ∣ ∣ \geq ∣ ∣ ∣ \frac{1}{x - 2} ∣ ∣ ∣$ $abs(5 / (2x-1))>= abs(1 / (x - 2))$ ?

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How do you solve $∣ ∣ ∣ \frac{5}{2 x - 1} ∣ ∣ ∣ \geq ∣ ∣ ∣ \frac{1}{x - 2} ∣ ∣ ∣$ $abs(5 / (2x-1))>= abs(1 / (x - 2))$ ?

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Answer 1 · 2018-07-16T19:34:08

$x < \frac{1}{2}$ $x<1/2$ or $\frac{1}{2} < x \leq \frac{11}{7}$ $1/2< x <=11/7$ or $x \geq 3$ $x >=3$

Explanation:

At first observe that $x \neq 2$ $xne 2$ and $x \neq \frac{1}{2}$ $xne 1/2$ by crossmultiplication we get

$5 | x - 2 | \geq | 2 x - 1 |$ $5|x-2|>=|2x-1|$ no we distinguish several cases:

a) $x > 2$ $x>2$ then we get

$5 x - 10 \geq 2 x - 1$ $5x-10>=2x-1$ and we get $x \geq 3$ $x>=3$

b) $\frac{1}{2} < x < 2$ $1/2<x<2$ and we get

$10 - 5 x \geq 2 x - 1$ $10-5x>=2x-1$ and we get $x < \frac{11}{7}$ $x<11/7$

$\frac{1}{2} < x \leq \frac{11}{7}$ $1/2<x<=11/7$

in our last case we have

c) $x < \frac{1}{2}$ $x<1/2$

and we get

$10 - 5 x \geq 1 - 2 x$ $10-5x>=1-2x$

so $x < \frac{1}{2}$ $x<1/2$ and we get the interval like above.

Answer 2 · 2018-07-17T07:04:33

The solution is $x \in (- \infty, \frac{1}{2}) \cup (\frac{1}{2}, \frac{11}{7}] \cup [3, + \infty)$ $x in (-oo,1/2) uu(1/2,11/7]uu[3,+oo)$

Explanation:

You cannot cross multiply when you have an inequality

The inequality is

$∣ ∣ ∣ \frac{5}{2 x - 1} ∣ ∣ ∣ \geq ∣ ∣ ∣ \frac{1}{x - 2} ∣ ∣ ∣$ $|5/(2x-1)|>=|1/(x-2)|$

$\Leftrightarrow$ $<=>$ , $\frac{5}{| 2 x - 1 |} \geq \frac{1}{| (x - 2) |}$ $5/|2x-1|>=1/|(x-2)|$

$\Leftrightarrow$ $<=>$ , $\frac{5}{| 2 x - 1 |} - \frac{1}{| (x - 2) |} \geq 0$ $5/|2x-1|-1/|(x-2)|>=0$

There are $2$ $2$ points to consider

$2 x - 1 = 0$ $2x-1=0$ , $\Rightarrow$ $=>$ , $x = \frac{1}{2}$ $x=1/2$ and

$x - 2 = 0$ $x-2=0$ , $\Rightarrow$ $=>$ , $x = 2$ $x=2$

There are $3$ $3$ intervals to consider

$I_{1} = (- \infty, \frac{1}{2})$ $I_1=(-oo, 1/2)$ and $I_{2} = (\frac{1}{2}, 2)$ $I_2=(1/2, 2)$ and $I_{3} = (2, + \infty)$ $I_3=(2,+oo)$

In the first interval $I_{1}$ $I_1$ , we have

$\frac{5}{- 2 x + 1} - \frac{1}{- x + 2} \geq 0$ $5/(-2x+1)-1/(-x+2)>=0$

$\frac{5 (2 - x) - 1 (1 - 2 x)}{(1 - 2 x) (2 - x)} \geq 0$ $(5(2-x)-1(1-2x))/((1-2x)(2-x))>=0$

$\frac{10 - 5 x - 1 + 2 x}{(1 - 2 x) (2 - x)} \geq 0$ $(10-5x-1+2x)/((1-2x)(2-x))>=0$

$\frac{9 - 3 x}{(1 - 2 x) (2 - x)} \geq 0$ $(9-3x)/((1-2x)(2-x))>=0$

$\frac{3 (3 - x)}{(1 - 2 x) (2 - x)} \geq 0$ $(3(3-x))/((1-2x)(2-x))>=0$

Let $f (x) = \frac{3 (3 - x)}{(1 - 2 x) (2 - x)}$ $f(x)=(3(3-x))/((1-2x)(2-x))$

Solving this equation with a sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a a a a$ $color(white)(aaaaaaa)$ $\frac{1}{2}$ $1/2$ $a a a a a a$ $color(white)(aaaaaa)$ $2$ $2$ $a a a a$ $color(white)(aaaa)$ $3$ $3$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $1 - 2 x$ $1-2x$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $∣ ∣$ $||$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $-$ $-$

$a a a a$ $color(white)(aaaa)$ $2 - x$ $2-x$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ #color(white)(aaaa)+# $a a$ $color(white)(aa)$ $∣ ∣$ $||$ $a$ $color(white)(a)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $-$ $-$

$a a a a$ $color(white)(aaaa)$ $3 - x$ $3-x$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ #color(white)(aaaa)+# $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a$ $color(white)(a)$ $0$ $0$ $a$ $color(white)(a)$ $-$ $-$

$a a a a$ $color(white)(aaaa)$ $f (x)$ $f(x)$ $a a a a a$ $color(white)(aaaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $∣ ∣$ $||$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a$ $color(white)(a)$ $∣ ∣$ $||$ $color(white)(a)$ $+$ $color(white)(a)$ $0$ $color(white)(a)$ $-$

Therefore,

In the interval $I_1$ , $f(x)>=0$ ,when $x in (-oo,1/2)$

In the second interval $I_2=(1/2,2)$

$5/(2x-1)-1/(-x+2)>=0$

$<=>$ , $(5(2-x)-1(2x-1))/((2x-1)(2-x))>=0$

$<=>$ , $(10-5x-2x+1)/((2x-1)(2-x))>=0$

$<=>$ , $(11-7x)/((2x-1)(2-x))>=0$

Let $g(x)=(11-7x)/((2x-1)(2-x))$

Solving this inequality with a sign chart,

$g(x)>=0$ when $x in (1/2,11/7]$

In the third interval $I_2=(2, +oo)$

$5/(2x-1)-1/(x-2)>=0$

$<=>$ , $(5(x-2)-1(2x-1))/((2x-1)(x-2))>=0$

$<=>$ , $((5x-10-2x+1))/((2x-1)(x-2))>=0$

$<=>$ , $((3x-9))/((2x-1)(x-2))>=0$

$<=>$ , $(3(x-3))/((2x-1)(x-2))>=0$

$h(x)=(3(x-3))/((2x-1)(x-2))$

Solving this inequality with a sign chart,

$h(x)>=0$ when $x in [3,+oo)$

graph{5/(|2x-1|)-1/(|x-2|) [-14.24, 14.24, -7.12, 7.12]}

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