How do you solve $\frac{12}{x + 4} \leq 4$ $12/(x+4)<=4$ ?

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How do you solve $\frac{12}{x + 4} \leq 4$ $12/(x+4)<=4$ ?

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Answer 1 · 2018-06-24T15:53:52

The solution is $x \in (- \infty, - 4) \cup [- 1, + \infty)$ $x in (-oo,-4)uu [-1,+oo)$

Explanation:

You cannot do crossing over.

The inequality is

$\frac{12}{x + 4} \leq 4$ $(12)/(x+4)<=4$

$\Leftrightarrow$ $<=>$ , $\frac{12}{x + 4} - 4 \leq 0$ $(12)/(x+4)-4<=0$

$\Leftrightarrow$ $<=>$ , $\frac{12 - 4 (x + 4)}{x + 4} \leq 0$ $(12-4(x+4))/(x+4)<=0$

$\Leftrightarrow$ $<=>$ , $\frac{12 - 16 - 4 x}{x + 4} \leq 0$ $(12-16-4x)/(x+4)<=0$

$\Leftrightarrow$ $<=>$ , $\frac{- 4 - 4 x}{x + 4} \leq 0$ $(-4-4x)/(x+4)<=0$

$\Leftrightarrow$ $<=>$ , $\frac{4 (1 + x)}{x + 4} \geq 0$ $(4(1+x))/(x+4)>=0$

Let $f (x) = \frac{4 (1 + x)}{x + 4}$ $f(x)=(4(1+x))/(x+4)$

Build a sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- 4$ $-4$ $a a a a$ $color(white)(aaaa)$ $- 1$ $-1$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $x + 4$ $x+4$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a$ $color(white)(aa)$ $∣ ∣$ $||$ $a a$ $color(white)(aa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x + 1$ $x+1$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a$ $color(white)(aa)$ #color(white)(aaa)-# $a a$ $color(white)(aa)$ $0$ $0$ $a a$ $color(white)(aa)$ $+$ $+$

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

Therefore,

$f (x) \geq 0$ $f(x)>=0$ when $x \in (- \infty, - 4) \cup [- 1, + \infty)$ $x in (-oo,-4)uu [-1,+oo)$

Answer 2 · 2018-06-24T17:12:36

$\frac{12}{x + 4} \leq 4$ $12/(x+4)<=4$ for $x < - 4$ $x<-4$ and $x \geq - 1$ $x>=-1$

Explanation:

As the expression is undefined for $x = - 4$ $x=-4$ , we want to stay away from that value.

Before we work on the expression algebraically, let's draw a graph:
graph{-(x+1)/(x+4) [-13.21, 6.79, -5.72, 4.28]}

Based on the graph we can see that the unequality is fulfilled for
$x < - 4$ $x<-4$ and $x \geq - 1$ $x>=-1$

Let us clean up the expression to make it easier to work with:
An equivalent expression is

$\frac{3}{x + 4} \leq 1$ $3/(x+4)<=1$

$\frac{3}{x + 4} - 1 \leq 0$ $3/(x+4)-1<=0$

$\frac{3 - (x + 4)}{x + 4} \leq 0$ $(3-(x+4))/(x+4)<=0$

$- \frac{x + 1}{x + 4} \leq 0$ $-(x+1)/(x+4)<=0$

As this is undefined for $x = - 4$ $x=-4$ , we need to consider two situations: $x > - 4$ $x> -4$ and $x < - 4$ $x<-4$

1) $x > - 4$ $x> -4$ : As $x + 4 > 0$ $x+4>0$ we can multiply both sides with the denominator $x + 4$ $x+4$ and still keep the sign of inequality:

$- \frac{(x + 1) (x + 4)}{x + 4} \leq 0$ $-((x+1)(x+4))/(x+4)<=0$

$- (x + 1) \leq 0$ $-(x+1)<=0$

$x + 1 \geq 0$ $x+1>=0$

$x \geq - 1$ $x>=-1$
Therefore $\frac{12}{x + 4} \leq 4$ $12/(x+4)<=4$ when $x \geq - 1$ $x>=-1$

2) $x < - 4$ $x< -4$ : Now the denominator $(x + 4)$ $(x+4)$ is negative, so if we multiply the unequality with the value of denominator, we have to turn the unequal sign around:

$- \frac{(x + 1) (x + 4)}{x + 4} \geq 0$ $-((x+1)(x+4))/(x+4)>=0$

$- (x + 1) \geq 0$ $-(x+1)>=0$

$x + 1 \leq 0$ $x+1<=0$

$x \leq - 1$ $x<=-1$

As the starting point was that $x < - 4$ $x<-4$ , this means that $\frac{12}{x + 4} \leq 4$ $12/(x+4)<=4$ for all $x < - 4$ $x<-4$

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