How do you solve $\frac{x + 68}{x + 8} \geq 5$ $(x+68)/(x+8)>=5$ ?

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How do you solve $\frac{x + 68}{x + 8} \geq 5$ $(x+68)/(x+8)>=5$ ?

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Answer 1 · 2017-05-07T14:21:46

$x \in (- 8, 7]$ $x in (-8, 7]$

Explanation:

Given:

$\frac{x + 68}{x + 8} \geq 5$ $(x+68)/(x+8) >= 5$

Subtract $\frac{x + 68}{x + 8}$ $(x+68)/(x+8)$ from both sides to get:

$0 \geq 5 - \frac{x + 68}{x + 8} = \frac{5 (x + 8) - (x + 68)}{x + 8} = \frac{4 x - 28}{x + 8} = \frac{4 (x - 7)}{x + 8}$ $0 >= 5-(x+68)/(x+8) = (5(x+8)-(x+68))/(x+8) = (4x-28)/(x+8) = (4(x-7))/(x+8)$

Note that the right hand side is a function which is continuous and non-zero except at $x = - 8$ $x=-8$ and $x = 7$ $x=7$ . The function changes sign at each of these two points.

When $x = - 8$ $x=-8$ the denominator is $0$ $0$ so the right hand side is undefined. So $- 8$ $-8$ is not part of the solution set.

When $x = 7$ $x=7$ the numerator is $0$ $0$ and the inequality is satisfied. So $7$ $7$ is part of the solution set.

When $x < - 8$ $x < -8$ or $x > 7$ $x > 7$ then the signs of the numerator and denominator are the same, so the quotient is positive and the inequality is not satisfied.

When $x \in (- 8, 7)$ $x in (-8, 7)$ the numerator is negative, the denominator is positive and the quotient is negative. So the inequality is satisfied.

So the solution set is:

$x \in (- 8, 7]$ $x in (-8, 7]$

Answer 2 · 2017-05-07T14:26:16

The solution is $x \in (- 8, 7]$ $x in (-8,7]$

Explanation:

We cannot do crossing over.

So, we simplify the inequality

$\frac{x + 68}{x + 8} \geq 5$ $(x+68)/(x+8)>=5$

$\frac{x + 68}{x + 8} - 5 \geq 0$ $(x+68)/(x+8)-5>=0$

$\frac{(x + 68) - 5 (x + 8)}{x + 8} \geq 0$ $((x+68)-5(x+8))/(x+8)>=0$

$\frac{x + 68 - 5 x - 40}{x + 8} \geq 0$ $(x+68-5x-40)/(x+8)>=0$

$\frac{28 - 4 x}{x + 8} \geq 0$ $(28-4x)/(x+8)>=0$

$\frac{4 (7 - x)}{x + 8} \geq 0$ $(4(7-x))/(x+8)>=0$

Let $f (x) = \frac{4 (7 - x)}{x + 8}$ $f(x)=(4(7-x))/(x+8)$

We can build the sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- 8$ $-8$ $a a a a a a a a$ $color(white)(aaaaaaaa)$ $7$ $7$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $x + 8$ $x+8$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $∣ ∣$ $||$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $7 - x$ $7-x$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $∣ ∣$ $||$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $-$ $-$

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

Therefore,

$f (x) \geq 0$ $f(x)>=0$ when $x \in (- 8, 7]$ $x in (-8,7]$

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How do you solve $\frac{x + 68}{x + 8} \geq 5$ $(x+68)/(x+8)>=5$ ?

2 Answers

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How do you solve (x+68)/(x+8)>=5x+68x+8≥5(x+68)/(x+8)>=5?

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How do you solve $\frac{x + 68}{x + 8} \geq 5$ $(x+68)/(x+8)>=5$ ?