How do you solve $\frac{2 x - 1}{x + 5} \geq 0$ $(2x-1)/(x+5)>=0$ using a sign chart?

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How do you solve $\frac{2 x - 1}{x + 5} \geq 0$ $(2x-1)/(x+5)>=0$ using a sign chart?

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Answer 1 · 2016-12-19T16:01:08

The answer is $x \in] - \infty, - 5 [\cup [\frac{1}{2}, + \infty [$ $x in ] -oo,-5 [ uu [1/2, +oo[$

Explanation:

Let $f (x) = \frac{2 x - 1}{x + 5}$ $f(x)=(2x-1)/(x+5)$

Let's do the sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $a a a a$ $color(white)(aaaa)$ $- 5$ $-5$ $a a a a$ $color(white)(aaaa)$ $a a a a$ $color(white)(aaaa)$ $\frac{1}{2}$ $1/2$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $x + 5$ $x+5$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a$ $color(white)(aa)$ $a a a$ $color(white)(aaa)$ $∥$ $∥$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ #color(white)(aa)+#

$a a a a$ $color(white)(aaaa)$ $2 x - 1$ $2x-1$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a$ $color(white)(aa)$ $a a$ $color(white)(aa)$ $∥$ $∥$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ #color(white)(aa)+#

$a a a a$ $color(white)(aaaa)$ $f (x)$ $f(x)$ $a a a a a a a a$ $color(white)(aaaaaaaa)$ $+$ $+$ $a a$ $color(white)(aa)$ $a a$ $color(white)(aa)$ $∥$ $∥$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ #color(white)(aa)+#

Therefore,

$f (x) \geq 0$ $f(x)>=0$ , when $x \in] - \infty, - 5 [\cup [\frac{1}{2}, + \infty [$ $x in ] -oo,-5 [ uu [1/2, +oo[$

graph{y-(2x-1)/(x+5)=0 [-52, 52.07, -26, 26]}

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How do you solve $\frac{2 x - 1}{x + 5} \geq 0$ $(2x-1)/(x+5)>=0$ using a sign chart?

1 Answer

Explanation:

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How do you solve $\frac{2 x - 1}{x + 5} \geq 0$ $(2x-1)/(x+5)>=0$ using a sign chart?