How do you solve $3/(x-2)<5/(x+2)$ using a sign chart?

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How do you solve $3/(x-2)<5/(x+2)$ using a sign chart?

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Answer 1 · 2017-03-06T10:50:36

The solution is $x in ]-2,2 [ uu ]8, +oo [$

Explanation:

We cannot do crossing over

Let's simplify the inequality

$3/(x-2)<5/(x+2)$

$3/(x-2)-5/(x+2)<0$

$(3(x+2)-5(x-2))/((x-2)(x+2))<0$

$(3x+6-5x+10)/((x-2)(x+2))<0$

$(16-2x)/((x-2)(x+2))<0$

$(2(8-x))/((x-2)(x+2))<0$

Let $f(x)=(2(8-x))/((x-2)(x+2))$

We, now, build the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-2$ $color(white)(aaaaaaaaa)$ $2$ $color(white)(aaaaaa)$ $8$ $color(white)(aaaaaaa)$ $+oo$

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

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

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

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

Therefore,

$f(x)<0$ when $x in ]-2,2 [ uu ]8, +oo [$

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How do you solve $3/(x-2)<5/(x+2)$ using a sign chart?

1 Answer

Explanation:

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