Question

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

Answer 1

The answer is `x in ]-7.77, -5[uu ]0.77,3[`

Explanation:

We cannot do crossing over

We rewrite the equation

`(x)/(3-x)>(2)/(x+5)`

`(x)/(3-x)-(2)/(x+5)>0`

`(x(x+5)-2(3-x))/((3-x)(x+5))>0`

`(x^2+5x-6+2x)/((3-x)(x+5))=(x^2+7x-6)/((3-x)(x+5))`

Let `f(x)=(x^2+7x-6)/((3-x)(x+5))`

The roots of `x^2+7x-6` are

`x=(-7+-sqrt(49+4*6))/2`

`x=(-7+-sqrt73)/2`

`x_1=(-7+8.54)/2=0.77`

`x_2=(-7-8.54)/2=-7.77`

Now, we can build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaaa)` `-oo` `color(white)(aaaa)` `-7.77` `color(white)(aaaa)` `-5` `color(white)(aaaa)` `0.77` `color(white)(aaaa)` `3` `color(white)(aaaaa)` `+oo`

`color(white)(aaaa)` `x+7.77` `color(white)(aaaaa)` `-` `color(white)(aaaaaa)` `+` `color(white)(aa)` `||` `color(white)(aa)` `+` `color(white)(aaa)` `+` `color(white)(aa)` #`||` #`color(white)(aaa)` `+`

`color(white)(aaaa)` `x+5` `color(white)(aaaaaaaa)` `-` `color(white)(aaaaaa)` `-` `color(white)(aa)` `||` `color(white)(aa)` `+` `color(white)(aaa)` `+` `color(white)(aa)` #`||` #`color(white)(aaa)` `+`

`color(white)(aaaa)` `x-0.77` `color(white)(aaaaaa)` `-` `color(white)(aaaaaa)` `-` `color(white)(aa)` `||` `color(white)(aa)` `-` `color(white)(aaa)` `+` `color(white)(aa)` #`||` #`color(white)(aaa)` `+`

`color(white)(aaaa)` `3-x` `color(white)(aaaaaaaaa)` `+` `color(white)(aaaaaa)` `+` `color(white)(aa)` `||` `color(white)(aa)` `+` `color(white)(aaa)` `+` `color(white)(aa)` #`||` #`color(white)(aaa)` `-`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaaaa)` `-` `color(white)(aaaaaa)` `+` `color(white)(aa)` `||` `color(white)(aa)` `-` `color(white)(aaa)` `+` `color(white)(aa)` #`||` #`color(white)(aaa)` `-`

Therefore,

`f(x)>0` when `x in ]-7.77, -5[uu ]0.77,3[`