Question

Question #78b61

Answer 1

`x<3`

Explanation:

`color(blue)((x+3)/x<2`

This is an inequality problem. Solve it by balancing both sides by the same operation

Simplify the left hand side

`rarrx/x+3/x<2`

`rarr1+3/x<2`

Subtract `1` from both sides

`rarr3/x<1`

Divide both sides by `3`

`rarr1/x<1/3`

Multiply both sides by `3x` (Reverse the sides)

`color(green)(rArrx<3`

Answer 2

The solution is `x in (-oo,0) uu (3,+oo)`

Explanation:

We cannot do crossing over.

So,

`(x+3)/x<2`

`(x+3)/x-2<0`

`(x+3-2x)/x<0`

`(3-x)/x<0`

Let,

`f(x)=(3-x)/x`

We build a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaaaa)` `0` `color(white)(aaaaaaaa)` `3` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x` `color(white)(aaaaaaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `3-x` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `||` `color(white)(aaaa)` `+` `color(white)(aaaa)` `-`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaaa)` `+` `color(white)(aaaa)` `-`

Therefore,

`f(x)<0` when `x in (-oo,0) uu (3,+oo)`
graph{(3-x)/x [-16.02, 16.02, -8.01, 8.01]}