# Solve the inequality : `1/x + 8 <= 8 - (5x)/43`

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### 1 Answer

You should reduce like terms both sides, hence the positive constant term 8 is found both sides and it needs to be reduced such that:

`1/x =lt -5x/43`

You need to move all terms containing x to the left side, such that:

`1/x + 5x/43 =lt ` 0

You need to bring the fractions to a common denominator, such that:

`(43 + 5x^2)/(43x) =lt` 0

You need to remember that the fraction is negative if the numerator and denominator possess different signs, hence you should discuss two situations:`(43 + 5x^2) =lt 0` and `43x gt 0` or `(43 + 5x^2)gt= 0` and `43xlt 0` .

Considering the first situation yields:

`43x gt 0 =gt xgt0 =gt x in (0,oo)`

`(43 + 5x^2) =lt 0 =gt (43 + 5x^2) = 0 =gt 5x^2 = -43 ` contradiction

Since the equation `43 + 5x^2 = 0` is never negative or zero, hence this situation needs to be disregarded.

Considering the situation `(43 + 5x^2)gt= 0 ` and `43xlt 0` yields:

`43x lt 0 =gt x lt 0 =gt x in (-oo,0)`

The expression (`43 + 5x^2` ) is always positive and never zero, hence the values of x are in `(-oo,0)U(0,oo).`

Notice that the values of x in `(0,oo)` fail to verify the equation 43x<0, hence the only values of x that check both numerator and denominator are in `(-oo,0)` .

**Hence, evaluating the values of x that check the inequality yields x in `(-oo,0)` .**