[Math] Solving inequalities with “x” in the denominator

algebra-precalculusinequality

Solving inequalities with "x" in the denominator has always been a stumbling block for me. Other than understanding how a particular expression, such as 1/x, works (in this case, x cannot be zero), how might I go about solving inequalities having such expressions mathematically.

For instance:

1/x < 0
x(1/x) < x(0)
1 < 0 // This is where things seem to break down.
      // Probably because of the vertical asymptote.

As a slightly longer example:

1/x < 4
1/x - 4 < 0
x(1/x - 4) < x(0)
1 - 4x < 0
-4x < -1
x > 1/4  // Makes sense up to this point (Only covers x > 0)
         // How might I now solve for a negative x value?

I attempted changing the LessThan operator to a GreaterThan operator as per the rule:

if a < b and c < 0, then ac > bc

yet this resulted in an incorrect answer.

I already know that the answer to the inequality is (-infinity, 0) union (1/4, infinity). I would just like to know how to solve this algebraically.

Best Answer

Solving for when $\frac{1}{x}<4$.

First, we do not allow $x=0$. Let's split into cases.

Case 1: $x>0$. Then $\frac{1}{x}<4$ becomes $1< 4x$ or $\frac{1}{4}<x$. This means that whenever both $x>0$ and $x>\frac{1}{4}$ the inequality holds. Combining these inequalities, we get that it holds for $x>\frac{1}{4}$. (I know it does nothing here, but sometimes this is important)

Case 2: $x<0$. Then $\frac{1}{x}<4$ becomes $1> 4x$ or $\frac{1}{4}>x$. This means that whenever both $x<0$ and $x<\frac{1}{4}$ the inequality holds. Combining these inequalities, we get that it holds for $x<0$.

Now, combining both cases we get that it holds for $x<0$ or $x>\frac{1}{4}$.

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