# Solving inequalities with “x” in the denominator

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.

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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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Aha and Eureka! You've just helped me tremendously. Understading how each case holds true only when BOTH inequalities are true was the clarifying point. Thank you! – Bradford Fisher Jul 18 '11 at 23:36

Here's one way. $1/x\lt4$, multiply both sides by $x^2$, $x\lt4x^2$, $4x^2-x\gt0$, $x(4x-1)\gt0$, so either $x\gt0$ and $4x-1\gt0$, or $x\lt0$ and $4x-1\lt0$, etc.

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Breaking into cases is unnecessary and unwieldy. Consider the inequality: $x<\frac{1}{x}$. We subtract $x$ from both sides: $0< \frac{1}{x} -x$. Now we find a common denominator on the right:$0 < \frac{1-x^2}{x}$. Factor the numerator: $0<\frac{(1-x)(1+x)}{x}.$ The expression may change signs at $x=-1,0,$ or at $x=1$. The numerator represents a parabola that spills water. The totality of signs goes like this:

$$\begin{array}{ccccccc} + & u & - & 0 & + & u& - \\ x<-1 & x=-1 & -1<x<0 & 0 & 0<x<1 & x=1 & 1<x \end{array}$$ In general, you make the expression comparable to $0$, factor numerator and denominator, and examine how the expression changes at the intercepts (numerator =0) and asymptotes (denominator =0) See youtube.com/ProfessorElvisZap for more details. Specifically, here .

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Never, Never, Never multiply through by a variable quantity, unless you are sure that the quantity is positive. For the inequality, $4<1/x$, we get $0<1/x - 4$. So $0<(1-4x)/x$. The expression may change signs at $0$ and at $1/4$. To the left of $0$ the expression is negative. Between $0$ and $1/4$, the expr. is positive. To the right of $1/4$ it is postive. The solution is $0<x<1/4$. – Scott Carter Jul 18 '11 at 23:37
Note, however, that the original inequality was $4\gt1/x$, not $4\lt1/x$. – Gerry Myerson Jul 19 '11 at 4:03
I don't believe in greater than. So it's true that I solved the wrong inequality, but my technique solves both inequalities. With $1/x<4$, you get $0<4-1/x$, and $0<(4x-1)/x$. Now the expression might change signs at $x=0$ and at $x=1/4$. Which it does. The expression $(4x-1)/x$ is positive to the outside of these values. Then $x<0$ or $1/4 < x$. The reason, I don't believe in greater than is that the sentence structure mimics the number line. – Scott Carter Jul 19 '11 at 18:59

It is generally a good idea to multiply both sides by a positive number. In this case solving $\frac{1}{x} \gt 4$, we multiply by $x^2$ which is always positive. thus we get $$4x^2-x \lt 0$$ upon rearranging. the solution is thus $$0\lt x \lt \frac{1}{4}.$$

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Note, however, that the original problem was $1/x\lt4$. Also, I suspect that anyone having difficulty with this sort of problem is going to need some intermediate steps to get from $4x^2-x\lt0$ to $0\lt x\lt1/4$. – Gerry Myerson Jul 19 '11 at 4:00
I know. The intermediate steps are very well handled by the other answers. – Nana Jul 19 '11 at 4:05

Concerning the first example: If $1/x < 0$, then $x < 0$. Now, multiplying by a negative number changes the direction of the inequality sign, and hence $x(1/x) > x (0)$ ($1 > 0$).

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See my answer to this question for a description of a general method to solve inequalities between rational functions. It also describes the relationship between multiplying through by a least common denominator vs. the square of such.

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