28. $\frac{6x-5}{4x+1}<0$

Key takeaway: This inequality compares the signs of a linear numerator and denominator. The expression is negative when the two parts have opposite signs.

Step 1: Find the critical points

Set numerator and denominator equal to zero:

• $6x-5=0 \Rightarrow x=\frac{5}{6}$
• $4x+1=0 \Rightarrow x=-\frac{1}{4}$

These points divide the number line into three intervals.

Step 2: Test the sign on each interval

• For $x<-\frac14$: numerator is negative, denominator is negative, so the quotient is positive.
• For $-\frac14<x<\frac56$: numerator is negative, denominator is positive, so the quotient is negative.
• For $x>\frac56$: numerator is positive, denominator is positive, so the quotient is positive.

Step 3: Apply the inequality

We need the quotient to be strictly less than zero, so take the interval where the expression is negative:

$\left(-\frac14,\frac56\right)$

Final answer

$\left(-\frac14,\frac56\right)$

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Pitfalls the pros know 👇 Do not include the endpoints. At $x=-\frac14$, the expression is undefined, and at $x=\frac56$, the expression equals zero, which does not satisfy $<0$.

What if the problem changes? If the sign were $\le 0$, then the interval would still be $\left(-\frac14,\frac56\right)$, but you would also consider whether any zero of the numerator is allowed. Here $x=\frac56$ gives zero and still would not be included for $<0$ but would be included for $\le 0$.

Tags: rational inequality, sign chart, endpoints