Inequalities

The four symbols. For real numbers $a$ and $b$, $a<b$ means $a$ is strictly less than $b$, $a>b$ means $a$ is strictly greater than $b$, while $a \le b$ and $a \ge b$ allow equality as well as the strict case. The strict symbols ($<,>$) exclude the boundary value; the slack symbols ($\le,\ge$) include it — a distinction that matters the moment you mark an endpoint on a number line or write an interval.

Linear inequality in one variable. This chapter's central object has the form

where: $x$ is the unknown, and $a,b,c$ are real constants with $a \neq 0$. Its solution set is every value of $x$ that makes the statement true. Unlike $ax+b=c$, which usually has exactly one root, a linear inequality typically has infinitely many solutions, filling an entire stretch of the real line.

From inequality to interval. The solution set $\{x \in \mathbb{R} : a < x < b\}$ is written as the open interval $(a,b)$; the parentheses signal that $a$ and $b$ themselves are excluded. The set $\{x \in \mathbb{R}: a \le x \le b\}$ is the closed interval $[a,b]$, and a mixed case like $\{x \in \mathbb{R}: a < x \le b\}$ is the half-open interval $(a,b]$. A compact expression such as $a < x < b$ is really two conditions joined into one line — $x>a$ and $x<b$ simultaneously — and this double inequality notation reappears constantly once systems of inequalities enter the picture.

Rule 1 — shifting. Adding or subtracting the same number on both sides preserves the inequality: if $a<b$, then $a+k<b+k$ and $a-k<b-k$ for any real $k$.

Rule 2 — scaling by a positive number. Multiplying or dividing both sides by the same positive number preserves the inequality: if $a<b$ and $k>0$, then $ak<bk$.

Rule 3 — scaling by a negative number. Multiplying or dividing both sides by the same negative number reverses the inequality: if $a<b$ and $k<0$, then $ak>bk$. This third rule is the single most important — and most often forgotten — fact in the whole chapter, and it is the one proved carefully below.

If $-3<2$ is a true statement, apply Rule 3 to find the correct inequality after multiplying both sides by $-4$, then use that result to solve $-2x<6$ for $x$.

Step 1. Start with the true inequality $-3<2$ and multiply both sides by $-4$, a negative number, so by Rule 3 the inequality must reverse: $(-3)(-4) \ \boxed{?}\ (2)(-4) \implies 12\ \boxed{?}\ -8$, and the correct relation is $12 > -8$.

Step 2. Check it numerically: $12$ is indeed greater than $-8$, confirming the reversal was necessary — had the sign been left as $12<-8$, that would be a false statement, exactly the trap Rule 3 prevents.

Step 3. Now apply the same rule to solve $-2x<6$. Divide both sides by $-2$ (a negative number), so the inequality must reverse from $<$ to $>$: $\dfrac{-2x}{-2} \boxed{?} \dfrac{6}{-2} \implies x \boxed{?} -3$, giving $x>-3$.

Step 4. Verify with a test value: pick $x=0$, which satisfies $x>-3$. Substituting into the original inequality gives $-2(0)=0$, and indeed $0<6$ is true. Pick $x=-3$ itself: $-2(-3)=6$, and $6<6$ is false, confirming $x=-3$ is correctly excluded. The solution set is $x>-3$, written in interval notation as $(-3,\infty)$.

• Forgetting to reverse the sign when multiplying or dividing by a negative number — the single most common error in the chapter, especially when a variable's coefficient is negative, as in $-2x<6$.
• Treating $\le$ and $<$ as identical — $x\le5$ includes $5$ as a solution while $x<5$ does not, which changes whether an interval or number-line endpoint is open or closed.
• Assuming an inequality has one unique solution like an equation does — a linear inequality in one variable almost always has infinitely many solutions forming an interval, not a single point.
• Squaring both sides without checking signs — squaring preserves direction only when both sides are known to be non-negative; this becomes relevant once inequalities combine with other operations later in the chapter.

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