Solve linear inequalities in one variable and graph the solution set on a number line, reversing the inequality when multiplying or dividing by a negative (LA A1: A-REI.B.3).

What this topic is asking

Standard A1: A-REI.B.3 also covers linear inequalities in one variable: solve them and graph the solution set on a number line. On LEAP 2025 these are Type I Major Content items, and they appear on the no-calculator Session 1a. The one rule that makes inequalities different from equations is the sign flip on a negative multiply or divide.

Solving like an equation, with one exception

The steps mirror equation solving: simplify each side, gather the variable on one side and constants on the other, then divide by the coefficient.

When the coefficient is positive, nothing special happens, the inequality reads the same way throughout.

The sign-flip rule

The one rule unique to inequalities: multiplying or dividing both sides by a negative reverses the inequality.

Adding or subtracting never flips the sign; only a negative multiply or divide does. This is why gathering the variable on the side where its coefficient stays positive can avoid the flip entirely.

Graphing the solution on a number line

The solution to a one-variable inequality is a ray (or segment) on the number line:

• Open circle at the endpoint for $<$ or $>$ (the endpoint is not a solution).
• Closed circle at the endpoint for $\le$ or $\ge$ (the endpoint is a solution).
• Shade toward the values that work: right for "greater than," left for "less than."

For $x \ge -5$: a closed circle at $-5$, shaded to the right. The shaded ray represents infinitely many solutions, every value in that direction, which is what makes an inequality different from an equation that has a single answer. To check a solution, pick any value in the shaded region and confirm it satisfies the original inequality: for $x \ge -5$, testing $x = 0$ in $-3x + 4 \le 19$ gives $4 \le 19$, true, so the shading direction is right.

Reading "between" with a compound inequality

Some LEAP items describe a value that lies between two bounds, written as a compound inequality such as $-2 \le x < 5$. This is shorthand for two conditions at once: $x \ge -2$ and $x < 5$. On a number line it is a segment with a closed circle at $-2$ (included) and an open circle at $5$ (excluded), shaded in between. Solve a compound inequality by performing each operation on all three parts at the same time, and remember the sign-flip rule still applies if you multiply or divide the whole chain by a negative.

How LEAP examines this topic

• Equation response. Solve and enter the inequality (for example $x \ge -5$).
• Graphing item. Place the endpoint circle and shade the correct direction on a number line.
• Multiple choice. Match a solution to its number-line graph, with open-versus-closed and direction distractors.

Why a negative flip is necessary

The sign flip is not an arbitrary rule, it keeps the inequality true, which is the reasoning A-REI.B.3 wants you to understand. Start with a true statement like $3 < 5$. Multiply both sides by $-1$: the left becomes $-3$ and the right becomes $-5$. But $-3 > -5$, not $-3 < -5$, because on the number line $-3$ sits to the right of $-5$. Multiplying by a negative reflects every number across zero, which reverses their order, so the only way to keep a true statement true is to reverse the inequality sign as well. Equations do not have this issue because $=$ has no direction to reverse: $3 = 3$ stays true when both sides are negated to $-3 = -3$. Seeing the flip as "negation reverses order on the number line" makes it automatic, and it explains why the rule applies to multiplying and dividing (which can negate) but never to adding or subtracting (which only shift).

Try this

Q1. Solve $5 - 2x \ge 11$. [2 points]

• Cue. $-2x \ge 6$, divide by $-2$ and flip: $x \le -3$.

Q2. How is $x \le 2$ graphed? [1 point]

• Cue. Closed circle at $2$, shaded to the left.