Solve multi-step linear inequalities in one variable, graph the solution set on a number line, and interpret it in a real-world context (MA.912.AR.2.4, MA.912.AR.2.5).

What this topic is asking

MA.912.AR.2 asks you to solve linear inequalities in one variable, graph the solution on a number line, and interpret it in context. Solving is almost identical to solving equations, with one extra rule, and the B.E.S.T. Algebra 1 EOC tests it with equation-editor, number-line GRID, and context items.

Solving: same as equations, with one twist

Use the balance method, distribute, combine, isolate $x$, with one rule added:

Adding, subtracting, and multiplying or dividing by a positive number do not change the sign.

Graphing on a number line

• $<$ or $>$: open circle at the endpoint (not included), shade the ray.
• $\le$ or $\ge$: closed (filled) circle at the endpoint (included), shade the ray.
• Shade toward the values that satisfy the inequality ($x > -3$ shades right; $x \le -3$ shades left).

How the B.E.S.T. EOC examines this topic

• Equation editor. Solve and type the inequality (e.g. $x > -4$).
• GRID and hot-spot. Place the open or closed circle and the shading on a number line.
• Multiple choice and context. Find the maximum or minimum whole-number value in a budget or constraint problem.

A clarifying idea: an inequality describes a range of solutions, not one value, so the graph is a ray and the answer is a set. The single point where the related equation would be solved is just the boundary of that range.

Why dividing by a negative flips the sign

The flip rule looks arbitrary but follows from how negatives reorder numbers. Start with a true inequality such as $3 < 5$. Multiply both sides by $-1$: the values become $-3$ and $-5$, and on the number line $-3$ is now to the right of $-5$, so $-3 > -5$. The relationship reversed because multiplying by a negative reflects every number across zero, turning "smaller" into "larger." The same reflection happens for any negative multiplier, which is why the sign must flip whenever you multiply or divide by a negative. Testing a value after solving, substitute a number from your solution back into the original, is the reliable check that you flipped (or did not flip) correctly.

Interpreting in context

Many EOC inequalities model a budget ("at most \$20") or a requirement ("at least 60 points"). After solving, respect the context: if the variable counts objects, the answer is a whole number, and a budget cap means you **round down** to the largest whole number that still satisfies$\le$. State the answer in words ("at most 8 games") to capture the meaning.

Compound inequalities

Some EOC items join two inequalities. A conjunction ("and") like $-2 \le x < 5$ means $x$ is between $-2$ (included) and $5$ (excluded), graphed as a single segment with a closed circle on the left and an open circle on the right. A disjunction ("or") like $x < -1$ or $x \ge 4$ means $x$ is in either piece, graphed as two rays pointing away from each other. To solve a three-part "and" inequality such as $-7 < 2x - 1 \le 9$, do the same operation to all three parts: add $1$ everywhere to get $-6 < 2x \le 10$, then divide by $2$ everywhere to get $-3 < x \le 5$. Keeping the operations balanced across all parts, and flipping every sign together if you divide the whole chain by a negative, is what keeps a compound inequality correct.

Try this

Q1. Solve $-5x + 2 > -13$. [2 points]

• Cue. $-5x > -15 \Rightarrow x < 3$ (flip when dividing by $-5$).

Q2. Should the endpoint of $x \ge 7$ be an open or closed circle? [1 point]

• Cue. Closed (filled), because $\ge$ includes $7$.