Solving Inequalities and Absolute Value Equations on the SAT

Published on February 11, 2026
Solving Inequalities and Absolute Value Equations on the SAT

Solving Linear and Compound Inequalities

Solving inequalities is similar to solving equations, with one critical difference: when you multiply or divide by a negative number, flip the inequality sign. x+3<5 gives x<2. -2x>6 gives x<-3 (flip the sign when dividing by -2). Compound inequalities use "and" or "or." For "and," both conditions must be true simultaneously. -3-3 AND x<5; the solution set is all x between -3 and 5. Solve compound "and" inequalities by working on both sides simultaneously. For "or," at least one condition must be true. x<-3 OR x>5; the solution set includes all numbers less than -3 or greater than 5 (disjoint regions). A visual difference: "and" compounds (like -35) produce multiple separate intervals. Recognizing this difference helps you avoid confusion when solving and graphing.

When solving compound "and" inequalities, rewrite as two separate inequalities and solve each, then find the overlap. -3-3 (so x>-5) AND x+2<7 (so x<5). The solution is -5

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Absolute Value Equations and Cases

The absolute value of x, written |x|, is the distance from x to 0 on the number line. |5|=5, |-5|=5. To solve |x|=3, consider both cases: x=3 or x=-3 (both are 3 units from 0). Absolute value equations like |2x+1|=5 require splitting into cases: Case 1: 2x+1=5 gives x=2. Case 2: 2x+1=-5 gives x=-3. Check both: |2(2)+1|=|5|=5 ✓ and |2(-3)+1|=|-5|=5 ✓. For |x|<3, the solution is -33, the solution is x<-3 OR x>3 (x is more than 3 units from 0). Key pattern: |expression|=a becomes two cases: expression=a OR expression=-a. |expression|a becomes expression<-a OR expression>a. Memorizing these three patterns covers most absolute value problems on the SAT.

A practical checklist: When you see an absolute value, (1) Determine if it is an equation, <, or > problem; (2) Apply the appropriate pattern; (3) Solve each case; (4) Check both solutions in the original equation; (5) Graph on a number line to visualize the solution set. This methodical approach prevents confusion.

Graphing Inequalities on Number Lines and the Coordinate Plane

On a number line, an open circle at x=2 means x≠2 (used for < or >). A closed circle means x=2 is included (used for ≤ or ≥). An arrow shows the solution extends infinitely. For x>-2, draw an open circle at -2 with an arrow pointing right. For -3≤x≤5, draw closed circles at -3 and 5 with a line between them. On the coordinate plane, inequalities like y<2x+3 shade the region below the line y=2x+3 (dashed line because < does not include the line). For y≥2x+3, shade above the line with a solid line. Checklist for graphing: (1) Solve the inequality to get the boundary line or value; (2) Determine if the boundary is included (solid line/closed circle) or excluded (dashed line/open circle); (3) Shade or show the direction that satisfies the inequality; (4) Verify a test point: pick any point in the shaded region and confirm it satisfies the inequality.

For systems of inequalities, graph each inequality separately, then shade only the region satisfying ALL inequalities (the intersection). This visual approach helps you identify the solution set clearly.

Practice and Common Mistakes

A 1-week drill addresses inequality fluency. Day 1-2: Solve linear and compound inequalities; practice flipping signs when multiplying/dividing by negatives. Day 3-4: Solve absolute value equations (cases method) and inequalities. Day 5-6: Graph inequalities on number lines and coordinate planes. Day 7: Mix all types and identify any patterns in your mistakes. Most student errors fall into these categories: (1) Forgetting to flip the inequality sign when multiplying/dividing by negative; (2) Misidentifying when to use "and" vs. "or" in solutions; (3) Graphing the wrong region or boundary type (open vs. closed). Track which mistakes you make most often and drill that specific skill.

On test day, when solving an inequality, note whether you multiplied or divided by a negative (flip if you did). After solving, test a point from your solution in the original inequality to verify. This 10-second check prevents costly errors.

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