Solving Inequalities and Systems of Inequalities on the SAT

An inequality compares two expressions using symbols >, <, >=, or <=. Solving inequalities is almost identical to solving equations, with one critical exception: when you multiply or divide both sides by a negative number, you must flip the inequality sign. For example, -2x>6 becomes x<-3, not x>-3. This single rule trips up many students and costs them points on otherwise straightforward problems. To avoid this mistake, build the habit of circling your inequality sign every time you multiply or divide by a negative. The best way to prevent inequality sign errors is to avoid negative multiplication entirely by adding the variable term to both sides until it is positive, then proceeding as usual. For -2x>6, add 2x to both sides: 0>6+2x, then subtract 6: -6>2x, which becomes -3>x (same as x<-3). This approach requires extra steps but eliminates the sign-flip error for many students.

The solution to an inequality is a range of values rather than a single value. For x+3<7, the solution is x<4, which represents all numbers less than 4. Graphing this on a number line uses an open circle at 4 (since 4 is not included) and an arrow pointing left. A closed circle indicates inclusion (for >= or <=), while an open circle indicates exclusion (for > or <). Interval notation writes this solution as (-infinity, 4). Compound inequalities combine two inequalities, like -3

A system of inequalities consists of two or more inequalities that must all be satisfied simultaneously. The solution is the region on a coordinate plane where all inequalities are true at once. To graph a system, graph each inequality on the same coordinate plane, then identify the region where all shaded areas overlap. For each linear inequality like y<2x+3, the boundary is the line y=2x+3. If the inequality is < or >, the boundary is a dashed line (not included). If the inequality is <= or >=, the boundary is a solid line (included). Shade the region above the line for y> or y>=, and below the line for y< or y<=. When graphing multiple inequalities, the solution region is where all the shaded areas overlap, which may be a polygon, a ray, or a single point depending on the inequalities. Desmos is excellent for visualizing systems of inequalities, as you can type multiple inequalities and see the overlapping region immediately.

Some SAT questions present a system of inequalities and ask you to find the region it describes or to determine whether a point lies in that region. To check if a point satisfies all inequalities, substitute its coordinates into each inequality and verify all are true. For the system y>x and y<-x+4, the point (1,2) satisfies both inequalities: 2>1 is true, and 2<-1+4=3 is true. So (1,2) is in the solution region. Linear programming problems, where you want to maximize or minimize a quantity subject to constraints, use systems of inequalities to define the feasible region and then find the optimal point. These are rare on the SAT but worth understanding if they appear. Graph the system, identify the vertices of the feasible region, and evaluate the objective function at each vertex to find the maximum or minimum.

Absolute value inequalities like |x-3|<5 represent a distance condition: the distance from x to 3 is less than 5. Solve these by translating to a compound inequality. For |x-3|<5, the equivalent compound inequality is -5, like |x-3|>5, the equivalent is x-3>5 or x-3<-5 (note the "or", not "and"). Solve both parts: x>8 or x<-2. The key difference is that < translates to "and" (a compound inequality) while > translates to "or" (two separate inequalities). This is a mechanical rule that is easy to mix up, so build the habit of writing out the translation explicitly before solving. For absolute value inequalities with more complex expressions inside, like |2x+1|<=3, translate and solve the resulting compound inequality the same way.

Common mistakes on absolute value inequalities include forgetting to flip the inequality when "or" appears, or incorrectly setting up the equivalent compound inequality. The mnemonic "less than means in between, greater than means split" can help you remember the pattern. Graph absolute value inequalities on a number line to visualize the solution. The solution to |x-3|<5 is an interval centered at 3 with radius 5, shown as -25 is two rays pointing away from 3, shown as x>8 or x<-2. Practicing these translations and graphing until they are automatic prevents errors on test day. If you struggle to remember the rules, Desmos can graph absolute value inequalities and show you the solution region visually, which is a reliable fallback when you are unsure.

On test day, the most common errors with inequalities are forgetting to flip the sign when multiplying or dividing by negatives, incorrectly translating absolute value inequalities, and confusing the inequality direction when graphing. To prevent these, slow down on inequality problems and review your work before submitting. Verify your solution by testing a value within your range to confirm it satisfies the original inequality. For x<4, test x=0: if the original was x+3<7, then 0+3=3<7 is true, confirming x=0 satisfies the inequality. For compound inequalities like -2Testing values is a quick verification step that takes 15 seconds and catches most errors without requiring you to re-solve from scratch. If a problem involves graphing and Desmos is available, graph the inequality to visualize the solution region rather than relying purely on algebra. Visual verification is often faster and more reliable than re-solving algebraically.

For multiple-choice inequality questions, you can sometimes use answer choices to check your work. If you are unsure whether a range is x<4 or x<=4 (open vs. closed dot), look at the answer choices to see if x=4 is included in one option but not another. Choose the option that matches your solution. If you are uncertain whether a point is in a feasible region defined by multiple inequalities, substitute its coordinates into the answer options to see which one is consistent. This strategy works when you are confident about the point but unsure about the inequality direction. On test day, flexibility in your approach and willingness to use multiple verification methods prevent small errors from costing you points on otherwise solvable problems.