2.3 Linear Inequalities

2.3 Linear Inequalities

Table 11 groups simplifying linear equations and inequalities with the linear-equations content area (10–15% of AAF, 2–3 CAT items). A linear inequality looks like a linear equation with = replaced by <, >, ≤, or ≥. The solution is usually an interval of real numbers, not a single value. You still distribute, combine like terms, and isolate the variable — with one extra law: if you multiply or divide both sides by a negative number, reverse the inequality symbol.

Adding the same number to both sides, or subtracting the same number, never reverses the symbol. Multiplying or dividing by a positive number never reverses it either. The reverse happens only for a negative multiplier or divisor. That single rule is the highest-yield inequality fact on AAF.

Two-variable inequality graphs (y ≥ 2x − 1, shading a half-plane) are not this section. They belong with linear applications and graphs; you will meet them in chapter 3 when you graph lines and then shade. Here the variable is one-dimensional and the picture is a number line.

Worked example: −2(x − 3) ≥ 8

Distribute first, exactly as in Section 2.2:

−2(x − 3) ≥ 8

−2x + 6 ≥ 8.

Subtract 6 from both sides (no reverse — this is subtraction):

−2x ≥ 2.

Divide both sides by −2. The divisor is negative, so flip ≥ to ≤:

x ≤ −1.

Check a boundary point and a test point. At x = −1: −2(−1 − 3) = −2(−4) = 8, and 8 ≥ 8 is true, so the endpoint is included (the original symbol was ≥). At x = −2 (left of −1): −2(−2 − 3) = −2(−5) = 10, and 10 ≥ 8 is true. At x = 0 (right of −1): −2(0 − 3) = 6, and 6 ≥ 8 is false. The solution is all real numbers at or left of −1.

If you forget to reverse, you report x ≥ −1, which is the complementary ray and is wrong on every check except the endpoint. Build the reverse into the same breath as the division: “divide by −2, flip the sign.”

Why the reverse is required

Start from the true comparison 2 < 6. Multiply both sides by −1 and you get −2 and −6. On the number line, −2 is to the right of −6, so −2 > −6. The inequality reversed because multiplication by a negative reflects the line through 0. Division by a negative is multiplication by a negative reciprocal, so it reverses for the same reason. This is not a style preference; it is the definition of order on the reals.

Compound inequalities

A compound inequality chains two comparisons on the same expression, or joins two inequalities with “and” / “or.”

And / sandwich form. Solve −3 < 2x + 1 ≤ 7.

Work all three parts at once. Subtract 1: −4 < 2x ≤ 6.

Divide by 2 (positive, so do not reverse): −2 < x ≤ 3.

In interval notation that is (−2, 3] — open on the left because of <, closed on the right because of ≤.

Or form. Solve x + 4 ≤ 1 or 3x > 12.

First piece: x ≤ −3. Second piece: x > 4. The solution is the union (−∞, −3] ∪ (4, ∞). You cannot collapse an “or” into a single sandwich; −3 and 4 have a gap between them.

When an AAF item says “x is at least 2 and at most 9,” write 2 ≤ x ≤ 9 or [2, 9]. “At least” is ≥; “at most” is ≤; “more than” is >; “less than” is <; “no more than” is ≤; “no less than” is ≥. Those English phrases are as testable as the algebra.

Interval notation versus inequality notation

Parentheses match open circles and strict inequalities. Brackets match closed circles and inclusive inequalities. Infinity always takes a parenthesis: there is no number called ∞ to include. Do not write [−∞, 4] — that is not standard interval notation.

Graphing on a number line is a one-minute sketch: mark the critical number, choose open or closed, and shade the direction your test point confirmed. AAF multiple-choice graphs will differ by exactly one of those three features (endpoint, open/closed, direction), so check all three before you click.

Another fully worked chain

Solve 1 − 3(2x − 4) < 10 and write the answer in interval notation.

Distribute the −3: 1 − 6x + 12 < 10.

Combine: 13 − 6x < 10.

Subtract 13: −6x < −3.

Divide by −6 and reverse: x > (−3)/(−6) → x > 1/2.

Interval: (1/2, ∞). Check x = 1: 1 − 3(2 − 4) = 1 − 3(−2) = 1 + 6 = 7 < 10, true. Check x = 0: 1 − 3(0 − 4) = 1 − 3(−4) = 13 < 10, false. The open ray to the right of 1/2 is correct.

The easy miss is reversing too early (when subtracting 13) or not reversing at the division by −6. Subtracting never flips; dividing by a negative always does.

Connecting one-variable inequalities to later graphing

If x ≤ −1 is the solution of a one-variable inequality, its graph is a closed ray on a number line. If the same relationship is rewritten in two variables as, say, y ≥ −2x + 6 after you treat y as a free output, the graph becomes a half-plane bounded by the line y = −2x + 6. That two-variable picture — solid versus dashed boundary, test point, shading — is a linear-applications skill, not a linear-equations skill. Do not shade a plane on a Table 11 “simplifying inequalities” item that only mentions x.

Keep the skills separate on purpose:

• This section: isolate x, reverse when dividing by a negative, report an interval, mark a number line.
• Chapter 3: graph Ax + By ≥ C in the coordinate plane, including systems of inequalities.

Both appear on AAF; they are scored in adjacent content areas (linear equations vs linear applications and graphs), each 10–15%. Mixing the pictures is a classification error, not just a drawing error.

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations