ORIGINAL WORKED EXAMPLE
Reverse exactly once
Solve 7 − 3x ≤ 19.
- Subtract 7 from both sides: −3x ≤ 12.
- Divide by −3 and reverse the comparison: x ≥ −4.
- The endpoint works: 7 − 3(−4) = 19. A larger value such as 0 also works.
Answer x ≥ −4
SAT · Algebra · TOPIC LESSON
Linear inequalities in one or two variables
Describe every allowed value or point, including the boundary and the context.
BY THE END, YOU CAN
- Translate limits into inequalities and solve them correctly.
- Connect number lines and half-planes to algebraic solution sets.
- Test feasibility under several simultaneous restrictions.
BEFORE YOU START
- Linear equations and coordinate graphs.
- Order of signed numbers.
Original educational lesson. Work through the examples before revealing the practice answers. This lesson supports SAT Math and Reading and Writing skill study; it does not reproduce official questions or guarantee a score.
LEARN THROUGH ORIGINAL PRACTICE
Original SAT-style worked examples and practice
The teaching passages, worked examples and exercises in this lesson are original. They are not official College Board questions or past-exam reproductions. Blueprint and observed-practice analytics describe separate factual metadata; those source questions are not displayed here.
01 / UNDERSTAND & APPLY
An inequality describes a range
An inequality compares quantities using <, >, ≤ or ≥. A strict symbol excludes equality; an inclusive symbol allows it. “At most” means ≤, “at least” means ≥, “less than” means <, and “more than” means >. The solution is generally a set of possible values rather than a single value.
Read the comparison from the quantity named. “The load exceeds 40 kg” gives L > 40. The reversed statement 40 > L describes something different.
02 / UNDERSTAND & APPLY
Why a negative multiplier reverses order
Adding the same quantity to both sides preserves order. Multiplication or division by a positive number also preserves it. Multiplication or division by a negative number reverses it: because 2 < 5, reflecting the numbers through zero gives −2 > −5.
Do not reverse the sign merely because a negative term is present or because you subtract. Reverse it when you multiply or divide the whole inequality by a negative number.
03 / UNDERSTAND & APPLY
Endpoints and compound conditions
A number-line graph uses an open endpoint for a strict inequality and a closed endpoint for an inclusive inequality. Shade left for smaller values and right for larger values. A compound “and” condition is the intersection of allowed sets; an “or” condition is their union.
When a variable is a whole-number count, first solve over the reals and then take the allowed integers. A lower bound such as n ≥ 4.2 requires at least 5 items, whereas an upper bound n ≤ 4.2 allows at most 4.
ORIGINAL WORKED EXAMPLE
An interval with two boundaries
Solve −2 < 3x + 1 ≤ 10.
- Subtract 1 from all three parts: −3 < 3x ≤ 9.
- Divide all parts by positive 3: −1 < x ≤ 3.
- The left endpoint −1 is excluded, while 3 is included.
Answer −1 < x ≤ 3
04 / UNDERSTAND & APPLY
A limit becomes an inequality
Translate the total, not just a single charge. A resource budget usually gives a maximum; a required performance target gives a minimum. Include contextual constraints such as nonnegative quantities.
ORIGINAL WORKED EXAMPLE
A budget and a whole-number maximum
A workshop pays a $25 setup fee and $8 per participant. Its budget is at most $100. What is the largest possible number of participants?
- Let n be a nonnegative integer. The budget gives 25 + 8n ≤ 100.
- Subtract 25 and divide by 8: n ≤ 75/8 = 9.375.
- The largest allowed integer is 9. Ten would cost $105, exceeding the budget.
Answer 9 participants
05 / UNDERSTAND & APPLY
Two-variable inequalities fill one side of a line
For y > mx + b, points above the boundary line are solutions; for y < mx + b, points below are solutions. Use a dashed line if equality is excluded and a solid line if it is included. For Ax + By ≤ C, either solve for y with the correct sign reversal or test a point not on the boundary.
The origin is a convenient test point only if it is not on the boundary. Substitution determines which side to shade; the visual position alone can mislead when the inequality has not been rearranged.
ORIGINAL WORKED EXAMPLE
Test a point and identify the boundary
Is (2,3) a solution of 2x + y < 8? Is (2,4)?
- At (2,3), the left side is 2(2) + 3 = 7; 7 < 8 is true.
- At (2,4), the left side is 8; 8 < 8 is false.
- The boundary 2x + y = 8 is dashed, and the solution region lies below y = 8 − 2x.
Answer (2,3) is a solution; (2,4) is not
06 / UNDERSTAND & APPLY
A feasible point satisfies every restriction
For a system of inequalities, the feasible region is the overlap of all allowed half-planes. Boundary intersections help describe its corners, but a corner may be excluded by a strict inequality. Testing only one condition cannot prove that a point is feasible.
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ORIGINAL WORKED EXAMPLE
Overlapping restrictions
Determine whether (2,2) and (5,3) satisfy x ≥ 0, y ≥ 1 and x + y ≤ 6.
- At (2,2), x is nonnegative, y is at least 1, and x + y = 4 ≤ 6. All conditions hold.
- At (5,3), the first two conditions hold, but x + y = 8 exceeds 6.
- A single failed condition excludes a point from the intersection.
Answer (2,2) is feasible; (5,3) is not
07 / UNDERSTAND & APPLY
Connect candidate tables and solution regions
A table of ordered pairs can be checked row by row against each inequality. It usually lists only selected solutions or candidates, not the entire continuous solution region. If the boundary equation and one shaded-side point are given, recover the comparison sign from that point and use the boundary style to determine whether equality belongs.
For example, a solid boundary y = 2x + 1 with the point (0,0) shaded represents y ≤ 2x + 1, because 0 ≤ 1 and the boundary is included.
08 / UNDERSTAND & APPLY
Check exactness and context
A calculator can show shaded inequalities or evaluate candidate points. Keep the strict or inclusive symbol intact: rounded coordinates near a boundary can conceal whether equality holds. If division would involve an unknown-sign parameter, split into positive, negative and zero cases rather than using one universal sign rule.
Common errors are shading a union instead of an intersection, using an open boundary for “at most,” and rounding a maximum upward beyond the limit.
09 / UNDERSTAND & APPLY
Original practice: every boundary matters
Explain whether endpoints and candidate points belong.
ORIGINAL PRACTICE · 1
Solve −4x > 20.
Reveal answer & reasoning
x < −5
Dividing by −4 reverses > to <.
ORIGINAL PRACTICE · 2
Solve 2x + 7 ≥ 15.
Reveal answer & reasoning
x ≥ 4
Subtract 7 and divide by positive 2; the symbol does not reverse.
ORIGINAL PRACTICE · 3
Translate “at least 18 tickets.”
Reveal answer & reasoning
n ≥ 18
At least includes the threshold 18 and all larger counts.
ORIGINAL PRACTICE · 4
For x < 3, is the endpoint open or closed?
Reveal answer & reasoning
Open
The strict symbol excludes x = 3.
ORIGINAL PRACTICE · 5
Is (1,5) a solution of y ≤ 3x + 1?
Reveal answer & reasoning
No
At x = 1 the maximum allowed y is 4, but the proposed y is 5.
ORIGINAL PRACTICE · 6
What inequality has a dashed boundary y = −x + 4 and shading above it?
Reveal answer & reasoning
y > −x + 4
Above gives > and the dashed line excludes equality.
ORIGINAL PRACTICE · 7
A bus holds at most 48 people; 7 seats are occupied. What is the maximum additional number?
Reveal answer & reasoning
41 people
7 + n ≤ 48 gives n ≤ 41, with n a nonnegative integer.
ORIGINAL PRACTICE · 8
Solve 1 ≤ 2x − 3 < 9.
Reveal answer & reasoning
2 ≤ x < 6
Add 3 to all parts and divide by 2. The lower bound is included and upper bound excluded.
KEEP THE BIG IDEAS
Quick revision
- Reverse an inequality only when multiplying or dividing by a negative quantity.
- Strict and inclusive boundaries describe different solution sets.
- For simultaneous constraints, check every condition and the contextual domain.
READ WITH CONTEXT
References & scope
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- College Board: SAT Math skill descriptions ↗
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