Learning Objectives
Solve a linear inequality and justify any reversal of its sign.
Combine restrictions using and or or.
Handle a parameter whose sign controls division.
Describe a region satisfying inequalities in two variables.
Prerequisites: linear equations, signed arithmetic and number-line intervals. All practice questions in this lesson are original, not official TMUA questions.
What This Lesson Is About
An inequality asks which values make one quantity larger or smaller than another. Unlike an equation with a few isolated solutions, it often describes a whole interval of inputs.
This lesson explains how algebraic operations affect order, how to combine conditions using “and” and “or”, and how to handle coefficients with an unknown sign. It then extends the same ideas to regions in the coordinate plane.
1. Solving a Linear Inequality
Start by interpreting the comparison symbol, then isolate the variable while preserving the meaning of the comparison.
1.1 Understanding strict and inclusive boundaries
An inequality compares quantities rather than requiring equality. The solution is usually an interval or a union of intervals. Use < or > to exclude equality and ≤ or ≥ to include it.
1.2 Reversing order when multiplying or dividing by a negative number
Adding the same quantity preserves order. Multiplying or dividing by a positive number also preserves order, while doing so by a negative number reverses it. For example, 2 < 5 becomes −2 > −5 after multiplication by −1.
Worked Example: A Negative Coefficient
Question: Solve 7 − 3x ≥ 16.
Method: Isolate the term containing x, then reverse the comparison when dividing by its negative coefficient.
Subtract 7: −3x ≥ 9. Divide by −3 and reverse the comparison: x ≤ −3.
At the boundary x = −3 both sides are 16. An interior value −4 gives 19 ≥ 16, while x = 0 fails.
2. Combining Inequalities
A compound statement contains more than one comparison. Its connecting word determines whether you keep common solutions or combine separate solution sets.
2.1 Solving a three-part inequality
A compound inequality such as −1 < 2x + 3 ≤ 9 requires both comparisons simultaneously. Apply the same operation to all three parts, or solve the two restrictions separately and keep their common values.
2.2 Interpreting “and” as intersection and “or” as union
“And” asks for inputs satisfying both conditions. “Or” asks for inputs satisfying at least one, including inputs satisfying both. Intersecting disjoint intervals gives no solution; joining them with or keeps both intervals.
Worked Example: Two Endpoints
Question: Solve −1 < 2x + 3 ≤ 9.
Method: Apply each operation to all three parts and retain the strict or inclusive status of each endpoint.
Subtract 3 throughout to get −4 < 2x ≤ 6. Divide by positive 2: −2 < x ≤ 3.
The lower endpoint fails the strict comparison, while x = 3 is allowed. A central value x = 0 satisfies both original conditions.
Worked Example: A Union Is Different
Question: Solve x < −1 or x ≥ 2, then compare with x < −1 and x ≥ 2.
Method: Interpret “or” by taking both regions, then interpret “and” by asking whether any input belongs to both.
The or statement includes both separate regions: all numbers below −1 and all numbers at least 2.
The and statement has no solution, since no number can be both below −1 and at least 2.
3. Handling a Coefficient with an Unknown Sign
A parameter is fixed within a problem, but its value may not be specified. Since division depends on its sign, different parameter values can require different answers.
3.1 Separating positive, negative and zero cases
If the coefficient contains a parameter, its sign matters. Split into positive, negative and zero cases before division. The zero case becomes a constant comparison that is either always true or always false.
3.2 Why unconditional division can give the wrong direction
For example, a statement such as ax > 1 cannot be converted to x > 1/a without knowing a > 0. When a < 0 the direction reverses; when a = 0 division is impossible.
Worked Example: All Parameter Cases
Question: Solve ax ≤ 2 for fixed real a.
Method: Split according to the sign of a before dividing, and test the original constant comparison when a = 0.
If a > 0, x ≤ 2/a. If a < 0, x ≥ 2/a. If a = 0, the statement is 0 ≤ 2 and every real x works.
For a = −2 the answer is x ≥ −1. Testing x = 0 gives 0 ≤ 2; testing x = −2 gives 4 ≤ 2, false.
4. Describing Regions with Two Variables
With two variables, a solution is a coordinate pair rather than a single point on a number line. A linear inequality selects one side of a boundary line.
4.1 Finding the boundary and testing a side
The boundary of a linear inequality in x and y is a straight line found by replacing the inequality by equality. One side of the line satisfies the inequality. A test point not on the boundary identifies the correct side.
4.2 Including edges and overlapping several conditions
Include the boundary for ≤ or ≥, and exclude it for < or >. When several conditions must hold, keep their overlapping region. A point lying on one allowed boundary can still fail another restriction.
Worked Example: A Feasible Triangle
Question: Describe the region x ≥ 0, y ≥ 0 and x + y ≤ 4.
Method: Identify each boundary and allowed side, then take their overlap and check which edges are included.
The first two inequalities require the first quadrant including its axes. The last condition keeps the side of x + y = 4 containing (0, 0).
Together these form the filled triangle with vertices (0, 0), (4, 0), (0, 4), including its edges. The point (1, 2) is inside; (3, 2) fails because its coordinate sum is 5.
Common Mistakes
Subtracting a negative number is still addition; it does not itself reverse an inequality.
A negative multiplication or division reverses every comparison affected.
And and or generally produce different solution sets.
Never divide by a parameter before checking its sign and zero case.
5. Practice
For each answer, state the complete solution set and check its boundaries. When a parameter appears in a divisor, separate its sign cases; when several conditions are joined, identify whether they require intersection or union.
1. Solve 5 − 2x < 11.
2. Solve 2 ≤ 3x − 1 < 8.
3. Solve 2x + 1 > 5 and 3x − 2 ≤ 10.
4. Solve x ≤ 1 or x > 4. Is x = 4 allowed?
5. Solve (k − 1)x > 0 for all real k.
6. Which of (1, 1), (2, 2), (0, 3) satisfies x ≥ 0, y > 0, x + 2y ≤ 5?
Worked Solutions
1. Reversing the comparison after negative division
−2x < 6, so x > −3. At −3 equality holds and is excluded; x = 0 gives 5 < 11.
2. Tracking both endpoints in a compound inequality
Add 1 and divide by 3: 1 ≤ x < 3. The value 1 is included and 3 excluded.
3. Taking the intersection of two conditions
The restrictions are x > 2 and x ≤ 4, so 2 < x ≤ 4. At x = 4, the first is 9 > 5 and the second is 10 ≤ 10.
4. Taking a union and checking an endpoint
Keep both intervals: x ≤ 1 or x > 4. The value 4 satisfies neither, so it is excluded.
5. Splitting all parameter sign cases
If k > 1, x > 0. If k < 1, x < 0. If k = 1, the inequality is 0 > 0, false for every x, so there is no solution.
6. Testing a point against every regional condition
Only (1, 1) works: its weighted sum is 3 ≤ 5. The other points have sums 6 and 6 respectively, so they fail despite satisfying the first two restrictions.
6. Summary
- Track the sign of any multiplier or divisor.
- Use interval intersections for and and unions for or.
- Split parameter problems into positive, negative and zero cases.
- For two-variable inequalities, test the correct side and boundary inclusion.