Solving linear inequalities and solution sets
Solve inequalities in one variable with the balancing steps used for equations, flip the sign when multiplying or dividing by a negative, and show the answer on a number line.
What a learner can do afterwards
- Solve linear inequalities, including brackets and fractional boundaries, and describe their number-line solution sets
- Choose valid balancing steps and reverse the inequality sign when multiplying or dividing both sides by a negative number
- Solve a double inequality such as -3 < 2x + 1 ≤ 7 and list the integers in it
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You have already used balancing steps and described inequality solutions on a number line. Adding or subtracting the same number preserves order. Multiplying or dividing by a positive number also preserves order. A negative multiplier or divisor reverses order, so reverse the inequality sign. Use nonzero numbers for reversible multiplication and division. Multiplying by zero loses information; dividing by zero is undefined.
A solution set contains all values that make the inequality true. On paper, draw an open circle for < or > and a filled circle for ≤ or ≥. Shade left for smaller values and right for larger values. The boundary separates included and excluded values. A statement with 'or' allows either condition: x < -3 or x ≥ 4 includes values below -3 and values from 4 upwards. Draw two rays with a gap between them.
Solve 3(x - 2) < 12. Expand: 3x - 6 < 12. Add 6 to both sides: 3x < 18. Divide by positive 3: x < 6. Keep the exact quotient, even when it is a fraction. On paper, draw an open circle at 6 and shade left. Whole numbers are 0, 1, 2, ... . The largest whole-number solution here is 5 because the strict sign excludes 6.
Positive division keeps the comparison pointing the same way.
Solve 8 - 3x ≤ -4. Subtract 8: -3x ≤ -12. Divide by -3 and reverse the sign: x ≥ 4. Keep a noninteger quotient as an exact fraction. On paper, draw a filled circle at 4 and shade right. The final ≥ includes 4. Check x = 5: 8 - 15 = -7, and -7 ≤ -4. For x = 3, the comparison is -1 ≤ -4, which is false.
The final ≥ sign includes the boundary value.
A double inequality requires both comparisons to hold. Solve -3 < 2x + 1 ≤ 7. Subtract 1 from all three parts: -4 < 2x ≤ 6. Divide throughout by 2: -2 < x ≤ 3. Integers include negative numbers, zero and positive numbers. Here they are -1, 0, 1, 2, 3. Whole numbers exclude negatives, so the whole-number solutions are 0, 1, 2, 3. Check both boundaries before listing or counting.
On paper, shade between an open circle at -2 and a filled circle at 3.
To check a candidate, substitute it into the original inequality and compare the two sides. For 7 - 2x ≥ 15, x = -4 gives 15 ≥ 15, which is true. The value -3 gives 13 ≥ 15, which is false. One value that disagrees with a proposed set disproves that set. Passing a few checks does not prove the whole set; use reversible balancing steps to find all solutions.
Keep fractional boundaries exact. Solve -6 < 4x + 1 ≤ 7. Subtract 1: -7 < 4x ≤ 6. Divide by positive 4: -7/4 < x ≤ 3/2. Since -2 < -7/4 < -1 and 1 < 3/2 < 2, the integer solutions are -1, 0, 1. Only 0 and 1 are whole numbers. Locate each boundary between neighbouring integers before making the list.
The dots show only the integer solutions, not every value in the interval.
Split a chain into two inequalities when x occurs in more than one part. For 2 - 3x ≤ -4 < 8 - 3x, solve 2 - 3x ≤ -4: subtract 2 and divide by -3 to get x ≥ 2. Solve -4 < 8 - 3x: subtract 8 and divide by -3 to get 4 > x, or x < 4. Both conditions must hold, so keep their overlap: 2 ≤ x < 4.
The overlap satisfies both comparisons.
When x is on both sides, collect its terms before dividing. Solve -2(x - 3) > x + 9. Multiply both bracket terms by -2: -2x + 6 > x + 9, since (-2)(-3) = 6. Subtract x: -3x + 6 > 9. Subtract 6: -3x > 3. Divide by -3 and reverse the sign: x < -1.
The outside multiplier acts on every term inside the bracket.
Negative division reverses both signs in a double inequality. Solve -8 < 4 - 2x ≤ 10. Subtract 4 throughout: -12 < -2x ≤ 6. Divide throughout by -2: 6 > x ≥ -3. Rewrite the same comparisons in increasing order: -3 ≤ x < 6. Its integer solutions run from -3 through 5.
Reordering the written comparisons keeps the same solution set.
Solve an inequality with the same balancing steps as an equation, flip the sign when you multiply or divide by a negative, and show the solution set on a number line with an open or closed endpoint.
Lesson, questions, worked solutions and flashcards written for this dataset.
Readings and extra questions written for this dataset. Video: One-step inequalities examples, Khan Academy. Video licensing terms are not recorded here.
More to read
Which balancing steps preserve the solutions?
Each step in solving an inequality should preserve exactly the same solutions. Addition and subtraction work when you apply the same change to both sides. Multiplication and division need attention to the sign of the number used.
Consider 5x - 8 ≤ 17. Adding 8 gives 5x ≤ 25. Dividing by positive 5 gives x ≤ 5. The boundary is included because equality is allowed.
A negative multiplier reverses order. Starting from 3 < 7, multiplying both sides by -2 gives -6 > -14. Negative division also requires reversing the sign.
Zero cannot be used for a reversible multiplication step. For example, x < 5 allows x = 2. Multiplication by zero produces 0 < 0, which is false even for that allowed input. Division by zero has no defined result. Use nonzero numbers when multiplying or dividing to obtain an equivalent inequality.
From a solution to a paper number line
The solved comparison tells you how to draw its solution set. Mark the boundary first. An open circle excludes that value; a filled circle includes it.
For x > -4, leave -4 open and shade to the right. Values such as -3.5 and 0 belong to the set, as do larger values beyond the edge of the drawing. For x ≤ 1, fill the circle at 1 and shade left.
A range such as -4 < x ≤ 1 needs both endpoint decisions. Shade between an open circle at -4 and a filled circle at 1. Every value in that stretch satisfies both comparisons.
An 'or' statement allows separate stretches. For x ≤ -4 or x > 1, shade left from a filled circle at -4 and right from an open circle at 1. The middle gap is excluded.
A row of dots showing selected integer solutions is different from a shaded interval. The interval also contains the allowed fractions and decimals between integers.
Exact boundaries and integer lists
A fraction can be a solution boundary. Keep it exact while deciding which integers lie inside the range.
For -10 < 3x + 2 ≤ 6, subtract 2 throughout: -12 < 3x ≤ 4. Divide by positive 3 to obtain -4 < x ≤ 4/3. The boundary 4/3 lies between 1 and 2. The integers are therefore -3, -2, -1, 0 and 1. The whole-number solutions are only 0 and 1.
Negative division reverses both comparisons in a double inequality. Solve -7 ≤ 5 - 2x < 9. Subtract 5: -12 ≤ -2x < 4. Divide by -2: 6 ≥ x > -2. Reading the comparisons in increasing order gives -2 < x ≤ 6.
Check a strict boundary separately from an inclusive boundary. The integer -2 is excluded in this last range, while 6 is included. Rounding a fractional boundary can change a solution set, so retain its exact fraction.
What a substitution check tells you
A substitution check answers one question: does this particular value satisfy the original inequality?
Take 12 - 3x > 0. At x = 3, the left side is 3, so the comparison is true. At x = 4, the left side is 0, and 0 > 0 is false.
Suppose someone claims that the solution is x ≤ 4. Their proposed set contains 4, but 4 fails the original comparison. This single counterexample disproves that proposed set. Solving with valid steps gives -3x > -12 and then x < 4.
Checking a few values cannot establish every part of an infinite solution set. Use balancing steps to obtain the full set, then test candidates to look for calculation errors, reversed directions or endpoint mistakes.
A value can satisfy one part of a double inequality and fail the other. To belong to its solution set, it must satisfy both comparisons.
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