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Lesson 2 of 3

Linear Inequalities · Lesson 2 of 3

Compound Inequalities and Applications

“Find the values that satisfy several constraints together and interpret their ranges in practical models.”

Learning Objectives

• Read a double inequality as two simultaneous conditions. • Solve chains while handling negative multiplication correctly. • Find intersections of solution intervals and graph their common region. • Translate temperature ranges using a linear conversion formula. • Build concentration models from substance amount and total volume. • Interpret a formula-based range with correct units and endpoints.

Two bounds on the same quantity

A single inequality gives one restriction. A compound inequality combines restrictions, often supplying both a lower and an upper bound. In this lesson the conditions are connected by AND: a solution must satisfy every condition. The word between needs care because ordinary language may leave endpoint inclusion unclear; the mathematical symbols must settle it.

Definition
Double inequality

A chain such as a ≤ f(x) < b abbreviates the two requirements a ≤ f(x) and f(x) < b. Here f(x) stands for the expression being bounded, such as 5x − 3. It does not introduce a new solving rule.

You may solve the two requirements separately and keep their common values. Alternatively, apply the same valid operation to all three parts of the chain. Adding or subtracting a number preserves both comparisons. Multiplying or dividing by a positive number also preserves them; a negative number reverses both. Working on only one part of a chain changes the problem.

Example — A lower inclusive and upper strict bound

Problem
Solve −8 ≤ 5x − 3 < 7.

  1. 1.Add 3 to all three parts: −5 ≤ 5x < 10.
  2. 2.Divide all three parts by positive 5: −1 ≤ x < 2.
  3. 3.The interval is [−1, 2). At x = −1 the expression equals −8 and is allowed. At x = 2 it equals 7 and is excluded.
  4. 4.Any interior value, such as x = 0, satisfies both original comparisons because −8 ≤ −3 < 7.
Example — Negative division in a chain

Problem
Solve −5 ≤ (5 − 3x)/2 ≤ 8.

  1. 1.Multiply every part by positive 2: −10 ≤ 5 − 3x ≤ 16.
  2. 2.Subtract 5 everywhere: −15 ≤ −3x ≤ 11.
  3. 3.Divide everywhere by −3, reversing both comparisons: 5 ≥ x ≥ −11/3.
  4. 4.Read from the smaller endpoint to the larger endpoint: −11/3 ≤ x ≤ 5. The interval is [−11/3, 5].
  5. 5.At x = 5 the middle expression is −5; at x = −11/3 it is 8. Both endpoints are permitted.
Both comparisons reverse

From −15 ≤ −3x ≤ 11, writing 5 ≤ x ≤ −11/3 is incorrect: dividing by a negative reverses each comparison. A descending chain such as 5 ≥ x ≥ −11/3 is valid. Rewrite it in ascending order afterward to make the interval easy to read.

Systems and the common solution region

A system lists two or more inequalities for the same variable. Solve each one first, then ask which values survive all the restrictions. This common set is the intersection. The graph is useful because an algebraic result can look plausible while accidentally keeping values that satisfy only one condition.

Definition
Intersection of solution sets

The values belonging to every required solution set. For an AND system, the intersection is the system’s solution set. If the sets have no common values, the system has no solution.

Example — Intersecting two rays

Problem
Solve 3x − 7 < 5 + x and 11 − 5x ≤ 1.

  1. 1.For the first inequality, subtract x and add 7: 2x < 12, hence x < 6.
  2. 2.For the second, subtract 11: −5x ≤ −10. Divide by −5 and reverse: x ≥ 2.
  3. 3.A value must satisfy x < 6 and x ≥ 2. Their intersection is 2 ≤ x < 6, or [2, 6).
  4. 4.At 2 both conditions hold; at 6 the first becomes equality and fails. Values below 2 fail the second, while values above 6 fail the first.
First condition: x less than 6012345678Second condition: x at least 2012345678Common solutions: from 2 inclusive to 6 exclusive012345678
The intersection of two constraints— Read each coloured region as the entire solution set. A hollow endpoint is excluded; a filled endpoint is included. Arrows indicate continuation.

When two lower bounds are present, keep the stronger one: x > 1 and x ≥ 3 together give x ≥ 3. When two upper bounds are present, keep the smaller boundary with its appropriate strictness. At a shared boundary, check every condition before including it. For example, x ≥ 2 and x < 2 have no common value, while x ≥ 2 and x ≤ 2 have the single solution x = 2.

Example — Checking whether an overlap exists

Problem
Compare the systems x > 4 with x ≤ 1, and x ≥ 4 with x ≤ 4.

  1. 1.In the first system, a number would have to lie to the right of 4 and at or to the left of 1. No number can do both, so the solution set is empty.
  2. 2.In the second, 4 satisfies both comparisons and every other number fails one. The solution set is {4}.
  3. 3.An intersection may be an interval, a ray, a single point, or the empty set. Do not assume two restrictions always create a nonempty interval.

Converting a temperature range

A conversion formula relates two descriptions of the same temperature. Substitute the formula into the bounds, then transform all three parts together. Because the temperature scale factor here is positive, conversion keeps the order of the endpoints. Units still matter: the answer is a range on the target scale.

Write C for the numerical temperature in degrees Celsius and F for its value in degrees Fahrenheit. Their relationship can be written in either direction. The constant 32 shifts the origin, while the factor 9/5 describes how the scale intervals compare.

Temperature conversionLaTeX
C and F refer to the same physical temperature on different scales.
Example — Strict temperature limits

Problem
A solution must stay strictly above 30°C and strictly below 35°C. Find the Fahrenheit range.

  1. 1.Write 30 < C < 35, and substitute C = (5/9)(F − 32).
  2. 2.This gives 30 < (5/9)(F − 32) < 35. Multiply every part by positive 9/5: 54 < F − 32 < 63.
  3. 3.Add 32 everywhere: 86 < F < 95. The required range is (86, 95)°F.
  4. 4.Both endpoints are excluded because the Celsius endpoints were excluded. If the original instruction allowed 30°C and 35°C, the converted answer would instead be 86 ≤ F ≤ 95.
Example — The reverse conversion

Problem
Take 68°F and 77°F as excluded limits. Find the corresponding Celsius range.

  1. 1.Write 68 < F < 77 and substitute F = (9/5)C + 32.
  2. 2.Subtract 32 everywhere: 36 < (9/5)C < 45.
  3. 3.Multiply every part by positive 5/9: 20 < C < 25.
  4. 4.Thus the strict Celsius range is (20, 25)°C. Including either Fahrenheit endpoint would include its corresponding Celsius endpoint.

Mixtures: amount first, percentage second

A concentration is a ratio: the amount of a substance divided by the total mixture amount. Mixing two concentrations does not usually give their simple arithmetic average because the volumes may differ. First calculate the substance amount in each portion, then add those amounts and divide by the combined volume.

In the models here, volumes are treated as additive and percentages are applied consistently to those volumes. If x litres of a second solution are added to V litres of the first, the total is V + x. The domain x ≥ 0 ensures V + x is positive when V > 0. This positivity must be established before multiplying an inequality by the total volume.

Concentration of an additive-volume mixtureLaTeX
V and x are volumes in the same units; p and q are concentration fractions, such as 0.12 for 12%. The numerator is substance amount, not a percentage. Multiply the resulting fraction by 100 to express it as a percentage.
Example — Adding a stronger solution

Problem
600 L of 12% acid solution is mixed with x L of 30% solution. Require a final acid content strictly between 15% and 18%. Find x.

  1. 1.The initial acid amount is 0.12 × 600 = 72 L in this model. The added acid amount is 0.30x L. Total solution volume is 600 + x L.
  2. 2.Write 0.15 < (72 + 0.30x)/(600 + x) < 0.18 with x ≥ 0. The denominator is positive.
  3. 3.The lower bound gives 72 + 0.30x > 0.15(600 + x) = 90 + 0.15x. Subtract 72 and 0.15x: 0.15x > 18, so x > 120.
  4. 4.The upper bound gives 72 + 0.30x < 0.18(600 + x) = 108 + 0.18x. Subtract 72 and 0.18x: 0.12x < 36, so x < 300.
  5. 5.Combining gives 120 < x < 300 L. At 120 L the mixture is exactly 15%; at 300 L it is exactly 18%, so both are excluded.
  6. 6.For an interior check choose x = 200: acid amount is 132 L and volume is 800 L, giving 16.5%, which lies in the required range.
Example — Diluting with a weaker solution

Problem
640 L of 8% boric acid solution is diluted with x L of 2% solution. Require a final concentration more than 4% and less than 6%.

  1. 1.The boric acid amount is 0.08 × 640 + 0.02x = 51.2 + 0.02x. The total volume is 640 + x, positive for x ≥ 0.
  2. 2.The lower bound gives 51.2 + 0.02x > 0.04(640 + x) = 25.6 + 0.04x.
  3. 3.Subtract 25.6 and 0.02x: 25.6 > 0.02x, so x < 1280. Adding too much weaker solution would lower the concentration too far.
  4. 4.The upper bound gives 51.2 + 0.02x < 0.06(640 + x) = 38.4 + 0.06x.
  5. 5.Subtract 38.4 and 0.02x: 12.8 < 0.04x, so x > 320. Enough weaker solution must be added to get below 6%.
  6. 6.Therefore 320 < x < 1280 L. The lower endpoint gives 6% and the upper gives 4%; the endpoint order in x is opposite the concentration order because adding weaker solution lowers the concentration.
Example — Diluting with water

Problem
How much water must be added to 1125 L of 45% acid solution so that acid content is more than 25% and less than 30%?

  1. 1.Let x ≥ 0 be the added water volume. Water contributes no acid, so the acid amount stays 0.45 × 1125 = 506.25 L.
  2. 2.The concentration condition is 0.25 < 506.25/(1125 + x) < 0.30. The denominator is positive.
  3. 3.The lower bound gives 506.25 > 281.25 + 0.25x. Thus 225 > 0.25x and x < 900.
  4. 4.The upper bound gives 506.25 < 337.5 + 0.30x. Thus 168.75 < 0.30x and x > 562.5.
  5. 5.Combine: 562.5 < x < 900 L. Fractional litres are meaningful, so do not round to integers. Check x = 750: 506.25/1875 = 0.27, or 27%.
A percentage is not an amount

Do not add 12% and 30% to obtain a mixture concentration, and do not multiply an inequality by a denominator before checking its sign. Track substance amount and total solution volume separately. When adding water, only the denominator changes; the original substance amount stays constant.

A range from a stated formula

Sometimes a question supplies a formula and asks what a bound on its output says about its input. Treat the formula as a mathematical model, identify its constants, substitute, and solve. The interpretation must stay within the meaning and limitations of that model.

The source uses the historical age-ratio formula IQ = 100 × MA/CA, where MA is mental age and CA is chronological age, in the same age units. We use it only as a supplied algebraic relationship; it is not a method for making a modern clinical assessment of an individual.

Given age-ratio modelLaTeX
MA and CA are expressed in the same units; CA must be positive. IQ is the numerical output of this stated model.
Example — An inclusive formula range

Problem
Using the given age-ratio model, find MA when CA = 12 years and 80 ≤ IQ ≤ 140.

  1. 1.Substitute CA = 12: 80 ≤ (100/12)MA ≤ 140.
  2. 2.Multiply all three parts by positive 12/100: 9.6 ≤ MA ≤ 16.8.
  3. 3.The model gives an inclusive range [9.6, 16.8] years. Check MA = 12: the output is 100, within the stated range.
  4. 4.The calculation establishes a range in the supplied formula; it does not establish what any particular child understands or can do.

Check your understanding

Use the common region rather than either condition in isolation. For applied questions, identify the positive quantities before clearing fractions and check the endpoint meaning against the original requirement.

Quiz

Quick check

Solve 2 ≤ 3x − 4 ≤ 5.

Quick check

Solve 6 ≤ −3(2x − 4) < 12.

Quick check

What is the common set for x < 6 and x ≥ 2?

Quick check

Which system has no solution?

Quick check

Which Fahrenheit interval corresponds to 30 < C < 35?

Quick check

Why can we multiply a mixture inequality by 600 + x without reversing it when x ≥ 0?

Quick check

Adding water to a solution changes which quantities in the additive-volume model?

Practice Problems

Practice Problems
  1. Solve −3 ≤ 4 − 7x/2 ≤ 18 and check both endpoints.
  2. Solve −15 < 3(x − 2)/5 ≤ 0.
  3. Solve −12 < 4 − 3x/(−5) ≤ 2, simplifying the negative denominator first.
  4. Solve 7 ≤ (3x + 11)/2 ≤ 11.
  5. Solve 5x + 1 > −24 and 5x − 1 < 24; draw the intersection.
  6. Solve 2(x − 1) < x + 5 and 3(x + 2) > 2 − x; give the interval.
  7. Solve 3x − 7 > 2(x − 6) and 6 − x > 11 − 2x. Explain which lower bound is stronger.
  8. Solve 5(2x − 7) − 3(2x + 3) ≤ 0 and 2x + 19 ≤ 6x + 47.
  9. Convert an inclusive range 10 ≤ C ≤ 20 to Fahrenheit.
  10. Explain why the boric-acid dilution has both a minimum and a maximum added volume.
  11. Using the given age-ratio model with CA = 10, find MA when 90 ≤ IQ < 120.

Key Takeaways

Key Takeaways

• A double inequality abbreviates two requirements that must both hold. • Apply chain operations to every part, reversing every comparison for a negative multiplier or divisor. • The solution of an AND system is the intersection, which may be empty or a single point. • Positive temperature scale factors preserve endpoint order and inclusion. • Mixture concentration comes from total substance amount divided by total volume; use the domain to establish denominator positivity. • Interpret bounds in their original units and context before accepting an answer.

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Understanding and Solving Linear Inequalities

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Chapter Summary and Practice