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C2.4 · Solve and graph two-operation whole-number inequalities

Learn to solve and graph two-operation whole-number inequalities through clear examples and targeted practice.

Ontario Grade 6 Mathematics

Algebra

Use inverse operations to find and show every whole-number solution

An equation says that two amounts are equal. An inequality compares amounts that may not be equal. For example, a number of counters could be less than 20. In this lesson, you will use two operations to describe a whole-number amount, find which values make the comparison true, and graph those values. A whole number is 0, 1, 2, 3, and so on. We will use only whole-number values for the unknown number.

What you will learn

  • Explain what an inequality tells us.
  • Use inverse operations to solve a two-operation whole-number inequality.
  • Check whether a number makes an inequality true.
  • Graph the whole-number solutions on a number line.

1. Read an inequality

Imagine there are 3 identical bags. Each bag holds the same whole number of counters. There are also 4 loose counters. The total is less than 22 counters. The unknown number of counters in each bag is represented by nn.
The phrase “less than” tells us to compare the total with 22. Three bags of nn counters make 3n3n counters. Adding the 4 loose counters gives 3n+43n+4. The complete comparison is an inequality.
An inequality uses a comparison sign. The sign << means “is less than,” and >> means “is greater than.” The signs ≤\leq and ≥\geq mean “is less than or equal to” and “is greater than or equal to.” The line under the sign means that equality is included.
This situation uses two operations on nn: first multiply by 3, then add 4. That is why it is called a two-operation inequality. The numbers and operations describe the situation; the sign describes which totals are allowed.
3n+4<223n+4<22
  • Whole-number values begin at 0 and continue by ones.
  • The comparison sign tells which values are allowed.
  • A two-operation inequality has two operations applied to the unknown.

2. Undo the operations in reverse order

To solve, find which whole numbers make the inequality true. Undo the operations that act on the unknown. Use the opposite operation each time: addition is undone by subtraction, and multiplication is undone by division.
Undo the operations in reverse order. In 3n+43n+4, addition by 4 happens after multiplication by 3. So first subtract 4 from both sides of the comparison. Then divide both sides by 3. Doing the same operation to both sides keeps the comparison fair.
For the whole-number examples in this lesson, we subtract or divide by positive numbers. The comparison direction stays the same. After solving, list or graph the whole-number values that fit. Do not include values that are not whole numbers.
3n+4<22  ⇒  3n<18  ⇒  n<63n+4<22\;\Rightarrow\;3n<18\;\Rightarrow\;n<6
  • Use the opposite operation to undo an operation.
  • Undo the last operation first.
  • Check which whole numbers satisfy the result.

3. Graph the whole-number solutions

A number line places numbers in order. For an inequality about whole numbers, show the allowed whole-number values as separate points. A point at an allowed value is filled in. A value that is not allowed is not shown as a solution.
For a result such as n<6n<6, the whole-number solutions are 0, 1, 2, 3, 4, and 5. Six is not included because the sign means strictly less than. The number line in the first example shows each allowed value as a filled point.
For a result such as n≥5n\geq5, the whole-number solutions begin at 5 and continue to the right: 5, 6, 7, and so on. The number line shows the start of this set and an arrow direction through the larger whole numbers. A graph is a way to display the same answer as the inequality.
n<6⇒n∈{0,1,2,3,4,5}n<6 \Rightarrow n∈\{0,1,2,3,4,5\}
  • A filled point means that value is included.
  • For whole-number solutions, show whole-number points, not every point between them.
  • The graph and the inequality must describe the same values.

4. Check a solution

A solution is a value that makes the original inequality true. Check by putting the value in place of the unknown and comparing the two sides.
It is useful to check a value near the edge of the solution set. If the inequality says n<6n<6, then 5 should work and 6 should not. This helps catch a mistake about whether the endpoint is included.
Keep the original situation in mind. If the unknown stands for a number of counters, a negative number or a fraction of a counter does not belong to the whole-number solutions. The inequality may describe a wider comparison, but the situation restricts the answers to whole numbers.
  • Substitute a value into the original inequality to check it.
  • Check the boundary value to confirm whether it is included.
  • Use the whole-number restriction when listing or graphing answers.

Worked example

Three bags and loose counters

Three bags hold the same whole number of counters. There are 4 more loose counters. The total is less than 22. How many counters could be in each bag?
Whole-number solutions for n < 6
Whole-number solutions for n < 6Filled points show every allowed whole-number value from 0 through 5. Six is not included.0123456+5 whole-number counter

Filled points show every allowed whole-number value from 0 through 5. Six is not included.

  1. Write the comparison
    Let nn be the whole number of counters in one bag. The three bags hold 3n3n counters. Adding 4 gives the total, and “less than 22” sets the comparison.
    3n+4<223n+4<22
  2. Undo addition
    Subtract 4 from both sides. This removes the 4 loose counters from the total while keeping the comparison fair.
    3n<183n<18
  3. Undo multiplication
    Divide both sides by 3. This finds the amount in one bag. Since 18 divided by 3 is 6, the unknown must be less than 6.
    n<6n<6
  4. List and check
    The whole numbers less than 6 are 0 through 5. For example, 5 gives 19 counters altogether, which is less than 22. Six gives 22, which is not less than 22.
    3(5)+4=19<223(5)+4=19<22
Answer: There could be 0, 1, 2, 3, 4, or 5 counters in each bag.
Check: The whole-number solutions stop at 5 because 6 makes the total equal to 22, not less than 22.

Worked example

A number doubled, then reduced

A whole number is doubled, then 3 is taken away. The result is at least 7. Find and graph the whole-number solutions.
Whole-number solutions for x ≥ 5
Whole-number solutions for x ≥ 5The filled point at 5 is included. The arrow to the right stands for all greater whole numbers.45678+3 whole numbers

The filled point at 5 is included. The arrow to the right stands for all greater whole numbers.

  1. Write the inequality
    Let xx be the whole number. Doubling it gives 2x2x. Taking away 3 gives 2x−32x-3. “At least 7” includes 7 and every greater value.
    2x−3≥72x-3\geq7
  2. Undo subtraction
    Add 3 to both sides to undo taking away 3. The comparison remains greater than or equal to.
    2x≥102x\geq10
  3. Undo doubling
    Divide both sides by 2 to find the values of xx. The result includes 5 and all greater whole numbers.
    x≥5x\geq5
  4. Check the first allowed value
    At 5, doubling and subtracting 3 gives 7, so the “at least 7” condition is met. Values 6 and above also work because doubling a larger whole number gives a larger result.
    2(5)−3=72(5)-3=7
Answer: The solutions are 5, 6, 7, and every greater whole number.
Check: Five is included because the comparison is “at least,” shown by ≥\geq.

Worked example

A whole number multiplied and increased

Four times a whole number, increased by 2, is no more than 26. Find the whole-number solutions.
Whole-number solutions for m ≤ 6
Whole-number solutions for m ≤ 6Filled points show the whole-number values from 0 through 6. Seven is not included.01234567+6 whole numbers

Filled points show the whole-number values from 0 through 6. Seven is not included.

  1. Turn the words into symbols
    Let mm be the whole number. Four times the number is 4m4m. “Increased by 2” means add 2. “No more than 26” means 26 or less.
    4m+2≤264m+2\leq26
  2. Undo addition
    Subtract 2 from both sides. This leaves the multiplied amount by itself on the left.
    4m≤244m\leq24
  3. Undo multiplication
    Divide both sides by 4. The number must be no more than 6, so 6 is included along with smaller whole numbers.
    m≤6m\leq6
  4. Check both ends
    Zero is allowed because 4(0)+2=24(0)+2=2, which is no more than 26. Six is also allowed because 4(6)+2=264(6)+2=26. Seven is not allowed because the result would be 30.
    2≤26,26≤26,30>262\leq26,\quad26\leq26,\quad30>26
Answer: The solutions are 0, 1, 2, 3, 4, 5, and 6.
Check: The endpoint 6 is included by ≤\leq, and 7 fails the original comparison.

Common mistakes and how to avoid them

Undoing the first operation you see instead of the last one.
Correction: Look at the operations in the expression. Undo the operation closest to the outside first. For 3n+43n+4, subtract 4 before dividing by 3.
Changing a less-than sign to less than or equal to when solving.
Correction: Keep the original comparison meaning. A strict “less than” does not include its boundary value.
Listing only the number at the end of the solving steps.
Correction: An inequality usually has many solutions. List or graph every whole-number value that fits.
Including fractions or negative numbers when the unknown is a whole-number amount.
Correction: Use only whole numbers starting at 0 when listing the solutions for these situations.

Lesson summary

  • Translate words such as “less than,” “at least,” and “no more than” into comparison signs.
  • Undo the operations in reverse order, using the opposite operation on both sides.
  • Keep the comparison meaning and identify whether the endpoint is included.
  • Check the answer in the original inequality and graph all allowed whole-number values.

Check your understanding

Question 1

Solve 3p+1<163p+1<16 for whole-number values of pp. Which set is correct?
  1. p=0,1,2,3,4p=0,1,2,3,4
  2. p=0,1,2,3,4,5p=0,1,2,3,4,5
  3. p=5,6,7,8p=5,6,7,8
  4. p=0,1,2,3,4,5,6p=0,1,2,3,4,5,6
Show answer and explanation
p=0,1,2,3,4p=0,1,2,3,4
Subtract 1 to get 3p<153p<15, then divide by 3 to get p<5p<5. The whole numbers less than 5 are 0 through 4.

Question 2

Solve 2q+5≥152q+5\geq15. Which statement is true?
  1. q≥5q\geq5, so 5 and greater whole numbers work.
  2. q>5q>5, so 5 does not work.
  3. q≤5q\leq5, so 0 through 5 work.
  4. q≥10q\geq10, so 10 and greater whole numbers work.
Show answer and explanation
q≥5q\geq5, so 5 and greater whole numbers work.
Subtract 5 to get 2q≥102q\geq10, then divide by 2 to get q≥5q\geq5. Five is included.

Question 3

For 4r−2≤144r-2\leq14, which whole-number value is the greatest solution?
  1. 3
  2. 4
  3. 5
  4. 6
Show answer and explanation
4
Add 2 to get 4r≤164r\leq16, then divide by 4 to get r≤4r\leq4. The greatest whole-number solution is 4.

Key terms

Inequality
A comparison that shows one amount is less than, greater than, or possibly equal to another amount.
Whole number
One of the numbers 0, 1, 2, 3, and so on.
Unknown
A number represented by a letter when its value is not yet known.
Solution
A value that makes an equation or inequality true.
Number line
A line that shows numbers in order from smaller to larger.
Endpoint
The boundary value named in an inequality, such as 5 in x≥5x\geq5.

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