Solving Multi-Step Inequalities

A multi-step inequality requires more than one operation to isolate the variable. Solving these inequalities combines several algebra skills:

• simplifying expressions
• combining like terms
• using the distributive property
• performing inverse operations
• working with variables on both sides
• reversing the inequality when multiplying or dividing by a negative number
• representing the final solution using a number line and interval notation

For example:

3x + 5 < 20

Subtract 5:

3x < 15

Divide by 3:

x < 5

The solution represents all real numbers less than 5, not just one value.

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Review of Inequality Symbols

The four main inequality symbols are:

The symbols ≤ and ≥ include the boundary value.

For example:

x < 4

does not include 4.

But:

x ≤ 4

does include 4.

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Solving Inequalities Like Equations

Most algebraic operations used to solve equations also work with inequalities.

You can:

• add the same number to both sides
• subtract the same number from both sides
• multiply both sides by the same positive number
• divide both sides by the same positive number

For example:

2x + 7 > 15

Subtract 7:

2x > 8

Divide by 2:

x > 4

The inequality direction remains unchanged because we divided by a positive number.

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The Negative-Number Rule

There is one critical difference between equations and inequalities:

When multiplying or dividing both sides by a negative number, reverse the inequality symbol.

For example:

−3x > 12

Divide by −3.

Because −3 is negative:

> becomes <

Therefore:

x < −4

This rule is essential when solving multi-step inequalities.

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Why the Inequality Reverses

Start with a true statement:

2 < 5

Multiply both sides by −1:

−2 and −5

On the number line:

−2 > −5

Therefore:

2 < 5

becomes:

−2 > −5

Multiplication by a negative reflects numbers across zero and reverses their order.

The same effect occurs when dividing by a negative number.

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Combining Like Terms

Before solving an inequality, simplify each side whenever possible.

Like terms have the same variable raised to the same power.

For example:

3x + 5x = 8x

and:

7 − 2 = 5

Consider:

3x + 2x + 4 > 19

Combine like terms:

5x + 4 > 19

Subtract 4:

5x > 15

Divide by 5:

x > 3

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Combining Constants

Consider:

6x + 7 − 3 > 28

Combine the constants:

6x + 4 > 28

Subtract 4:

6x > 24

Divide by 6:

x > 4

Simplifying first makes the remaining steps easier to see.

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Combining Several Like Terms

Solve:

4x + 3 + 2x − 7 ≤ 20

Combine variable terms:

4x + 2x = 6x

Combine constants:

3 − 7 = −4

So:

6x − 4 ≤ 20

Add 4:

6x ≤ 24

Divide by 6:

x ≤ 4

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Order of Operations

When simplifying an inequality, follow the usual order of operations.

A useful sequence is:

1. simplify parentheses
2. combine like terms on each side
3. move variable terms if necessary
4. move constants
5. isolate the variable
6. check whether the final multiplication or division uses a negative number
7. reverse the symbol if required

This systematic approach reduces errors.

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Two-Step Inequalities

Consider:

4x − 7 ≥ 13

Add 7:

4x ≥ 20

Divide by 4:

x ≥ 5

Number line:

• closed circle at 5
• shade right

Interval notation:

[5, ∞)

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Another Two-Step Example

Solve:

5x + 8 < 28

Subtract 8:

5x < 20

Divide by 5:

x < 4

Number line:

• open circle at 4
• shade left

Interval notation:

(−∞, 4)

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Multi-Step Inequality with a Negative Coefficient

Solve:

−4x + 7 > 19

Subtract 7:

−4x > 12

Divide by −4.

Reverse > to <:

x < −3

Number line:

• open circle at −3
• shade left

Interval notation:

(−∞, −3)

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Another Negative-Coefficient Example

Solve:

−5x − 8 ≤ 17

Add 8:

−5x ≤ 25

Divide by −5.

Reverse ≤ to ≥:

x ≥ −5

Interval notation:

[−5, ∞)

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The Distributive Property

Some multi-step inequalities contain parentheses.

Use the distributive property:

a(b + c) = ab + ac

For example:

3(x + 4) > 21

Distribute:

3x + 12 > 21

Subtract 12:

3x > 9

Divide by 3:

x > 3

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Distributing a Negative Number

Extra care is needed when the number outside the parentheses is negative.

Solve:

−2(x + 3) ≥ 10

Distribute:

−2x − 6 ≥ 10

Add 6:

−2x ≥ 16

Divide by −2 and reverse:

x ≤ −8

The inequality reverses during the division by −2, not simply because a negative number appeared earlier.

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Distributing Across Subtraction

Solve:

4(2x − 3) < 20

Distribute:

8x − 12 < 20

Add 12:

8x < 32

Divide by 8:

x < 4

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Combining Like Terms After Distributing

Solve:

2(x + 5) + 3x > 25

Distribute:

2x + 10 + 3x > 25

Combine like terms:

5x + 10 > 25

Subtract 10:

5x > 15

Divide by 5:

x > 3

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Variables on Both Sides

Multi-step inequalities may contain variables on both sides.

Consider:

5x + 3 > 2x + 12

Subtract 2x from both sides:

3x + 3 > 12

Subtract 3:

3x > 9

Divide by 3:

x > 3

The goal is to collect variable terms on one side.

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Choosing Which Variable Term to Move

Consider:

3x + 10 ≤ 7x − 6

We could subtract 3x:

10 ≤ 4x − 6

Add 6:

16 ≤ 4x

Divide by 4:

4 ≤ x

Usually it is clearer to write the variable first:

x ≥ 4

Notice that:

4 ≤ x

and:

x ≥ 4

mean exactly the same thing.

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Avoiding Negative Coefficients

Sometimes choosing which variable term to move can make the problem easier.

Consider:

2x + 8 > 5x − 4

Subtract 2x:

8 > 3x − 4

Add 4:

12 > 3x

Divide by 3:

4 > x

Rewrite:

x < 4

This approach avoids creating a negative coefficient.

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Example Producing a Negative Coefficient

The same problem can be solved another way:

2x + 8 > 5x − 4

Subtract 5x:

−3x + 8 > −4

Subtract 8:

−3x > −12

Divide by −3 and reverse:

x < 4

Both methods give the same answer.

This demonstrates that different correct algebraic approaches should produce the same solution set.

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Variables on Both Sides with a Negative Result

Solve:

7x + 4 ≤ 3x − 12

Subtract 3x:

4x + 4 ≤ −12

Subtract 4:

4x ≤ −16

Divide by 4:

x ≤ −4

Interval notation:

(−∞, −4]

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Simplifying Both Sides

Consider:

3x + 4 + 2x > x + 16

Combine like terms on the left:

5x + 4 > x + 16

Subtract x:

4x + 4 > 16

Subtract 4:

4x > 12

Divide by 4:

x > 3

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Parentheses on Both Sides

Solve:

2(x + 3) ≤ 4(x − 1)

Distribute:

2x + 6 ≤ 4x − 4

Subtract 2x:

6 ≤ 2x − 4

Add 4:

10 ≤ 2x

Divide by 2:

5 ≤ x

Therefore:

x ≥ 5

Interval notation:

[5, ∞)

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A More Challenging Example

Solve:

−3(2x − 4) > 2(x + 5)

Distribute:

−6x + 12 > 2x + 10

Subtract 2x:

−8x + 12 > 10

Subtract 12:

−8x > −2

Divide by −8 and reverse:

x < 1/4

Interval notation:

(−∞, 1/4)

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Inequalities with Fractions

Fractions do not change the basic method.

Solve:

x/3 + 4 > 9

Subtract 4:

x/3 > 5

Multiply by 3:

x > 15

Because 3 is positive, the inequality does not reverse.

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Clearing Fractions

Consider:

x/4 + x/2 ≤ 9

The least common denominator is 4.

Multiply every term by 4:

x + 2x ≤ 36

3x ≤ 36

Divide by 3:

x ≤ 12

Interval notation:

(−∞, 12]

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Clearing Fractions with a Negative Multiplier

Be careful if you choose to multiply an inequality by a negative number.

For example:

−x/3 > 4

Multiply by −3.

Because −3 is negative, reverse the symbol:

x < −12

A safer alternative is sometimes to first multiply by positive 3:

−x > 12

Then divide by −1 and reverse:

x < −12

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Decimal Inequalities

Decimals can be solved in the same way.

Solve:

0.5x + 2 > 7

Subtract 2:

0.5x > 5

Divide by 0.5:

x > 10

You can also multiply the entire inequality by 2:

x + 4 > 14

x > 10

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Interval Notation

Interval notation is a compact way to represent a set of numbers.

For example:

x > 3

can be written:

(3, ∞)

This means all real numbers greater than 3.

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Parentheses in Interval Notation

Parentheses indicate that an endpoint is not included.

For:

x > 5

interval notation is:

(5, ∞)

For:

x < 5

interval notation is:

(−∞, 5)

These correspond to open circles on a number line.

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Square Brackets in Interval Notation

Square brackets indicate that an endpoint is included.

For:

x ≥ 5

interval notation is:

[5, ∞)

For:

x ≤ 5

interval notation is:

(−∞, 5]

These correspond to closed circles at the finite endpoint.

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Infinity Always Uses Parentheses

Infinity is not a number that can be reached or included.

Therefore:

∞ and −∞ always use parentheses.

Correct:

[4, ∞)

Correct:

(−∞, 7]

Incorrect:

[4, ∞]

Incorrect:

[−∞, 7]

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Inequality, Number Line, and Interval Notation

These are three different ways of representing the same solution set.

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Representing a Solution on a Number Line

Suppose the solution is:

x ≤ −2

The number-line graph has:

• a closed circle at −2
• shading extending left

Interval notation:

(−∞, −2]

Now consider:

x > −2

The graph has:

• an open circle at −2
• shading extending right

Interval notation:

(−2, ∞)

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Verifying by Substitution

Substitution can be used to check whether particular values satisfy the original inequality.

Suppose:

3x + 4 < 19

We solve:

3x < 15

x < 5

Choose a value from the solution set:

x = 4

Substitute into the original inequality:

3(4) + 4 < 19

16 < 19

True.

Therefore, x = 4 is a valid solution.

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Testing a Value Outside the Solution Set

Using the same inequality:

3x + 4 < 19

with solution:

x < 5

try:

x = 6

3(6) + 4 < 19

22 < 19

False.

This is exactly what we expect because 6 is outside the solution set.

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Testing the Boundary

For:

3x + 4 < 19

the boundary is:

x = 5

Substitute:

3(5) + 4 < 19

19 < 19

False.

Therefore, 5 is correctly excluded.

If the original inequality had been:

3x + 4 ≤ 19

then:

19 ≤ 19

would be true, so 5 would be included.

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Why Testing One Value Is Not a Complete Proof

Testing one number can show that a particular number works, but it does not prove that every number in the proposed interval works.

The algebraic solution establishes the complete solution set.

Substitution is useful for:

• checking the boundary
• checking a typical solution
• detecting a reversed inequality
• finding arithmetic mistakes

It should support the algebra rather than replace it.

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Special Case: All Real Numbers

Sometimes the variable disappears during simplification.

Consider:

2(x + 3) < 2x + 10

Distribute:

2x + 6 < 2x + 10

Subtract 2x:

6 < 10

This statement is always true.

Therefore, every real number is a solution.

We can write:

all real numbers

Interval notation:

(−∞, ∞)

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Why Every Number Works

Test any value.

For x = 100:

2(100 + 3) < 2(100) + 10

206 < 210

True.

For x = −50:

2(−50 + 3) < 2(−50) + 10

−94 < −90

True.

The inequality is true regardless of x.

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Special Case: No Solution

Consider:

3(x + 2) > 3x + 10

Distribute:

3x + 6 > 3x + 10

Subtract 3x:

6 > 10

This statement is always false.

Therefore:

no solution

No real value of x can make the original inequality true.

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Recognizing Special Cases

If the variable disappears, examine the remaining statement.

If the statement is true:

all real numbers

Example:

4 < 9

If the statement is false:

no solution

Example:

12 < 5

Do not try to continue solving for x after the variable has disappeared.

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Number-Line Representation of Special Cases

For all real numbers, the entire number line is shaded.

Interval notation:

(−∞, ∞)

For no solution, no part of the number line is shaded.

The solution set may also be represented by:

∅

which means the empty set.

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Real-World Multi-Step Inequalities

Multi-step inequalities are useful for situations involving:

• budgets
• fees
• wages
• transportation
• production
• capacity
• temperature ranges
• minimum requirements

For example, suppose a taxi charges:

• $5 starting fee
• $2 per kilometre

A passenger has no more than $25.

If d is the distance:

5 + 2d ≤ 25

Subtract 5:

2d ≤ 20

Divide by 2:

d ≤ 10

The passenger can travel at most 10 km under this simplified model.

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Budget Example

A student has:

$80

to spend.

There is a fixed delivery charge of:

$8

and each item costs:

$12.

If n is the number of items:

12n + 8 ≤ 80

Subtract 8:

12n ≤ 72

Divide by 12:

n ≤ 6

Because n represents a number of items, realistic solutions are non-negative whole numbers up to 6.

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Minimum Earnings Example

A worker receives:

$50

as a fixed payment plus:

$18 per hour.

The worker wants to earn at least:

$230.

Let h be the number of hours:

50 + 18h ≥ 230

Subtract 50:

18h ≥ 180

Divide by 18:

h ≥ 10

The worker must work at least 10 hours.

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Comparing Two Plans

Suppose Plan A costs:

20 + 5x

and Plan B costs:

50 + 2x

We want to know when Plan A is cheaper:

20 + 5x < 50 + 2x

Subtract 2x:

20 + 3x < 50

Subtract 20:

3x < 30

Divide by 3:

x < 10

Plan A is cheaper when x is less than 10, assuming x represents values allowed by the context.

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A Complete Multi-Step Example

Solve:

4(2x − 3) + 5 ≤ 3x + 18

Distribute:

8x − 12 + 5 ≤ 3x + 18

Combine like terms:

8x − 7 ≤ 3x + 18

Subtract 3x:

5x − 7 ≤ 18

Add 7:

5x ≤ 25

Divide by 5:

x ≤ 5

Interval notation:

(−∞, 5]

Number line:

• closed circle at 5
• shade left

Verification using x = 5:

4[2(5) − 3] + 5 ≤ 3(5) + 18

4(7) + 5 ≤ 33

33 ≤ 33

True.

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Complete Example with a Negative Coefficient

Solve:

3(x − 4) − 5x > 8

Distribute:

3x − 12 − 5x > 8

Combine:

−2x − 12 > 8

Add 12:

−2x > 20

Divide by −2 and reverse:

x < −10

Interval notation:

(−∞, −10)

Check x = −11:

3(−11 − 4) − 5(−11) > 8

3(−15) + 55 > 8

−45 + 55 > 8

10 > 8

True.

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Complete Example with Variables on Both Sides

Solve:

5(x + 2) ≤ 3(x + 6)

Distribute:

5x + 10 ≤ 3x + 18

Subtract 3x:

2x + 10 ≤ 18

Subtract 10:

2x ≤ 8

Divide by 2:

x ≤ 4

Interval notation:

(−∞, 4]

Check x = 4:

5(4 + 2) ≤ 3(4 + 6)

30 ≤ 30

True.

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Complete Example with Fractions

Solve:

x/2 + x/4 − 3 > 6

Multiply every term by 4:

2x + x − 12 > 24

Combine:

3x − 12 > 24

Add 12:

3x > 36

Divide by 3:

x > 12

Interval notation:

(12, ∞)

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Complete Example Producing All Real Numbers

Solve:

4(x − 2) + 3 < 4x

Distribute:

4x − 8 + 3 < 4x

Combine:

4x − 5 < 4x

Subtract 4x:

−5 < 0

This is always true.

Therefore:

all real numbers

Interval notation:

(−∞, ∞)

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Complete Example Producing No Solution

Solve:

2(x + 5) > 2x + 14

Distribute:

2x + 10 > 2x + 14

Subtract 2x:

10 > 14

This is false.

Therefore:

no solution

or:

∅

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A Reliable Multi-Step Strategy

When solving a multi-step inequality:

1. Simplify both sides.
   Distribute and combine like terms.
2. Move variable terms to one side.
   Add or subtract variable terms.
3. Move constants to the other side.
   Use inverse operations.
4. Isolate the variable.
   Multiply or divide as necessary.
5. Check for a negative multiplication or division.
   Reverse the inequality if required.
6. Write the solution clearly.
7. Convert to interval notation.
8. Represent it on a number line.
9. Verify using substitution.
10. Interpret the answer in context if the problem describes a real situation.

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Common Mistakes

Not simplifying first

Combining like terms early usually makes the inequality easier to solve.

Distributing incorrectly

For:

−3(x − 4)

the correct result is:

−3x + 12

The negative must multiply every term.

Combining unlike terms

For example:

3x + 5

cannot be simplified to 8x.

Forgetting to perform operations on both sides

An inequality must remain balanced just like an equation.

Reversing the inequality when adding or subtracting

The sign reverses only when multiplying or dividing by a negative number.

Forgetting to reverse after dividing by a negative

For:

−2x > 8

the correct solution is:

x < −4

Reversing because a negative number appears somewhere

A negative number appearing in the inequality does not automatically cause reversal.

Using brackets with infinity

Infinity always uses parentheses.

Confusing parentheses and brackets

Parentheses exclude a finite endpoint. Square brackets include it.

Forgetting special cases

If the variable disappears, the answer may be all real numbers or no solution.

Testing only the simplified inequality

When verifying, substitute into the original inequality.

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Key Terms

Multi-step inequality — inequality requiring several algebraic operations to isolate the variable.

Like terms — terms with identical variable parts that can be combined.

Distributive property — rule used to multiply a factor across terms inside parentheses.

Inverse operation — operation used to undo another operation.

Solution set — collection of all values satisfying an inequality.

Boundary value — value separating solutions from non-solutions.

Interval notation — compact notation used to describe continuous sets of real numbers.

Parenthesis — interval symbol showing that a finite endpoint is excluded.

Square bracket — interval symbol showing that a finite endpoint is included.

Infinity — concept describing values extending without bound; always written with a parenthesis in interval notation.

Substitution — replacing a variable with a particular value.

Verification — checking whether a proposed solution satisfies the original inequality.

All real numbers — solution set containing every real number, written (−∞, ∞).

No solution — situation in which no value satisfies an inequality.

Empty set (∅) — notation representing a solution set containing no elements.

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Key Takeaways

• Multi-step inequalities combine several algebraic skills.
• Simplify expressions before isolating the variable.
• Use the distributive property correctly when parentheses are present.
• Combine only like terms.
• Variables may need to be moved from one side of the inequality to the other.
• Addition and subtraction do not reverse the inequality.
• Multiplication or division by a positive number does not reverse the inequality.
• Multiplication or division by a negative number reverses the inequality.
• Different correct algebraic methods should produce the same solution set.
• Solutions can be represented using inequalities, number lines, and interval notation.
• Open endpoints correspond to < or >.
• Closed endpoints correspond to ≤ or ≥.
• Infinity and negative infinity always use parentheses.
• Substitution can help verify a solution.
• Checking the boundary is especially useful for determining whether it should be included.
• If the variable disappears and a true statement remains, the solution is all real numbers.
• If the variable disappears and a false statement remains, there is no solution.
• Real-world contexts may further restrict a mathematical solution set.
• A complete solution should clearly communicate the algebra, final inequality, interval notation, and graphical interpretation.

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Check Your Understanding

1. Simplify and solve: 3x + 5x − 4 > 20. Write the answer in interval notation.
2. Solve: 5x + 7 ≤ 32. Describe its number-line representation.
3. Solve: −4x + 6 > 22. Explain exactly when and why the inequality reverses.
4. Solve: 3(x + 5) ≥ 27. Give the inequality and interval notation.
5. Solve: −2(x − 4) < 16. Verify your answer using substitution.
6. Solve: 5x + 8 > 2x + 20. Represent the solution on a number line and in interval notation.
7. Solve: 3x − 7 ≤ 6x + 8.
8. Solve: 4(x − 2) + 3x > 20.
9. Solve: −3(2x + 5) ≥ 4x − 5. Write the answer in interval notation.
10. Solve: x/3 + x/6 ≤ 8.
11. Solve: 2(x + 4) < 2x + 12. Explain why the answer is all real numbers.
12. Solve: 5(x − 1) > 5x + 2. Explain why the inequality has no solution.
13. A taxi charges a $6 starting fee plus $3 per kilometre. You have no more than $42. Write and solve an inequality to determine the maximum distance you can travel.
14. Plan A costs $25 + $4x, while Plan B costs $55 + $2x. Determine when Plan A is cheaper than Plan B. Express the answer as an inequality and interval.
15. Solve and fully communicate the solution to 4(2x − 3) − 5x ≤ −2(x + 6) + 9. Show the simplification, explain any inequality reversal if needed, give the final inequality, write the solution in interval notation, describe its number-line graph, and verify one value by substitution.