1.3.7 Inequalities with a modulus sign (A-level only)

What you'll learn

How to interpret a modulus as a distance.
How to solve inequalities of the forms $∣ f (x) ∣ < a$ and $∣ f (x) ∣ > a$ .
How to handle inclusive inequalities, variable expressions on both sides and special cases.
How graphs and number lines can help you check your solutions.

What does a modulus mean?

The modulus or absolute value of a real number is its distance from zero on a number line. It is written using vertical bars.

For example, both 4 and $- 4$ are a distance of 4 from zero, so $∣4∣ = 4$ and $∣ - 4∣ = 4$ .

Definition

The modulus function

For any real number $x$ ,

\begin{cases} x, & x\ge 0,\\ -x, & x<0. \end{cases}

The output of a modulus is therefore always non-negative.

The expression $∣ x - a ∣$ represents the distance between $x$ and $a$ . This distance interpretation is the key to understanding modulus inequalities.

Example

Interpreting a modulus as a distance

Solve $∣ x - 5∣ = 2$ .

The expression $∣ x - 5∣$ is the distance between $x$ and 5.
A number at distance 2 from 5 could lie either to its left or to its right.
Therefore $x = 5 - 2$ or $x = 5 + 2$ , giving $x = 3$ or $x = 7$ .

Less-than modulus inequalities

Consider an inequality of the form

∣ x - a ∣ < b,

where $b > 0$ .

It says that the distance between $x$ and $a$ is less than $b$ . Therefore, $x$ must lie between the two boundary values $a - b$ and $a + b$ .

Key Idea

The inside rule

If $b > 0$ , then

\quad\Longleftrightarrow\quad a-b<x<a+b.

Equivalently,

\quad\Longleftrightarrow\quad -b<u<b.

A less-than modulus inequality produces an interval inside two boundaries.

Example

Solving a less-than inequality

Solve $∣2 x - 1∣ < 7$ .

Replace the modulus inequality with a double inequality:
$- 7 < 2 x - 1 < 7.$
Add 1 to all three parts:
$- 6 < 2 x < 8.$
Divide all three parts by 2:
$- 3 < x < 4.$

The inequality $∣ x - 2∣ < 3$ can also be understood graphically. The graph of $y = ∣ x - 2∣$ lies below $y = 3$ between their two intersection points.

Graph and number-line representations of the solution to the inequality absolute value of x minus 2 is less than 3

Greater-than modulus inequalities

Now consider

∣ x - a ∣ > b,

where $b > 0$ .

This says that the distance between $x$ and $a$ is greater than $b$ . The solution lies outside the two boundary values.

Key Idea

The outside rule

If $b > 0$ , then

\quad\Longleftrightarrow\quad x<a-b \quad\text{or}\quad x>a+b.

Equivalently,

\quad\Longleftrightarrow\quad u<-b \quad\text{or}\quad u>b.

A greater-than modulus inequality usually produces two separate regions joined by or.

Example

Solving a greater-than inequality

Solve $∣3 x + 2∣ > 8$ .

Split the inequality into its two possible cases:
$\quad\text{or}\quad 3x+2>8.$
Solve the first inequality:
$\quad\Rightarrow\quad x<-\frac{10}{3}.$
Solve the second inequality:
$\quad\Rightarrow\quad x>2.$
Combine the two regions:
$x<−103orx>2.x<-\frac{10}{3} \quad\text{or}\quad x>2.$

Common Mistake

Using and instead of or

For $∣ u ∣ > b$ , the conditions are $u < - b$ or $u > b$ . They cannot both hold at once, so joining them with “and” would give no solutions.

Inclusive inequalities

The same rules apply when the inequality includes equality.

If $b≥0b\ge 0$ , then

∣u∣≤b⟺−b≤u≤b,|u|\le b \quad\Longleftrightarrow\quad -b\le u\le b,

and

∣u∣≥b⟺u≤−boru≥b.|u|\ge b \quad\Longleftrightarrow\quad u\le-b \quad\text{or}\quad u\ge b.

On a number line, use a closed circle when an endpoint is included by $≤\le$ or $≥\ge$ . Use an open circle when it is excluded by $<$ or $>$ .

Example

Including the boundary values

Solve $∣5−2x∣≤9|5-2x|\le 9$ .

Write the equivalent double inequality:
$−9≤5−2x≤9.-9\le 5-2x\le 9.$
Subtract 5 throughout:
$−14≤−2x≤4.-14\le -2x\le 4.$
Divide throughout by $- 2$ . Dividing an inequality by a negative number reverses both inequality signs:
$7≥x≥−2.7\ge x\ge -2.$
Put the bounds in increasing order:
$−2≤x≤7.-2\le x\le 7.$

Common Mistake

Forgetting to reverse inequality signs

When you multiply or divide an inequality by a negative quantity, reverse its direction. In a double inequality, both signs must be reversed.

When the other side contains a variable

You may meet an inequality such as

∣ f (x) ∣ < g (x) .

The right-hand side needs careful attention because a modulus is always non-negative. In particular, $∣ f (x) ∣ < g (x)$ can hold only where $g (x) > 0$ .

When $g (x) > 0$ , you can use

- g (x) < f (x) < g (x) .

Example

Comparing a modulus with a variable expression

Solve $∣ x - 1∣ < x + 3$ .

Since $∣x−1∣≥0|x-1|\ge 0$ , the right-hand side must be positive:
$\quad\Rightarrow\quad x>-3.$
Apply the inside rule:
$- (x + 3) < x - 1 < x + 3.$
Solve the left-hand inequality:
$\quad\Rightarrow\quad -2<2x \quad\Rightarrow\quad x>-1.$
The right-hand inequality simplifies to $- 1 < 3$ , which is true for every $x$ . Combining this with the restrictions gives
$x > - 1.$

Common Mistake

Check the sign of the right-hand side

You may use $∣f(x)∣<g(x)⟺−g(x)<f(x)<g(x)|f(x)|<g(x)\Longleftrightarrow-g(x)<f(x)<g(x)$ only where $g (x) > 0$ . Ignoring this condition can introduce invalid solutions.

Squaring both sides

If both sides are known to be non-negative, squaring preserves their order. This gives

\quad\Longleftrightarrow\quad f(x)^2<g(x)^2,

provided $g (x) > 0$ .

This method can be useful when both sides contain expressions, although you must then solve the resulting polynomial inequality carefully.

Example

Solving by squaring

Solve $∣x+1∣≤2x+1|x+1|\le 2x+1$ .

The right-hand side must be non-negative:
$2x+1≥0⇒x≥−12.2x+1\ge 0 \quad\Rightarrow\quad x\ge-\frac12.$
Under this restriction, square both sides:
$(x+1)2≤(2x+1)2.(x+1)^2\le(2x+1)^2.$
Expand and collect terms:
$x2+2x+1≤4x2+4x+1⇒0≤3x2+2x.x^2+2x+1\le4x^2+4x+1 \quad\Rightarrow\quad 0\le3x^2+2x.$
Factorise:
$x(3x+2)≥0.x(3x+2)\ge0.$
This gives $x≤−23x\le-\frac23$ or $x≥0x\ge0$ .
Intersect this with the earlier restriction $x≥−12x\ge-\frac12$ . The first region is removed, leaving
$x≥0.x\ge0.$

Tip

Check with a test value

After finding an interval, substitute one value from inside it and one from outside it into the original inequality. This quickly detects reversed signs or an incorrect choice of region.

Special right-hand sides

Because $∣f(x)∣≥0|f(x)|\ge0$ , comparisons with zero or a negative number can often be decided immediately.

$∣ f (x) ∣ < 0$ has no solutions.
$∣f(x)∣≤0|f(x)|\le0$ means $f (x) = 0$ .
$∣ f (x) ∣ > 0$ means $f(x)≠0f(x)\neq0$ .
$∣f(x)∣≥0|f(x)|\ge0$ is true for every value where $f (x)$ is defined.
If $c < 0$ , then $∣ f (x) ∣ > c$ is always true, while $∣ f (x) ∣ < c$ is impossible.

These observations can save considerable algebra.

Using graphs

To solve $∣ f (x) ∣ < g (x)$ graphically, sketch or plot $y = ∣ f (x) ∣$ and $y = g (x)$ . The solution consists of the $x$ -values where the modulus graph lies below the other graph.

For $∣ f (x) ∣ > g (x)$ , select the regions where the modulus graph lies above. Intersection points give the boundary values, and the original inequality tells you whether those boundaries are included.

Exam technique

In the exam

Identify whether the solution should be inside an interval for $<$ or $≤\le$ , or outside for $>$ or $≥\ge$ .
If the right-hand side is not a positive constant, check its sign before removing the modulus or squaring.
Solve every resulting inequality carefully, reversing signs when multiplying or dividing by a negative quantity.
State separate regions using or, and include endpoints only when the original inequality includes equality.
Test a value in each proposed region in the original inequality.

Self review

Check yourself

Can you solve $∣4 x - 3∣ < 5$ and show the answer on a number line?
How does the solution of $∣x+2∣≥6|x+2|\ge6$ differ from that of $∣x+2∣≤6|x+2|\le6$ ?
Why must you check the sign of $x + 4$ before rewriting $∣2 x - 1∣ < x + 4$ ?

What you'll learn

How to interpret a modulus as a distance.
How to solve inequalities of the forms $∣ f (x) ∣ < a$ and $∣ f (x) ∣ > a$ .
How to handle inclusive inequalities, variable expressions on both sides and special cases.
How graphs and number lines can help you check your solutions.

What does a modulus mean?

The modulus or absolute value of a real number is its distance from zero on a number line. It is written using vertical bars.

For example, both 4 and $- 4$ are a distance of 4 from zero, so $∣4∣ = 4$ and $∣ - 4∣ = 4$ .

Definition

The modulus function

For any real number $x$ ,

\begin{cases} x, & x\ge 0,\\ -x, & x<0. \end{cases}

The output of a modulus is therefore always non-negative.

The expression $∣ x - a ∣$ represents the distance between $x$ and $a$ . This distance interpretation is the key to understanding modulus inequalities.

Example

Interpreting a modulus as a distance

Solve $∣ x - 5∣ = 2$ .

The expression $∣ x - 5∣$ is the distance between $x$ and 5.
A number at distance 2 from 5 could lie either to its left or to its right.
Therefore $x = 5 - 2$ or $x = 5 + 2$ , giving $x = 3$ or $x = 7$ .

Less-than modulus inequalities

Consider an inequality of the form

∣ x - a ∣ < b,

where $b > 0$ .

It says that the distance between $x$ and $a$ is less than $b$ . Therefore, $x$ must lie between the two boundary values $a - b$ and $a + b$ .

Key Idea

The inside rule

If $b > 0$ , then

\quad\Longleftrightarrow\quad a-b<x<a+b.

Equivalently,

\quad\Longleftrightarrow\quad -b<u<b.

A less-than modulus inequality produces an interval inside two boundaries.

Example

Solving a less-than inequality

Solve $∣2 x - 1∣ < 7$ .

Replace the modulus inequality with a double inequality:
$- 7 < 2 x - 1 < 7.$
Add 1 to all three parts:
$- 6 < 2 x < 8.$
Divide all three parts by 2:
$- 3 < x < 4.$

The inequality $∣ x - 2∣ < 3$ can also be understood graphically. The graph of $y = ∣ x - 2∣$ lies below $y = 3$ between their two intersection points.

Graph and number-line representations of the solution to the inequality absolute value of x minus 2 is less than 3

Greater-than modulus inequalities

Now consider

∣ x - a ∣ > b,

where $b > 0$ .

This says that the distance between $x$ and $a$ is greater than $b$ . The solution lies outside the two boundary values.

Key Idea

The outside rule

If $b > 0$ , then

\quad\Longleftrightarrow\quad x<a-b \quad\text{or}\quad x>a+b.

Equivalently,

\quad\Longleftrightarrow\quad u<-b \quad\text{or}\quad u>b.

A greater-than modulus inequality usually produces two separate regions joined by or.

Example

Solving a greater-than inequality

Solve $∣3 x + 2∣ > 8$ .

Split the inequality into its two possible cases:
$\quad\text{or}\quad 3x+2>8.$
Solve the first inequality:
$\quad\Rightarrow\quad x<-\frac{10}{3}.$
Solve the second inequality:
$\quad\Rightarrow\quad x>2.$
Combine the two regions:
$x<−103orx>2.x<-\frac{10}{3} \quad\text{or}\quad x>2.$

Common Mistake

Using and instead of or

For $∣ u ∣ > b$ , the conditions are $u < - b$ or $u > b$ . They cannot both hold at once, so joining them with “and” would give no solutions.

Inclusive inequalities

The same rules apply when the inequality includes equality.

If $b≥0b\ge 0$ , then

∣u∣≤b⟺−b≤u≤b,|u|\le b \quad\Longleftrightarrow\quad -b\le u\le b,

and

∣u∣≥b⟺u≤−boru≥b.|u|\ge b \quad\Longleftrightarrow\quad u\le-b \quad\text{or}\quad u\ge b.

On a number line, use a closed circle when an endpoint is included by $≤\le$ or $≥\ge$ . Use an open circle when it is excluded by $<$ or $>$ .

Example

Including the boundary values

Solve $∣5−2x∣≤9|5-2x|\le 9$ .

Write the equivalent double inequality:
$−9≤5−2x≤9.-9\le 5-2x\le 9.$
Subtract 5 throughout:
$−14≤−2x≤4.-14\le -2x\le 4.$
Divide throughout by $- 2$ . Dividing an inequality by a negative number reverses both inequality signs:
$7≥x≥−2.7\ge x\ge -2.$
Put the bounds in increasing order:
$−2≤x≤7.-2\le x\le 7.$

Common Mistake

Forgetting to reverse inequality signs

When you multiply or divide an inequality by a negative quantity, reverse its direction. In a double inequality, both signs must be reversed.

When the other side contains a variable

You may meet an inequality such as

∣ f (x) ∣ < g (x) .

The right-hand side needs careful attention because a modulus is always non-negative. In particular, $∣ f (x) ∣ < g (x)$ can hold only where $g (x) > 0$ .

When $g (x) > 0$ , you can use

- g (x) < f (x) < g (x) .

Example

Comparing a modulus with a variable expression

Solve $∣ x - 1∣ < x + 3$ .

Since $∣x−1∣≥0|x-1|\ge 0$ , the right-hand side must be positive:
$\quad\Rightarrow\quad x>-3.$
Apply the inside rule:
$- (x + 3) < x - 1 < x + 3.$
Solve the left-hand inequality:
$\quad\Rightarrow\quad -2<2x \quad\Rightarrow\quad x>-1.$
The right-hand inequality simplifies to $- 1 < 3$ , which is true for every $x$ . Combining this with the restrictions gives
$x > - 1.$

Common Mistake

Check the sign of the right-hand side

You may use $∣f(x)∣<g(x)⟺−g(x)<f(x)<g(x)|f(x)|<g(x)\Longleftrightarrow-g(x)<f(x)<g(x)$ only where $g (x) > 0$ . Ignoring this condition can introduce invalid solutions.

Squaring both sides

If both sides are known to be non-negative, squaring preserves their order. This gives

\quad\Longleftrightarrow\quad f(x)^2<g(x)^2,

provided $g (x) > 0$ .

This method can be useful when both sides contain expressions, although you must then solve the resulting polynomial inequality carefully.

Example

Solving by squaring

Solve $∣x+1∣≤2x+1|x+1|\le 2x+1$ .

The right-hand side must be non-negative:
$2x+1≥0⇒x≥−12.2x+1\ge 0 \quad\Rightarrow\quad x\ge-\frac12.$
Under this restriction, square both sides:
$(x+1)2≤(2x+1)2.(x+1)^2\le(2x+1)^2.$
Expand and collect terms:
$x2+2x+1≤4x2+4x+1⇒0≤3x2+2x.x^2+2x+1\le4x^2+4x+1 \quad\Rightarrow\quad 0\le3x^2+2x.$
Factorise:
$x(3x+2)≥0.x(3x+2)\ge0.$
This gives $x≤−23x\le-\frac23$ or $x≥0x\ge0$ .
Intersect this with the earlier restriction $x≥−12x\ge-\frac12$ . The first region is removed, leaving
$x≥0.x\ge0.$

Tip

Check with a test value

After finding an interval, substitute one value from inside it and one from outside it into the original inequality. This quickly detects reversed signs or an incorrect choice of region.

Special right-hand sides

Because $∣f(x)∣≥0|f(x)|\ge0$ , comparisons with zero or a negative number can often be decided immediately.

$∣ f (x) ∣ < 0$ has no solutions.
$∣f(x)∣≤0|f(x)|\le0$ means $f (x) = 0$ .
$∣ f (x) ∣ > 0$ means $f(x)≠0f(x)\neq0$ .
$∣f(x)∣≥0|f(x)|\ge0$ is true for every value where $f (x)$ is defined.
If $c < 0$ , then $∣ f (x) ∣ > c$ is always true, while $∣ f (x) ∣ < c$ is impossible.

These observations can save considerable algebra.

Using graphs

To solve $∣ f (x) ∣ < g (x)$ graphically, sketch or plot $y = ∣ f (x) ∣$ and $y = g (x)$ . The solution consists of the $x$ -values where the modulus graph lies below the other graph.

Exam technique

In the exam

Identify whether the solution should be inside an interval for $<$ or $≤\le$ , or outside for $>$ or $≥\ge$ .
If the right-hand side is not a positive constant, check its sign before removing the modulus or squaring.
Solve every resulting inequality carefully, reversing signs when multiplying or dividing by a negative quantity.
State separate regions using or, and include endpoints only when the original inequality includes equality.
Test a value in each proposed region in the original inequality.

Self review

Check yourself

Can you solve $∣4 x - 3∣ < 5$ and show the answer on a number line?
How does the solution of $∣x+2∣≥6|x+2|\ge6$ differ from that of $∣x+2∣≤6|x+2|\le6$ ?
Why must you check the sign of $x + 4$ before rewriting $∣2 x - 1∣ < x + 4$ ?

What you'll learn

What does a modulus mean?

Less-than modulus inequalities

Greater-than modulus inequalities

Inclusive inequalities

When the other side contains a variable

Squaring both sides

Special right-hand sides

Using graphs

Review 1.3.7 Inequalities with a modulus sign (A-level only) by teaching Genie

1.3.7 Inequalities with a modulus sign (A-level only)

What you'll learn

What does a modulus mean?

Less-than modulus inequalities

Greater-than modulus inequalities

Inclusive inequalities

When the other side contains a variable

Squaring both sides

Special right-hand sides

Using graphs

Review 1.3.7 Inequalities with a modulus sign (A-level only) by teaching Genie

1.3.7 Inequalities with a modulus sign (A-level only)

What you'll learn

What does a modulus mean?

Less-than modulus inequalities

Greater-than modulus inequalities

Inclusive inequalities

When the other side contains a variable

Squaring both sides

Special right-hand sides

Using graphs

Review 1.3.7 Inequalities with a modulus sign (A-level only) by teaching Genie

1.3.7 Inequalities with a modulus sign (A-level only) Revision Guide

1.3.7 Inequalities with a modulus sign (A-level only)

What you'll learn

What does a modulus mean?

Less-than modulus inequalities

Greater-than modulus inequalities

Inclusive inequalities

When the other side contains a variable

Squaring both sides

Special right-hand sides

Using graphs

Review 1.3.7 Inequalities with a modulus sign (A-level only) by teaching Genie

1.3.7 Inequalities with a modulus sign (A-level only) Revision Guide