example of problem solving in inequality

Solving Inequalities

Sometimes we need to solve Inequalities like these:

Our aim is to have x (or whatever the variable is) on its own on the left of the inequality sign:

We call that "solved".

Example: x + 2 > 12

Subtract 2 from both sides:

x + 2 − 2 > 12 − 2

x > 10

How to Solve

Solving inequalities is very like solving equations , we do most of the same things ...

... but we must also pay attention to the direction of the inequality .

Some things can change the direction !

< becomes >

> becomes <

≤ becomes ≥

≥ becomes ≤

Safe Things To Do

These things do not affect the direction of the inequality:

• Add (or subtract) a number from both sides
• Multiply (or divide) both sides by a positive number
• Simplify a side

Example: 3x < 7+3

We can simplify 7+3 without affecting the inequality:

But these things do change the direction of the inequality ("<" becomes ">" for example):

• Multiply (or divide) both sides by a negative number
• Swapping left and right hand sides

Example: 2y+7 < 12

When we swap the left and right hand sides, we must also change the direction of the inequality :

12 > 2y+7

Here are the details:

Adding or Subtracting a Value

We can often solve inequalities by adding (or subtracting) a number from both sides (just as in Introduction to Algebra ), like this:

Example: x + 3 < 7

If we subtract 3 from both sides, we get:

x + 3 − 3 < 7 − 3    

And that is our solution: x < 4

In other words, x can be any value less than 4.

What did we do?

And that works well for adding and subtracting , because if we add (or subtract) the same amount from both sides, it does not affect the inequality

Example: Alex has more coins than Billy. If both Alex and Billy get three more coins each, Alex will still have more coins than Billy.

What If I Solve It, But "x" Is On The Right?

No matter, just swap sides, but reverse the sign so it still "points at" the correct value!

Example: 12 < x + 5

If we subtract 5 from both sides, we get:

12 − 5 < x + 5 − 5    

That is a solution!

But it is normal to put "x" on the left hand side ...

... so let us flip sides (and the inequality sign!):

Do you see how the inequality sign still "points at" the smaller value (7) ?

And that is our solution: x > 7

Note: "x" can be on the right, but people usually like to see it on the left hand side.

Multiplying or Dividing by a Value

Another thing we do is multiply or divide both sides by a value (just as in Algebra - Multiplying ).

But we need to be a bit more careful (as you will see).

Positive Values

Everything is fine if we want to multiply or divide by a positive number :

Example: 3y < 15

If we divide both sides by 3 we get:

3y /3 < 15 /3

And that is our solution: y < 5

Negative Values

Well, just look at the number line!

For example, from 3 to 7 is an increase , but from −3 to −7 is a decrease.

See how the inequality sign reverses (from < to >) ?

Let us try an example:

Example: −2y < −8

Let us divide both sides by −2 ... and reverse the inequality !

−2y < −8

−2y /−2 > −8 /−2

And that is the correct solution: y > 4

(Note that I reversed the inequality on the same line I divided by the negative number.)

So, just remember:

When multiplying or dividing by a negative number, reverse the inequality

Multiplying or Dividing by Variables

Here is another (tricky!) example:

Example: bx < 3b

It seems easy just to divide both sides by b , which gives us:

... but wait ... if b is negative we need to reverse the inequality like this:

But we don't know if b is positive or negative, so we can't answer this one !

To help you understand, imagine replacing b with 1 or −1 in the example of bx < 3b :

• if b is 1 , then the answer is x < 3
• but if b is −1 , then we are solving −x < −3 , and the answer is x > 3

The answer could be x < 3 or x > 3 and we can't choose because we don't know b .

Do not try dividing by a variable to solve an inequality (unless you know the variable is always positive, or always negative).

A Bigger Example

Example: x−3 2 < −5.

First, let us clear out the "/2" by multiplying both sides by 2.

Because we are multiplying by a positive number, the inequalities will not change.

x−3 2 ×2 < −5  ×2  

x−3 < −10

Now add 3 to both sides:

x−3 + 3 < −10 + 3    

And that is our solution: x < −7

Two Inequalities At Once!

How do we solve something with two inequalities at once?

Example: −2 < 6−2x 3 < 4

First, let us clear out the "/3" by multiplying each part by 3.

Because we are multiplying by a positive number, the inequalities don't change:

−6 < 6−2x < 12

−12 < −2x < 6

Now divide each part by 2 (a positive number, so again the inequalities don't change):

−6 < −x < 3

Now multiply each part by −1. Because we are multiplying by a negative number, the inequalities change direction .

6 > x > −3

And that is the solution!

But to be neat it is better to have the smaller number on the left, larger on the right. So let us swap them over (and make sure the inequalities point correctly):

−3 < x < 6

• Many simple inequalities can be solved by adding, subtracting, multiplying or dividing both sides until you are left with the variable on its own.
• Multiplying or dividing both sides by a negative number
• Don't multiply or divide by a variable (unless you know it is always positive or always negative)

High Impact Tutoring Built By Math Experts

Personalized standards-aligned one-on-one math tutoring for schools and districts

In order to access this I need to be confident with:

Solving inequalities

Here you will learn about solving inequalities, including how to solve linear inequalities, identify integers in the solution set, and represent solutions on a number line.

Students will first learn about solving simple inequalities as part of expressions and equations in 6th grade math and expand that knowledge in 7th grade math.

What is solving inequalities?

Solving inequalities allows you to calculate the values of an unknown variable in an inequality.

Solving inequalities is similar to solving equations, but where an equation has one unique solution, an inequality has a range of solutions.

In order to solve an inequality, you need to balance the inequality on each side of the inequality sign in the same way as you would balance an equation on each side of the equal sign. Solutions can be integers, decimals, positive numbers, or negative numbers.

For example,

Common Core State Standards

How does this relate to 6th grade and 7th grade math?

• Grade 6 – Expressions and Equations (6.EE.B.8) Write an inequality of the form x > c or x < c to represent a constraint or condition in a real-world or mathematical problem. Recognize that inequalities of the form x > c or x < c have infinitely many solutions; represent solutions of such inequalities on number line diagrams.
• Grade 7 – Expressions and Equations (7.EE.4b) Solve word problems leading to inequalities of the form px + q > r or px + q < r, where p, q, and r are specific rational numbers. Graph the solution set of the inequality and interpret it in the context of the problem. For example: As a salesperson, you are paid \$50 per week plus \$3 per sale. This week you want your pay to be at least \$100. Write an inequality for the number of sales you need to make, and describe the solutions.

How to solve inequalities

In order to solve inequalities:

Rearrange the inequality so that all the unknowns are on one side of the inequality sign.

Rearrange the inequality by dividing by the \textbf{x} coefficient so that \textbf{‘x’} is isolated.

Write your solution with the inequality symbol.

[FREE] Solving Inequalities Worksheet (Grade 6 to 8)

Use this worksheet to check your grade 6 to 8 students’ understanding of solving inequalities. 15 questions with answers to identify areas of strength and support!

Solving linear inequalities examples

Example 1: solving linear inequalities.

4 x+6<26

In this case you are subtracting 6 from both sides.

\begin{aligned} 4x+6&<26\\ 4x&<20 \end{aligned}

This leaves 4x on the left side of the inequality sign and 20 on the right side.

2 Rearrange the inequality by dividing by the \textbf{x} coefficient so that \textbf{‘x’} is isolated.

In this case you need to divide both sides by 4.

\begin{aligned} 4x&<20\\ x&<5 \end{aligned}

This leaves x on the left side of the inequality sign and 5 on the right side.

3 Write your solution with the inequality symbol.

Any value less than 5 satisfies the inequality.

Example 2: solving linear inequalities

5x-4 \geq 26

In this case you need to add 4 to both sides.

\begin{aligned} 5x-4&\geq26\\ 5x&\geq30 \end{aligned}

This leaves 5x on the left side of the inequality sign and 30 on the right side.

In this case you need to divide both sides by 5.

\begin{aligned} 5x&\geq30\\ x&\geq6 \end{aligned}

This leaves x on the left side of the inequality sign and 6 on the right side.

Any value greater than or equal to 6 satisfies the inequality.

Example 3: solving linear inequalities with parentheses

3(x-4)\leq12

Let’s start by expanding the parentheses.

3x-12\leq12

Then you need to add 12 to both sides.

\begin{aligned} 3x-12&\leq12\\ 3x&\leq24 \end{aligned}

This leaves 3x on the left side of the inequality sign and 24 on the right side.

In this case you need to divide both sides by 3.

\begin{aligned} 3x&\leq24\\ x&\leq8 \end{aligned}

This leaves x on the left side of the inequality sign and 8 on the right side.

Any value less than or equal to 8 satisfies the inequality.

Example 4: solving linear inequalities with unknowns on both sides

5x-6 > 2x + 15

In this case you need to subtract 2x from both sides.

\begin{aligned} 5x-6&>2x+15\\ 3x-6&>15 \end{aligned}

This leaves 3x-6 on the left side of the inequality sign and 15 on the right side.

Rearrange the inequality so that \textbf{‘x’} s are on one side of the inequality sign and numbers on the other.

In this case you need to add 6 to both sides.

\begin{aligned} 3x-6&>15\\ 3x&>21 \end{aligned}

This leaves 3x on the left side of the inequality sign and 21 on the right side.

\begin{aligned} 3x&>21\\ x&>7 \end{aligned}

This leaves x on the left side of the inequality sign and 7 on the right side.

Any value greater than 7 satisfies the inequality.

Example 5: solving linear inequalities with fractions

\cfrac{x+3}{5}<2

Rearrange the inequality to eliminate the denominator.

In this case you need to multiply both sides by 5.

\begin{aligned} \cfrac{x+3}{5}&<2\\ x+3&<10 \end{aligned}

In this case you need to subtract 3 from both sides.

\begin{aligned} \cfrac{x+3}{5}&<2\\ x+3&<10\\ x&<7 \end{aligned}

Any value less than 7 satisfies the inequality.

Example 6: solving linear inequalities with non-integer solutions

In this case you need to subtract 6 from both sides.

\begin{aligned} 6x+1&\geq4\\ 6x&\geq3 \end{aligned}

In this case you need to divide both sides by 6.

\begin{aligned} 6x+1&\geq4\\ 6x&\geq3\\ x&\geq\cfrac{3}{6} \end{aligned}

This can be simplified to \, \cfrac{1}{2} \, or the decimal equivalent.

x\geq\cfrac{1}{2}

Any value greater than or equal to \, \cfrac{1}{2} \, satisfies the inequality.

Example 7: solving linear inequalities and representing solutions on a number line

Represent the solution on a number line:

2x-7 < 5

In this case you need to add 7 to both sides.

\begin{aligned} 2x-7&<5\\ 2x&<12 \end{aligned}

In this case you need to divide both sides by 2.

\begin{aligned} 2x-7& <5\\ 2x& <12\\ x& < 6 \end{aligned}

Represent your solution on a number line.

Any value less than 6 satisfies the inequality. An open circle is required at 6 and the values lower than 6 indicated with an arrow.

Example 8: solving linear inequalities with negative x coefficients

In this case you need to subtract 1 from both sides.

\begin{aligned} 1-2x & <7 \\ -2x & <6 \end{aligned}

In this case you need to divide both sides by negative 2.

6 \div-2=-3

Change the direction of the inequality sign.

Because you divided by a negative number, you also need to change the direction of the inequality sign.

\begin{aligned} 1-2x & <7 \\ -2x & <6 \\ x &>-3 \end{aligned}

Example 9: solving linear inequalities and listing integer values that satisfy the inequality

List the integer values that satisfy:

3 < x+1\leq8

In this case you need to subtract 1 from each part.

\begin{aligned} 3&<x+1\leq8\\ 2&<x\leq7\\ \end{aligned}

List the integer values satisfied by the inequality.

2<x\leq7

2 is not included in the solution set. 7 is included in the solution set. The integers that satisfy this inequality are:

3, 4, 5, 6, 7

Example 10: solving linear inequalities and listing integer values that satisfy the inequality

7\leq4x\leq20

In this case you need to divide each part by 4.

\begin{aligned} 7\leq \, & 4x\leq20\\ \cfrac{7}{4}\leq & \; x \leq5 \end{aligned}

\cfrac{7}{4} \leq x \leq 5

\cfrac{7}{4} \, is included in the solution set but it is not an integer.

The first integer higher is 2.

5 is also included in the solution set.

The integers that satisfy this inequality are:

Example 11: solving linear inequalities and representing the solution on a number line

-3<2x+5\leq7

In this case you need to subtract 5 from each part.

\begin{aligned} -3<2x+5&\leq7\\ -8<2x&\leq2 \end{aligned}

Rearrange the inequality so that \textbf{‘x’} is isolated. In this case you need to divide each part by \bf{2} .

\begin{aligned} -3<2x+5&\leq7\\ -8<2x&\leq2\\ -4<x&\leq1 \end{aligned}

Represent the solution set on a the number line

-4<x\leq1

-4 is not included in the solution set so requires an open circle. 1 is included in the solution set so requires a closed circle.

Put a solid line between the circles to indicate all the values that satisfy the solution set.

Teaching tips for solving inequalities

• Students should master solving simple inequalities (one-step inequalities) before moving on to two-step inequalities (or multi-step inequalities).
• Foster student engagement by practicing solving inequalities through classroom games such as BINGO rather than daily worksheets.
• Student practice problems should have a variety of inequalities such as inequalities with fractions, negative numbers, and parentheses. (see examples above)

Easy mistakes to make

• Solutions as inequalities Not including the inequality symbol in the solution is a common mistake. An inequality has a range of values that satisfy it rather than a unique solution so the inequality symbol is essential. For example, when solving x + 3 < 7 giving a solution of 4 or x = 4 is incorrect, the answer must be written as an inequality x < 4 .
• Balancing inequalities Errors can be made with solving equations and inequalities by not applying inverse operations or not balancing the inequalities. Work should be shown step-by-step with the inverse operations applied to both sides of the inequality. For example, when solving x + 3 < 7 , adding 3 to both sides rather than subtracting 3 from both sides.

Related inequalities lessons

• Inequalities
• Linear inequalities
• Inequalities on a number line
• Graphic inequalities
• Quadratic inequalities
• Greater than sign
• Less than sign

Practice solving inequalities questions

3x+7 < 31

\begin{aligned} 3x+7&<31\\ 3x&<24\\ x&<8 \end{aligned}

\begin{aligned} 4x-3&\geq25\\ 4x&\geq28\\ x&\geq7 \end{aligned}

2(x-5)\leq8

\begin{aligned} 2(x-5)&\leq8\\ 2x-10&\leq8\\ 2x&\leq18\\ x&\leq9 \end{aligned}

6x-5 > 4x + 1

\begin{aligned} 6x-5&>4x+1\\ 2x-5&>1\\ 2x&>6\\ x&>3 \end{aligned}

\cfrac{x-4}{2}>6

\begin{aligned} \cfrac{x-4}{2}&>6\\ x-4&>12\\ x&>16 \end{aligned}

\begin{aligned} 8x+1&\geq3\\ 8x&\geq2\\ x&\geq\cfrac{2}{8}\\ x&\geq\cfrac{1}{4} \end{aligned}

7. Represent the solution on a number line.

5x-2 < 28

\begin{aligned} 5x-2&<28\\ 5x&<30\\ x&<6 \end{aligned}

An open circle is required and all values less than 6 indicated.

2-3x > 14

\begin{aligned} 2-3x &>14 \\ -3x &>12 \\ x &< -4 \end{aligned}

Change the direction of the inequality sign as you have divided by a negative number.

9. List the integer values that satisfy:

2<x+3\leq5

\begin{aligned} 2<x&+3\leq5\\ -1< \, &x\leq2 \end{aligned}

-1 is not included in the solution set as is greater than -1.

2 is included in the solution set as x is less than or equal to 2.

10. List the integer values that satisfy:

4\leq3x\leq21

\begin{aligned} 4\leq3&x\leq21\\ \cfrac{4}{3} \, \leq \, &x\leq7 \end{aligned}

The first integer greater than \, \cfrac{4}{3} \, is 2.

7 is included in the solution set as x is less than or equal to 7.

11. List the integer values that satisfy:

-4<3x+2\leq5

\begin{aligned} -4<3x&+2\leq5\\ -6<3&x\leq3\\ -2< \, &x\leq1 \end{aligned}

-2 is not included in the solution set as x is greater than -2.

1 is included in the solution set as x is less than or equal to 1.

Solving inequalities FAQs

Solving inequalities is where you calculate the values that an unknown variable can be in an inequality.

To solve an inequality, you need to balance the inequality on each side of the inequality sign in the same way as you would balance an equation on each side of the equal sign. Solutions can be integers, decimals, positive numbers, or negative numbers.

The next lessons are

• Types of graphs
• Math formulas
• Coordinate plane

Still stuck?

At Third Space Learning, we specialize in helping teachers and school leaders to provide personalized math support for more of their students through high-quality, online one-on-one math tutoring delivered by subject experts.

Each week, our tutors support thousands of students who are at risk of not meeting their grade-level expectations, and help accelerate their progress and boost their confidence.

Find out how we can help your students achieve success with our math tutoring programs .

[FREE] Common Core Practice Tests (Grades 3 to 6)

Prepare for math tests in your state with these Grade 3 to Grade 6 practice assessments for Common Core and state equivalents.

40 multiple choice questions and detailed answers to support test prep, created by US math experts covering a range of topics!

Privacy Overview

Solving Inequalities

Related Pages Solving Equations Algebraic Expressions More Algebra Lessons

In these lessons, we will look at the rules, approaches, and techniques for solving inequalities.

The following figure shows how to solve two-step inequalities. Scroll down the page for more examples and solutions.

The rules for solving inequalities are similar to those for solving linear equations. However, there is one exception when multiplying or dividing by a negative number.

To solve an inequality, we 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.
• Multiply both sides by the same negative number and reverse the sign.
• Divide both sides by the same negative number and reverse the sign.

Inequalities Of The Form “x + a > b” or “x + a < b”

Example: Solve x + 7 < 15

Solution: x + 7 < 15 x + 7 – 7 < 15 – 7 x < 8

Inequalities Of The Form “x – a < b” or “x – a > b”

Example: Solve x – 6 > 14

Solution: x – 6 > 14 x – 6+ 6 > 14 + 6 x > 20

Example: Solve the inequality x – 3 + 2 < 10

Solution: x – 3 + 2 < 10 x – 1 < 10 x – 1 + 1 < 10 + 1 x < 11

Inequalities Of The Form “a – x < b” or “a – x > b”

Example: Solve the inequality 7 – x < 9

Solution: 7 – x < 9 7 – x – 7 < 9 – 7 – x < 2 x > –2 (remember to reverse the symbol when multiplying by –1)

Example: Solve the inequality 12 > 18 – y

Solution: 12 > 18 – y 18 – y < 12 18 – y – 18 < 12 –18 – y < –6 y > 6 (remember to reverse the symbol when multiplying by –1)

Inequalities Of The Form “ < b” or “ > b”

Solving linear inequalities with like terms.

If an equation has like terms, we simplify the equation and then solve it. We do the same when solving inequalities with like terms.

Example: Evaluate 3x – 8 + 2x < 12

Solution: 3x – 8 + 2x < 12 3x + 2x < 12 + 8 5x < 20 x < 4

Example: Evaluate 6x – 8 > x + 7

Solution: 6x – 8 > x + 7 6x – x > 7 + 8 5x > 15 x > 3

Example: Evaluate 2(8 – p) ≤ 3(p + 7)

Solution: 2(8 – p) ≤ 3(p + 7) 16 – 2p ≤ 3p + 21 16 – 21 ≤ 3p + 2p –5 ≤ 5p –1 ≤ p p ≥ –1 (a < b is equivalent to b > a)

An Introduction To Solving Inequalities

Solving One-Step Linear Inequalities In One Variable

The solutions to linear inequalities can be expressed several ways: using inequalities, using a graph, or using interval notation.

The steps to solve linear inequalities are the same as linear equations, except if you multiply or divide by a negative when solving for the variable, you must reverse the inequality symbol.

Example: Solve. Express the solution as an inequality, graph and interval notation. x + 4 > 7 -2x > 8 x/-2 > -1 x - 9 ≥ -12 7x > -7 x - 9 ≤ -12

Solving Two-Step Linear Inequalities In One Variable

Example: Solve. Express the solution as an inequality, graph and interval notation. 3x + 4 ≥ 10 -2x - 1 > 9 10 ≥ -3x - 2 -8 > 5x + 12

Solving Linear Inequalities

Main rule to remember: If you multiply or divide by a negative number, the inequality flips direction.

Examples of how to solve linear inequalities are shown:

Example: Solve: 3x - 6 > 8x - 7

Students learn that when solving an inequality, such as -3x is less than 12, the goal is the same as when solving an equation: to get the variable by itself on one side.

Note that when multiplying or dividing both sides of an inequality by a negative number, the direction of the inequality sign must be switched.

For example, to solve -3x is less than 12, divide both sides by -3, to get x is greater than -4.

And when graphing an inequality on a number line, less than or greater than is shown with an open dot, and less than or equal to or greater than or equal to is shown with a closed dot.

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

• Mathematicians
• Math Lessons
• Square Roots
• Math Calculators
• Solving Inequalities – Explanation & Examples

JUMP TO TOPIC

What is Inequality in Math?

Inequality symbols, operations on inequalities, solving linear inequalities with addition, solving linear inequalities with subtraction, solving linear inequalities with multiplication, solving linear inequalities with division, solving linear inequalities using the distributive property, practice questions, solving inequalities – explanation & examples.

The word inequality means a mathematical expression in which the sides are not equal to each other. Basically, an inequality compares any two values and shows that one value is less than, greater than, or equal to the value on the other side of the equation.

Basically, there are five inequality symbols used to represent equations of inequality.

These inequality symbols are: less than ( < ), greater than ( > ), less than or equal ( ≤ ), greater than or equal ( ≥ ) and the not equal symbol ( ≠ ).

Inequalities are used to compare numbers and determine the range or ranges of values that satisfy the conditions of a given variable.

Operations on linear inequalities involve addition, subtraction, multiplication, and division. The general rules for these operations are shown below.

Although we have used < symbol for illustration, you should note that the same rules apply to >, ≤, and ≥.

• The inequality symbol does not change when the same number is added on both sides of the inequality. For example, if a< b, then a + c < b +
• Subtracting both sides of the inequality by the same number does not change the inequality sign. For example, if a< b, then a – c < b – c.
• Multiplying both sides of an inequality by a positive number does not change the inequality sign. For example, if a< b and if c is a positive number, then a * c < b *
• Dividing both sides of an inequality by a positive number does not change the inequality sign. If a< b and if c is a positive number, then a/c < b/c
• Multiplying both sides of an inequality equation by a negative number changes the direction of the inequality symbol. For example, given that a < b and c is a negative number, then a * c > b *

How to Solve Inequalities?

Like linear equations, inequalities can be solved by applying similar rules and steps with a few exceptions. The only difference when solving linear equations is an operation that involves multiplication or division by a negative number. Multiplying or dividing an inequality by a negative number changes the inequality symbol.

Linear inequalities can be solved using the following operations:

• Subtraction
• Multiplication
• Distribution of property

Let’s see a few examples below to understand this concept.

Solve 3x − 5 ≤ 3 − x.

We start by adding both sides of the inequality by 5

3x – 5 + 5 ≤ 3 + 5 − x

Then add both sides by x.

3x + x ≤ 8 – x + x

Finally, divide both sides of the inequality by 4 to get;

Calculate the range of values of y, which satisfies the inequality: y − 4 < 2y + 5.

Add both sides of the inequality by 4.

y – 4 + 4 < 2y + 5 + 4

y < 2y + 9

Subtract both sides by 2y.

y – 2y < 2y – 2y + 9

Y < 9 Multiply both sides of the inequality by −1 and change the inequality symbol’s direction. y > − 9

Solve x + 8 > 5.

Isolate the variable x by subtracting 8 from both sides of the inequality.

x + 8 – 8 > 5 – 8 => x > −3

Therefore, x > −3.

Solve 5x + 10 > 3x + 24.

Subtract 10 from both sides of the inequality.

5x + 10 – 10 > 3x + 24 – 10

5x > 3x + 14.

Now we subtract both sides of the inequality by 3x.

5x – 3x > 3x – 3x + 14

Solve x/4 > 5

Multiply both sides of an inequality by the denominator of the fraction

4(x/4) > 5 x 4

Solve -x/4 ≥ 10

Multiply both sides of an inequality by 4.

4(-x/4) ≥ 10 x 4

Multiply both sides of the inequality by -1 and reverse the direction of the inequality symbol.

x ≤ – 40

Solve the inequality: 8x − 2 > 0.

First of all, add both sides of the inequality by 2

8x – 2 + 2 > 0 + 2

Now, solve by dividing both sides of the inequality by 8 to get;

Solve the following inequality:

−5x > 100

Divide both sides of the inequality by -5 and change the direction of the inequality symbol

= −5x/-5 < 100/-5

= x < − 20

Solve: 2 (x – 4) ≥ 3x – 5

2 (x – 4) ≥ 3x – 5

Apply the distributive property to remove the parentheses.

⟹ 2x – 8 ≥ 3x – 5

Add both sides by 8.

⟹ 2x – 8 + 8 ≥ 3x – 5 + 8

⟹ 2x ≥ 3x + 3

Subtract both sides by 3.

⟹ 2x – 3x ≥ 3x + 3 – 3x

⟹ x ≤ – 3

A student scored 60 marks in the first test and 45 marks in the second test of the terminal examination. How many minimum marks should the student score in the third test get a mean of least 62 marks?

Let the marks scored in the third test be x marks.

(60 + 45 + x)/3 ≥ 62 105 + x ≥ 196 x ≥ 93 Therefore, the student must score 93 marks to maintain a mean of at least 62 marks.

Justin requires at least $500 to hold his birthday party. If already he has saved $ 150 and 7 months are left to this date. What is the minimum amount he must save monthly?

Let the minimum amount saved monthly = x

150 + 7x ≥ 500

Solve for x

150 – 150 + 7x ≥ 500 – 150

Therefore, Justin should save $50 or more

Find two consecutive odd numbers which are greater than 10 and have the sum of less than 40.

Let the smaller odd number = x

Therefore, the next number will be x + 2

x > 10 ………. greater than 10

x + (x + 2) < 40 ……sum is less 40

Solve the equations.

2x + 2 < 40

x + 1< 20

Combine the two expressions.

10 < x < 19

Therefore, the consecutive odd numbers are 11 and 13, 13 and 15, 15 and 17, 17 and 19.

Inequalities and the Number Line

The best tool to represent and visualize numbers is the number line. A number line is defined as a straight horizontal line with numbers placed along at equal segments or intervals. A number line has a neutral point at the middle, known as the origin. On the right side of the origin on the number line are positive numbers, while the left side of the origin is negative numbers.

Linear equations can also be solved by a graphical method using a number line. For example, to plot x > 1, on a number line, you circle the number 1 on the number line and draw a line going from the circle in the direction of the numbers that satisfies the inequality statement.

If the inequality symbol is greater than or equal to or less than or equal to sign (≥ or ≤), draw the circle over the numerical number and fill or shade the circle. Finally, draw a line going from the shaded circle in the numbers’ direction that satisfies the inequality equation.

The same procedure is used to solve equations involving intervals.

 Example 15

–2 <  x  < 2

–1 ≤  x  ≤ 2

–1 <  x  ≤ 2

Previous Lesson  |  Main Page | Next Lesson

• Parallelogram
• Quadrilateral
• Parallelepiped
• Tetrahedron
• Dodecahedron
• Fraction Calculator
• Mixed Fraction Calculator
• Greatest Common Factor Calulator
• Decimal to Fraction Calculator
• Whole Numbers
• Rational Numbers
• Place Value
• Irrational Numbers
• Natural Numbers
• Binary Operation
• Numerator and Denominator
• Order of Operations (PEMDAS)
• Scientific Notation

Last modified on April 17th, 2024

Inequality Word Problems

Inequalities are common in our everyday life. They help us express relationships between quantities that are unequal. Writing and solving word problems involving them helps develop our problem-solving approach, understanding, logical reasoning, and analytical skills.

Here are the 4 main keywords commonly used to write mathematical expressions involving inequalities.   

• At least →  ‘greater than or equal to’
• More than → ‘greater than’
• No more than or at most → ‘less than or equal to’
• Less than → ‘less than’

To participate in the annual sports day, Mr. Adams would like to have nine students in each group. But fewer than 54 students are in class today, so Mr. Adams is unable to make as many full groups as he wants. How many full groups can Mr. Adams make? Write the inequality that describes the situation.

Let ‘x’ be the total number of groups Mr. Adams can make. Since each group has 9 students, the total number of students in ‘x’ groups is 9x As we know, fewer than 54 students are in a class today. Thus, the inequality that represents the situation is: 9x < 54 On dividing both sides by 9, the maximum number of groups Mr. Adams can make is  x < 6 Thus, Mr. Adams can make a maximum of 6 full groups.

 Bruce needs at least \$561 to buy a new tablet. He has already saved \$121 and earns \$44 per month as a part-timer in a company. Write the inequality and determine how long he has to work to buy the tablet.

Let ‘x’ be the number of months Bruce needs to work. As we know,  The amount already saved by Bruce is \$121 He earns \$44 per month The cost of the tablet is at least \$561 After ‘x’ months of work, Bruce will have \$(121 + 44x) Now, the inequality representing the situation is: 121 + 44x ≥ 561 On subtracting 121 from both sides, 121 + 44x – 121 ≥ 561 – 121 ⇒ 44x ≥ 440 On dividing both sides by 44, x ≥ 10 Thus, Bruce needs to work for at least 10 months to buy the new tablet.

A store is offering a \$26 discount on all women’s clothes. Ava is looking at clothes originally priced between \$199 and \$299. How much can she expect to spend after the discount?

Let ‘x’ be the original price of the clothes Ava chooses. As we know, the original price range is 199 ≤ x ≤ 299, and the discount is \$26 Now, Ava pays \$(x – 26) after the discount. Thus, the inequality is: 199 – 26 ≤ x – 26 ≤ 299 – 26 ⇒ 173 ≤ x – 26 ≤ 273 Thus, she can expect to spend between \$173 and \$273 after the discount.

A florist makes a profit of \$6.25 per plant. If the store wants to profit at least \$4225, how many plants does it need to sell?

Let ‘P’ be the profit, ‘p’ be the profit per plant, and ‘n’ be the number of plants.  As we know, the store wants a profit of at least \$4225, and the florist makes a profit of \$6.25 per plant. Here, P ≥ 4225 and p = 6.25 …..(i) Also, P = p × n Substituting the values of (i), we get 6.25 × n ≥ 4225 On dividing both sides by 6.25, we get ${n\geq \dfrac{4225}{6\cdot 25}}$ ⇒ ${n\geq 676}$ Thus, the store needs to sell at least 676 plants to make a profit of \$4225.

Daniel had \$1200 in his savings account at the start of the year, but he withdraws \$60 each month to spend on transportation. He wants to have at least \$300 in the account at the end of the year. How many months can Daniel withdraw money from the account?

As we know, Daniel had \$1200 in his savings account at the start of the year, but he withdrew \$60 for transportation each month.  Thus, after ‘n’ months, he will have \$(1200−60n) left in his account. Also, Daniel wants to have at least \$300 in the account at the end of the year.  Here, the inequality is: 1200 – 60n ≥ 300 ⇒ 1200 – 60n – 1200 ≥ 300 – 1200 (by subtraction property) ⇒ -60n ≥ -900 ⇒ 60n ≤ 900 (by inversion property) ⇒ n ≤ ${\dfrac{900}{60}}$ ⇒ n ≤ 15 Thus, Daniel can withdraw money from the account for at most 15 months.

Anne is a model trying to lose weight for an upcoming beauty pageant. She currently weighs 165 lb. If she cuts 2 lb per week, how long will it take her to weigh less than 155 lb?

Let ‘t’ be the number of weeks to weigh less than 155 lb. As we know, Anne initially weighs 165 lb After ‘t’ weeks of cutting 2 lb per week, her weight will be 165 – 2t Now, Anne’s weight will be less than 155 lb Here, the inequality from the given word problem is: 165 – 2t < 155 On subtracting 165 from both sides, we get 165 – 2t – 165 < 155 – 165 ⇒ – 2t < -10 On dividing by -2, the inequality sign is reversed. ${\dfrac{-2t}{-2} >\dfrac{-10}{-2}}$ ⇒ t > 5

Rory and Cinder are on the same debate team. In one topic, Rory scored 5 points more than Cinder, but they scored less than 19 together. What are the possible points Rory scored?

Let Rory’s score be ‘r,’ and Cinder’s score be ‘c.’ As we know, Rory scored 5 points more than Cinder. Thus, Rory’s score is r = c + 5 …..(i) Also, their scores sum up to less than 19 points. Thus, the inequality is: r + c < 19 …..(ii) Substituting (i) in (ii), we get (c + 5) + c < 19 ⇒ 2c + 5 < 19 On subtracting 5 from both sides, we get 2c + 5 – 5 < 19 – 5 ⇒ 2c < 14 On dividing both sides by 2, we get ${\dfrac{2c}{2} >\dfrac{14}{2}}$ ⇒ c < 7 means Cinder’s score is less than 7 points. Now, from (i), r = c + 5 ⇒ c = r – 5 Thus, c < 7 ⇒ r – 5 < 7 On adding 5 to both sides, we get r – 5 + 5 < 7 + 5 ⇒ r < 12 means Rory’s score is less than 12 points. Hence, Rory’s scores can be 6, 7, 8, 9, 10, or 11 points.

An average carton of juice cans contains 74 pieces, but the number can vary by 4. Find out the maximum and minimum number of cans that can be present in a carton.

Let ‘c’ be the number of juice cans in a carton. As we know, the average number of cans in a carton is 74, and it varies by 4 cans. Thus, the required inequality is |c – 74| ≤ 4 ⇒ -4 ≤ c – 74 ≤ 4 On adding 74 to each side, we get -4 + 74 ≤ c – 74 + 74 ≤ 4 + 74 ⇒ 70 ≤ c ≤ 78 Hence, the minimum number of cans in a carton is 70, and the maximum number is 78.

 Layla rehearses singing for at least 12 hours per week, for three-fourths of an hour each session. If she has already sung 3 hours this week, how many more sessions remain for her to exceed her weekly practice goal?

Let ‘p’ be Layla’s total hours of practice in a week, and ‘s’ be the number of sessions she needs to complete. As we know, Layla has already rehearsed 3 hours, then her remaining rehearsal time is (p – 3) Each session lasts for three-fourths of an hour. Thus, we have the inequality: ${\dfrac{3}{4}s >p-3}$ …..(i) As we know, Layla rehearses for at least 12 hours, which means p ≥ 12 …..(ii) From (i), ${\dfrac{3}{4}s >p-3}$ ⇒ ${s >\dfrac{4}{3}\left( p-3\right)}$ From (ii), substituting the value p = 12 in (i), we get ${s >\dfrac{4}{3}\left( 12-3\right)}$ ⇒ ${s >\dfrac{4}{3}\cdot 9}$ ⇒ s > 12 Thus, Layla must complete more than 12 sessions to exceed her weekly rehearsal goal.

• Privacy Policy
• Trigonometry

Join Our Newsletter

© 2024 Mathmonks.com . All rights reserved. Reproduction in whole or in part without permission is prohibited.

• HW Guidelines
• Study Skills Quiz
• Find Local Tutors
• Demo MathHelp.com
• Join MathHelp.com

Select a Course Below

• ACCUPLACER Math
• Math Placement Test
• PRAXIS Math
• + more tests
• 5th Grade Math
• 6th Grade Math
• Pre-Algebra
• College Pre-Algebra
• Introductory Algebra
• Intermediate Algebra
• College Algebra

Examples of Solving Harder Linear Inequalities

Intro & Formatting Worked Examples Harder Examples & Word Prob's

Once you'd learned how to solve one-variable linear equations, you were then given word problems. To solve these problems, you'd have to figure out a linear equation that modelled the situation, and then you'd have to solve that equation to find the answer to the word problem.

Content Continues Below

MathHelp.com

Solving Inequalities

So, now that you know how to solve linear inequalities — you guessed it! — they give you word problems.

• The velocity of an object fired directly upward is given by V = 80 − 32 t , where the time t is measured in seconds. When will the velocity be between 32 and 64 feet per second (inclusive)?

This question is asking when the velocity, V , will be between two given values. So I'll take the expression for the velocity,, and put it between the two values they've given me. They've specified that the interval of velocities is inclusive, which means that the interval endpoints are included. Mathematically, this means that the inequality for this model will be an "or equal to" inequality. Because the solution is a bracket (that is, the solution is within an interval), I'll need to set up a three-part (that is, a compound) inequality.

I will set up the compound inequality, and then solve for the range of times t :

32 ≤ 80 − 32 t ≤ 64

32 − 80 ≤ 80 − 80 − 32 t ≤ 64 − 80

−48 ≤ −32 t ≤ −16

−48 / −32 ≥ −32 t / −32 ≥ −16 / −32

1.5 ≥ t ≥ 0.5

Note that, since I had to divide through by a negative, I had to flip the inequality signs.

Advertisement

Note also that you might (as I do) find the above answer to be more easily understood if it's written the other way around, with "less than" inequalities.

And, because this is a (sort of) real world problem, my working should show the fractions, but my answer should probably be converted to decimal form, because it's more natural to say "one and a half seconds" than it is to say "three-halves seconds". So I convert the last line above to the following:

0.5 ≤ t ≤ 1.5

Looking back at the original question, it did not ask for the value of the variable " t ", but asked for the times when the velocity was between certain values. So the actual answer is:

The velocity will be between 32 and 64 feet per second between 0.5 seconds after launch and 1.5 seconds after launch.

Okay; my answer above was *extremely* verbose and "complete"; you don't likely need to be so extreme. You can probably safely get away with saying something simpler like, "between 0.5 seconds and 1.5 seconds". Just make sure that you do indeed include the approprioate units (in this case, "seconds").

Always remember when doing word problems, that, once you've found the value for the variable, you need to go back and re-read the problem to make sure that you're answering the actual question. The inequality 0.5 ≤  t  ≤ 1.5 did not answer the actual question regarding time. I had to interpret the inequality and express the values in terms of the original question.

• Solve 5 x + 7 < 3( x  + 1) .

First I'll multiply through on the right-hand side, and then solve as usual:

5 x + 7 < 3( x + 1)

5 x + 7 < 3 x + 3

2 x + 7 < 3

2 x < −4

x < −2

In solving this inequality, I divided through by a positive 2 to get the final answer; as a result (that is, because I did *not* divide through by a minus), I didn't have to flip the inequality sign.

• You want to invest $30,000 . Part of this will be invested in a stable 5% -simple-interest account. The remainder will be "invested" in your father's business, and he says that he'll pay you back with 7% interest. Your father knows that you're making these investments in order to pay your child's college tuition with the interest income. What is the least you can "invest" with your father, and still (assuming he really pays you back) get at least $1900 in interest?

First, I have to set up equations for this. The interest formula for simple interest is I = Prt , where I is the interest, P is the beginning principal, r is the interest rate expressed as a decimal, and t is the time in years.

Since no time-frame is specified for this problem, I'll assume that t  = 1 ; that is, I'll assume (hope) that he's promising to pay me at the end of one year. I'll let x be the amount that I'm going to "invest" with my father. Then the rest of my money, being however much is left after whatever I give to him, will be represented by "the total, less what I've already given him", so 30000 −  x will be left to invest in the safe account.

Then the interest on the business investment, assuming that I get paid back, will be:

I = ( x )(0.07)(1) = 0.07 x

The interest on the safe investment will be:

(30 000 − x )(0.05)(1) = 1500 − 0.05 x

The total interest is the sum of what is earned on each of the two separate investments, so my expression for the total interest is:

0.07 x + (1500 − 0.05 x ) = 0.02 x + 1500

I need to get at least $1900 ; that is, the sum of the two investments' interest must be greater than, or at least equal to, $1,900 . This allows me to create my inequality:

0.02 x + 1500 ≥ 1900

0.02 x ≥ 400

x ≥ 20 000

That is, I will need to "invest" at least $20,000 with my father in order to get $1,900 in interest income. Since I want to give him as little money as possible, I will give him the minimum amount:

I will invest $20,000 at 7% .

• An alloy needs to contain between 46% copper and 50% copper. Find the least and greatest amounts of a 60% copper alloy that should be mixed with a 40% copper alloy in order to end up with thirty pounds of an alloy containing an allowable percentage of copper.

This is similar to a mixture word problem , except that this will involve inequality symbols rather than "equals" signs. I'll set it up the same way, though, starting with picking a variable for the unknown that I'm seeking. I will use x to stand for the pounds of 60% copper alloy that I need to use. Then 30 −  x will be the number of pounds, out of total of thirty pounds needed, that will come from the 40% alloy.

Of course, I'll remember to convert the percentages to decimal form for doing the algebra.

How did I get those values in the bottom right-hand box? I multiplied the total number of pounds in the mixture ( 30 ) by the minimum and maximum percentages ( 46% and 50% , respectively). That is, I multiplied across the bottom row, just as I did in the " 60% alloy" row and the " 40% alloy" row, to get the right-hand column's value.

The total amount of copper in the mixture will be the sum of the copper contributed by each of the two alloys that are being put into the mixture. So I'll add the expressions for the amount of copper from each of the alloys, and place the expression for the total amount of copper in the mixture as being between the minimum and the maximum allowable amounts of copper:

13.8 ≤ 0.6 x + (12 − 0.4 x ) ≤ 15

13.8 ≤ 0.2 x + 12 ≤ 15

1.8 ≤ 0.2 x ≤ 3

9 ≤ x ≤ 15

Checking back to my set-up, I see that I chose my variable to stand for the number of pounds that I need to use of the 60% copper alloy. And they'd only asked me for this amount, so I can ignore the other alloy in my answer.

I will need to use between 9 and 15 pounds of the 60% alloy.

Per yoozh, I'm verbose in my answer. You can answer simply as " between 9 and 15 pounds ".

• Solve 3( x − 2) + 4 ≥ 2(2 x − 3)

First I'll multiply through and simplify; then I'll solve:

3( x − 2) + 4 ≥ 2(2 x − 3)

3 x − 6 + 4 ≥ 4 x − 6

3 x − 2 ≥ 4 x − 6

−2 ≥ x − 6            (*)

Why did I move the 3 x over to the right-hand side (to get to the line marked with a star), instead of moving the 4 x to the left-hand side? Because by moving the smaller term, I was able to avoid having a negative coefficient on the variable, and therefore I was able to avoid having to remember to flip the inequality when I divided through by that negative coefficient. I find it simpler to work this way; I make fewer errors. But it's just a matter of taste; you do what works for you.

Why did I switch the inequality in the last line and put the variable on the left? Because I'm more comfortable with inequalities when the answers are formatted this way. Again, it's only a matter of taste. The form of the answer in the previous line, 4 ≥ x , is perfectly acceptable.

As long as you remember to flip the inequality sign when you multiply or divide through by a negative, you shouldn't have any trouble with solving linear inequalities.

URL: https://www.purplemath.com/modules/ineqlin3.htm

Page 1 Page 2 Page 3

Standardized Test Prep

College math, homeschool math, share this page.

• Terms of Use
• Privacy / Cookies
• About Purplemath
• About the Author
• Tutoring from PM
• Advertising
• Linking to PM
• Site licencing

Visit Our Profiles

• school Campus Bookshelves
• menu_book Bookshelves
• perm_media Learning Objects
• login Login
• how_to_reg Request Instructor Account
• hub Instructor Commons

Margin Size

• Download Page (PDF)
• Download Full Book (PDF)
• Periodic Table
• Physics Constants
• Scientific Calculator
• Reference & Cite
• Tools expand_more
• Readability

selected template will load here

This action is not available.

3.7: Solving Systems of Inequalities with Two Variables

• Last updated
• Save as PDF
• Page ID 6400

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

Learning Objectives

• Check solutions to systems of inequalities with two variables.
• Graph solution sets of systems of inequalities.

Solutions to Systems of Inequalities

A system of inequalities 33 consists of a set of two or more inequalities with the same variables. The inequalities define the conditions that are to be considered simultaneously. For example,

\(\left\{ \begin{array} { l } { y > x - 2 } \\ { y \leq 2 x + 2 } \end{array} \right.\)

We know that each inequality in the set contains infinitely many ordered pair solutions defined by a region in a rectangular coordinate plane. When considering two of these inequalities together, the intersection of these sets will define the set of simultaneous ordered pair solutions. When we graph each of the above inequalities separately we have:

And when graphed on the same set of axes, the intersection can be determined.

The intersection is shaded darker and the final graph of the solution set will be presented as follows:

The graph suggests that \((3, 2)\) is a solution because it is in the intersection. To verify this, we can show that it solves both of the original inequalities as follows:

Points on the solid boundary are included in the set of simultaneous solutions and points on the dashed boundary are not. Consider the point \((−1, 0)\) on the solid boundary defined by \(y = 2x + 2\) and verify that it solves the original system:

Notice that this point satisfies both inequalities and thus is included in the solution set. Now consider the point \((2, 0)\) on the dashed boundary defined by \(y = x − 2\) and verify that it does not solve the original system:

This point does not satisfy both inequalities and thus is not included in the solution set.

Example \(\PageIndex{1}\)

Determine whether or not \((-3,3)\) is a solution to the following system:

\(\left\{ \begin{aligned} 2 x + 6 y \leq 6 \\ - \frac { 1 } { 3 } x - y \leq 3 \end{aligned} \right.\)

Substitute the coordinates of \((x, y) = (−3, 3)\) into both inequalities.

\((-3,3)\) is not a solution; it does not satisfy both inequalities.

We can graph the solutions of systems that contain nonlinear inequalities in a similar manner. For example, both solution sets of the following inequalities can be graphed on the same set of axes:

\(\left\{ \begin{array} { l } { y < \frac { 1 } { 2 } x + 4 } \\ { y \geq x ^ { 2 } } \end{array} \right.\)

And the intersection of both regions contains the region of simultaneous ordered pair solutions.

From the graph, we expect the ordered pair \((1, 3)\) to solve both inequalities.

Graphing Solutions to Systems of Inequalities

Solutions to a system of inequalities are the ordered pairs that solve all the inequalities in the system. Therefore, to solve these systems we graph the solution sets of the inequalities on the same set of axes and determine where they intersect. This intersection, or overlap, will define the region of common ordered pair solutions.

Example \(\PageIndex{2}\):

Graph the solution set: \(\left\{ \begin{array} { l } { - 2 x + y > - 4 } \\ { 3 x - 6 y \geq 6 } \end{array} \right.\).

To facilitate the graphing process, we first solve for \(y\).

\(\left\{ \begin{array} { l l } { - 2 x + y > - 4 } \\ { 3 x - 6 y \geq 6 } \end{array} \right. \quad\Rightarrow\quad \left\{ \begin{array} { l } { y > 2 x - 4 } \\ { y \leq \frac { 1 } { 2 } x - 1 } \end{array} \right.\)

For the first inequality, we use a dashed boundary defined by \(y = 2x − 4\) and shade all points above the line. For the second inequality, we use a solid boundary defined by \(y = \frac{1}{ 2} x − 1\) and shade all points below. The intersection is darkened.

Now we present our solution with only the intersection shaded.

Example \(\PageIndex{3}\):

Graph the solution set: \(\left\{ \begin{array} { l } { - 3 x + 2 y > 6 } \\ { 6 x - 4 y > 8 } \end{array} \right.\).

We begin by solving both inequalities for \(y\).

\(\left\{ \begin{array} { l } { - 3 x + 2 y > 6 } \\ { 6 x - 4 y > 8 } \end{array} \right. \quad\Rightarrow\quad \left\{ \begin{array} { l } { y > \frac { 3 } { 2 } x + 3 } \\ { y < \frac { 3 } { 2 } x - 2 } \end{array} \right.\)

Because of the strict inequalities, we will use a dashed line for each boundary. For the first inequality shade all points above the boundary and for the second inequality shade all points below the boundary.

As we can see, there is no intersection of these two shaded regions. Therefore, there are no simultaneous solutions.

\(\varnothing\)

Example \(\PageIndex{4}\):

Graph the solution set: \(\left\{ \begin{array} { l } { y \geq - 4 } \\ { y < x + 3 } \\ { y \leq - 3 x + 3 } \end{array} \right.\)

Begin by graphing the solution sets to all three inequalities.

After graphing all three inequalities on the same set of axes, we determine that the intersection lies in the triangular region pictured below.

The graph suggests that \((−1, 1)\) is a simultaneous solution. As a check, we could substitute that point into the inequalities and verify that it solves all three conditions.

Use the same technique to graph the solution sets to systems of nonlinear inequalities.

Example \(\PageIndex{5}\):

Graph the solution set: \(\left\{ \begin{array} { l } { y < ( x + 1 ) ^ { 2 } } \\ { y \leq - \frac { 1 } { 2 } x + 3 } \end{array} \right.\).

The first inequality has a parabolic boundary. This boundary is a horizontal translation of the basic function \(y = x^{2}\) to the left \(1\) unit. Because of the strict inequality, the boundary is dashed, indicating that it is not included in the solution set. The second inequality is linear and will be graphed with a solid boundary. Solution sets to both are graphed below.

After graphing the inequalities on the same set of axes, we determine that the intersection lies in the region pictured below.

Exercise \(\PageIndex{1}\)

Graph the solution set: \(\left\{ \begin{array} { l } { y \geq - | x + 1 | + 3 } \\ { y \leq 2 } \end{array} \right.\).

www.youtube.com/v/9z87e7Iw9JE

Key Takeaways

• To graph solutions to systems of inequalities, graph the solution sets of each inequality on the same set of axes and determine where they intersect.
• You can check your answer by choosing a few values inside and out of the shaded region to see if they satisfy the inequalities or not. While this is not a proof, doing so will give a good indication that you have graphed the correct region.

Exercise \(\PageIndex{2}\)

Determine whether or not the given point is a solution to the given system of inequalities.

1. \((-2,1)\);

\(\left\{ \begin{array} { l } { y > 3 x + 5 } \\ { y \leq - x + 1 } \end{array} \right.\)

2. \((-1,-3)\);

\(\left\{ \begin{array} { l } { y \geq 3 x - 1 } \\ { y < - 2 x } \end{array} \right.\)

3. \((-2,-1)\);

\(\left\{ \begin{array} { l } { x - 2 y > - 1 } \\ { 3 x - y < - 3 } \end{array} \right.\)

4. \((0,-5)\);

\(\left\{ \begin{array} { c } { 5 x - y \geq 5 } \\ { 3 x + 2 y < - 1 } \end{array} \right.\)

5. \((-\frac{1}{2} ,0)\);

\(\left\{ \begin{array} { l } { - 8 x + 5 y \geq 3 } \\ { 2 x - 3 y < 0 } \end{array} \right.\)

6. \((-1, \frac{1}{2})\);

\(\left\{ \begin{array} { l } { 2 x - 9 y < - 1 } \\ { 3 x - 6 y > - 2 } \end{array} \right.\)

7. \((-1,-2)\);

\(\left\{ \begin{array} { c } { 2 x - y \geq - 1 } \\ { x - 3 y < 6 } \\ { 2 x - 3 y > - 1 } \end{array} \right.\)

8. \((-5,2)\);

\(\left\{ \begin{array} { c } { - x + 5 y > 10 } \\ { 2 x + y < 1 } \\ { x + 3 y < - 2 } \end{array} \right.\)

9. \((0,3)\);

\(\left\{ \begin{array} { l } { y + 4 \geq 0 } \\ { \frac { 1 } { 2 } x + \frac { 1 } { 3 } y \leq 1 } \\ { - 3 x + 2 y \leq 6 } \end{array} \right.\)

10. \((1,1)\);

• \(\left\{ \begin{array} { l } { y \leq - \frac { 3 } { 4 } x + 2 } \\ { y \geq - 5 x + 2 } \\ { y \geq \frac { 1 } { 3 } x - 1 } \end{array} \right.\)

11. \((-1,2)\);

\( \left\{ \begin{array} { l } { y \geq x ^ { 2 } + 1 } \\ { y < - 2 x + 3 } \end{array} \right.\)

12. \((4,5)\);

\(\left\{ \begin{array} { l } { y < ( x - 1 ) ^ { 2 } - 1 } \\ { y > \frac { 1 } { 2 } x - 1 } \end{array} \right.\)

13. \((-2,-3)\);

\(\left\{ \begin{array} { l } { y < 0 } \\ { y \geq - | x | + 4 } \end{array} \right.\)

14. \((1,2)\);

\(\left\{ \begin{array} { l } { y < | x - 3 | + 2 } \\ { y \geq 2 } \end{array} \right.\)

15. \(\left( - \frac { 1 } { 2 } , - 5 \right)\);

\(\left\{ \begin{array} { l } { y \leq - 3 x - 5 } \\ { y > ( x - 1 ) ^ { 2 } - 10 } \end{array} \right.\)

16. \((-4,1)\)

\(\left\{ \begin{array} { l } { x \geq - 5 } \\ { y < ( x + 3 ) ^ { 2 } - 2 } \end{array} \right.\)

17. \(\left( - \frac { 3 } { 2 } , \frac { 1 } { 3 } \right)\);

\(\left\{ \begin{array} { l } { x - 2 y \leq 4 } \\ { y \leq | 3 x - 1 | + 2 } \end{array} \right.\)

18, \(\left( - 3 , - \frac { 3 } { 4 } \right)\);

\(\left\{ \begin{array} { l } { 3 x - 4 y < 24 } \\ { y < ( x + 2 ) ^ { 2 } - 1 } \end{array} \right.\)

19. \((4,2)\);

• \(\left\{ \begin{array} { l } { y < ( x - 3 ) ^ { 2 } + 1 } \\ { y < - \frac { 3 } { 4 } x + 5 } \end{array} \right.\)

20. \((\frac{5}{2}, 1)\)

• \(\left\{ \begin{array} { l } { y \geq - 1 } \\ { y < - ( x - 2 ) ^ { 2 } + 3 } \end{array} \right.\)

Exercise \(\PageIndex{3}\)

Graph the solution set.

• \(\left\{ \begin{array} { l } { y \geq \frac { 2 } { 3 } x - 3 } \\ { y < - \frac { 1 } { 3 } x + 3 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y \geq - \frac { 1 } { 4 } x + 1 } \\ { y < \frac { 1 } { 2 } x - 2 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y > \frac { 2 } { 3 } x + 1 } \\ { y > \frac { 4 } { 3 } x - 5 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y \leq - 5 x + 4 } \\ { y < \frac { 4 } { 3 } x - 2 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { x - y \geq - 3 } \\ { x + y \geq 3 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { 3 x + y < 4 } \\ { 2 x - y \leq 1 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { - x + 2 y \leq 0 } \\ { 3 x + 5 y < 15 } \end{array} \right.\)
• \(\left\{ \begin{array} { c } { 2 x + 3 y < 6 } \\ { - 4 x + 3 y \geq - 12 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { 3 x + 2 y > 1 } \\ { 4 x - 2 y > 3 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { x - 4 y \geq 2 } \\ { 8 x + 4 y \leq 3 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { 5 x - 2 y \leq 6 } \\ { - 5 x + 2 y < 2 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { 12 x + 10 y > 20 } \\ { 18 x + 15 y < - 15 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { x + y < 0 } \\ { y + 4 > 0 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { x > - 3 } \\ { y < 1 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { 2 x - 2 y < 0 } \\ { 3 x - 3 y > 3 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y + 1 \leq 0 } \\ { y + 3 \geq 0 } \end{array} \right.\)
• Construct a system of linear inequalities that describes all points in the first quadrant.
• Construct a system of linear inequalities that describes all points in the second quadrant.
• Construct a system of linear inequalities that describes all points in the third quadrant.
• Construct a system of linear inequalities that describes all points in the fourth quadrant.

15. \(\varnothing\)

17. \(\left\{ \begin{array} { l } { x > 0 } \\ { y > 0 } \end{array} \right.\)

19. \(\left\{ \begin{array} { l } { x < 0 } \\ { y < 0 } \end{array} \right.\)

Exercise \(\PageIndex{4}\)

• \(\left\{ \begin{array} { l } { y \geq - \frac { 1 } { 2 } x + 3 } \\ { y \geq \frac { 3 } { 2 } x - 3 } \\ { y \leq \frac { 3 } { 2 } x + 1 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { 3 x - 2 y > 6 } \\ { 5 x + 2 y > 8 } \\ { - 3 x + 4 y \leq 4 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { 3 x - 5 y > - 15 } \\ { 5 x - 2 y \leq 8 } \\ { x + y < - 1 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { 3 x - 2 y < - 1 } \\ { 5 x + 2 y > 7 } \\ { y + 1 > 0 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { 3 x - 2 y < - 1 } \\ { 5 x + 2 y < 7 } \\ { y + 1 > 0 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { 4 x + 5 y - 8 < 0 } \\ { y > 0 } \\ { x + 3 > 0 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y - 2 < 0 } \\ { y + 2 > 0 } \\ { 2 x - y \geq 0 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { \frac { 1 } { 2 } x + \frac { 1 } { 2 } y < 1 } \\ { x < 3 } \\ { - \frac { 1 } { 2 } x + \frac { 1 } { 2 } y \leq 1 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { \frac { 1 } { 2 } x + \frac { 1 } { 3 } y \leq 1 } \\ { y + 4 \geq 0 } \\ { - \frac { 1 } { 2 } x + \frac { 1 } { 3 } y \leq 1 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y < x + 2 } \\ { y \geq x ^ { 2 } - 3 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y \geq x ^ { 2 } + 1 } \\ { y > - \frac { 3 } { 4 } x + 3 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y \leq ( x + 2 ) ^ { 2 } } \\ { y \leq \frac { 1 } { 3 } x + 4 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y < - ( x + 1 ) ^ { 2 } - 1 } \\ { y < \frac { 3 } { 2 } x - 2 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y \leq \frac { 1 } { 3 } x + 3 } \\ { y \geq | x + 3 | - 2 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y \leq - x + 5 } \\ { y > | x - 1 | + 2 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y > - | x - 2 | + 5 } \\ { y > 2 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y \leq - | x | + 3 } \\ { y < \frac { 1 } { 4 } x } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y > | x | + 1 } \\ { y \leq x - 1 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y \leq | x | + 1 } \\ { y > x - 1 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y \leq | x - 3 | + 1 } \\ { x \leq 2 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y > | x + 1 | } \\ { y < x - 2 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y < x ^ { 3 } + 2 } \\ { y \leq x + 3 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y \leq 4 } \\ { y \geq ( x + 3 ) ^ { 3 } + 1 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y \geq - 2 x + 6 } \\ { y > \sqrt { x } + 3 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y \leq \sqrt { x + 4 } } \\ { x \leq - 1 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y \leq - x ^ { 2 } + 4 } \\ { y \geq x ^ { 2 } - 4 } \end{array} \right.\)
• \(\left\{ \begin{array} { l } { y \geq | x - 1 | - 3 } \\ { y \leq - | x - 1 | + 3 } \end{array} \right.\)

21. \(\varnothing\)

33 A set of two or more inequalities with the same variables.

• School Guide
• Mathematics
• Number System and Arithmetic
• Trigonometry
• Probability
• Mensuration
• Maths Formulas
• Class 8 Maths Notes
• Class 9 Maths Notes
• Class 10 Maths Notes
• Class 11 Maths Notes
• Class 12 Maths Notes
• Solving Absolute Value Equations
• Absolute Value
• Linear Inequalities
• Real-Life Applications of Absolute Value
• Compound Inequalities
• Inequalities
• Mean Absolute Deviation
• Absolute Value of a Complex Number
• Quadratic Inequalities
• Inequalities in LaTeX
• Ruby | BigDecimal absolute value
• Get the absolute values in Pandas
• Ruby | BigDecimal class absolute value
• Python | Absolute value of list elements
• Python - Absolute Tuple Summation
• GRE Algebra | Solving Linear Inequalities
• Program to find absolute value of a given number
• Getting square of absolute value in Julia - abs2() Method
• Minimize the maximum value of the absolute difference

Absolute Value Inequalities

Inequalities that involve algebraic expressions with absolute value symbols and inequality symbols are called Absolute Value Inequality. In this article, we will discuss inequalities and absolute value inequalities and others in detail.

Table of Content

What is Inequalities?

What is absolute value inequalities, solving absolute value inequalities, types of absolute value inequalities, intersection and union in absolute value inequalities, examples on absolute value inequalities.

Inequalities can be understood as the complement of equations. Inequalities are used to establish a relationship between two or more mathematical values based on the unequal relationship between them. Let us look at an example

• 5 + x > y + 10
• x > y + 5

The above relationship indicates that x is not equal to y+5 rather it is greater than the value of y after 5 is added to it.

Absolute Value Inequalities are a subcategory of inequalities that compare absolute values of mathematical quantities. These usually include symbols like >, < which denote unequal relationships.

What is Absolute Value?

Let us see the formal definition

An absolute value inequality is an expression that uses absolute value expression and variables to denote the relationship between quantities.

We can categorize the inequality into two major types:

• Lesser Than or Equal To
• Greater Than or Equal To
• Compound Inequalities with Absolute Values

Each of the given types is denoted using different symbols which will be discussed later.

Example of Absolute Value Inequalities

Here are some examples to understand the Absolute Value Inequalities

• |x + 5| < 8
• -13 < x < 3

We use number line approach to solve an inequality and follow the steps added below:

Step 1: Write down the inequality and assume it to be an equality making it an equation instead of inequation. Step 2: Draw a number line depending on the intervals and represent the equation on the number line. Step 3: From each interval, select a number and check if it satisfies the inequality. Step 4: Perform step 3 for every interval and the intervals for which a random number satisfies the inequality will be included in your final answer. Step 5: Take the union of all the intervals to get the answer.

Graphical Representation of Inequalities

We can use a graph to plot the inequalities and find the corresponding solution for the inequalities. Let us see how we can use the graph to plot the solution

Note: Open dot ◌ is used for representing an open interval whereas a closed dot ⚈ is used to represent a closed interval in the graph.

Here is the representation of different cases:

Representation of Inequalities

Depending on the type of sign in the inequality, there are different types of inequalities which are mentioned below:

Inequalities with Greater Than Condition

Inequalities with less than condition, compound inequalities involving absolute values.

These inequalities generally use a greater than sign i.e. the number is greater than the value of some other mathematical value. Here are some examples of such inequalities

• x+ y > 7 + 3y
• 65y > x + 22

These inequalities generally use a less-than sign i.e. the number is less than the value of some other mathematical value. Here are some examples of such inequalities

• x + y < 89
• 5y + 6x < 0

As the name suggests, Compound Inequalities involve both greater than and less than cases i.e. the number is less than and greater than the value of some other mathematical value. These inequalities use the absolute value. Here are some examples of such inequalities

• |x – 5| < 7
• |x + 6y| > 89
• |4x + 2| <= 24

Let us understand how we can take Intersection and Union in Absolute Value Inequalities.

Union of Inequalities

For a given set of values, if the inequality is x>=a or x<b then we need to find the union of the values of x which can be given by

Case 1: x >= a or x < b

{x: x < b U x ≥ a}

Case 2: x < a or x >= b

{x: x <a U x ≥ b} = {x: x < a}∪{x: x ≥ b}.

The solution i.e. the union can be calculated using graph. Consider the example x <= 3 || x >= -4 , then the union of the inequalities will give an overlapping interval which will include all real numbers as shown below.

Intersection of Inequalities

For a given set of values, if the inequality is x >= a and x < b then we need to find the intersection of the values of x which can be given by

Case 1: a <= x < b

{x: a≤x < b}

Case 2: a <= x U b > x

{x: a ≤ x U x < b}

The solution i.e. the intersection can be calculated using graph. Consider the example x <= 4 U x >= -5 , then the intersection of the inequalities will give an interval which will include all real numbers from -5 and 4 as shown below.

Triangle Inequality Compound Inequalities Word Problems of Linear Inequalities

Example 1: Solve for inequality |x+24|>-5 using the formula-based approach.

Given Inequality |x + a| > b -∞ < x + a < -b b < x + a < +∞ Solving both of them individually Case 1: -∞ < x < -a – b Case 2: b – a < x < ∞ x ⋿ (-∞,-a-b) ⋃ (b-a, ∞)

Example 2: Solve this less than equal to absolute inequality |y + 5| <= 3y

Given Inequality |y + 5| <= 3y Case 1: y + 5 <= 3y 5 <= 2y 5/2 <= y y ϵ [5/2, ∞)…(i) Case 2: -3y <= y + 5 -4y <= 5 y >= -1.25 y ϵ (-∞, -5/4]…(ii) From eq. (i) and eq. (ii) y ϵ [5/2, -5/4]

Practice Problems on Absolute Value Inequalities

P1. Use the union and intersection method to find the solution for x given |x+7|<1001 and |x+2|>24.

P2. Use the graphical representation method to find the solution of |2x+5|+y>7x

P3. Solve the inequality ∣2x + 3∣ < 5.

P4. Find all values of such that ∣x − 4∣ ≥ 2.

FAQs Absolute Value Inequalities

When do we use curved and square brackets while writing the solution of inequality.

Curved bracket i.e. ( ) is used to denote a quantity that is not included in the range of possible values of the variable whereas the square bracket i.e. [ ] is used when the number is included in the range of values of the variable.

What are the equivalents of curved and square brackets in graphical representation?

Open dot denotes the curved bracket. Closed dot is used instead of the square bracket.

When is the sign switched from negative to positive and vice-versa in an inequality?

Sign in an inequality changes when we multiply the quantity on both sides of the equation by a negative sign or divide the quantity on both sides by a negative value.

Can an absolute value inequality be negative?

No, absolute value inequality can never be negative.

Please Login to comment...

Similar reads.

• linear algebra
• School Learning

Improve your Coding Skills with Practice

What kind of Experience do you want to share?

Artificial Intelligence and gender equality

• Share to Facebook
• Share to Twitter
• Share to LinkedIn
• Share to E-mail

The world has a gender equality problem, and Artificial Intelligence (AI) mirrors the gender bias in our society.

Although globally more women are accessing the internet every year , in low-income countries, only 20 per cent are connected . The gender digital divide creates a data gap that is reflected in the gender bias in AI. 

Who creates AI and what biases are built into AI data (or not), can perpetuate, widen, or reduce gender equality gaps.

What is AI gender bias? 

A study by the Berkeley Haas Center for Equity, Gender and Leadership analysed 133 AI systems across different industries and found that about 44 per cent of them showed gender bias , and 25 per cent exhibited both gender and racial bias.

Beyza Doğuç, an artist from Ankara, Turkey, encountered gender bias in Generative AI when she was researching for a novel and prompted it to write a story about a doctor and a nurse. Generative AI creates new content (text, images, video, etc.) inspired by similar content and data that it was trained on, often in response to questions or prompts by a user.

The AI made the doctor male and the nurse female. Doğuç continued to give it more prompts, and the AI always chose gender stereotypical roles for the characters and associated certain qualities and skills with male or female characters. When she asked the AI about the gender bias it exhibited, the AI explained it was because of the data it had been trained on and specifically, “word embedding” – which means the way certain words are encoded in machine learning to reflect their meaning and association with other words – it’s how machines learn and work with human language. If the AI is trained on data that associates women and men with different and specific skills or interests, it will generate content reflecting that bias.

“Artificial intelligence mirrors the biases that are present in our society and that manifest in AI training data,” said Doğuç, in a recent interview with UN Women.

Who develops AI, and what kind of data it is trained on, has gender implications for AI-powered solutions.

Sola Mahfouz, a quantum computing researcher at Tufts University, is excited about AI, but also concerned. “Is it equitable? How much does it mirror our society’s patriarchal structures and inherent biases from its predominantly male creators,” she reflected. 

Mahfouz was born in Afghanistan, where she was forced to leave school when the Taliban came to her home and threatened her family. She eventually escaped Afghanistan and immigrated to the U.S. in 2016 to attend college.

As companies are scrambling for more data to feed AI systems, researchers from Epoch claim that tech companies could run out of high-quality data used by AI by 2026 .

Natacha Sangwa is a student from Rwanda who participated in the first coding camp organized under the African Girls Can Code Initiative last year. “I have noticed that [AI] is mostly developed by men and trained on datasets that are primarily based on men,” said Sangwa, who saw first-hand how that impacts women’s experience with the technology. “When women use some AI-powered systems to diagnose illnesses, they often receive inaccurate answers, because the AI is not aware of symptoms that may present differently in women.” 

If current trends continue, AI-powered technology and services will continue lacking diverse gender and racial perspectives, and that gap will result in lower quality of services, biased decisions about jobs, credit, health care and more. 

How to avoid gender bias in AI?

Removing gender bias in AI starts with prioritizing gender equality as a goal, as AI systems are conceptualized and built. This includes assessing data for misrepresentation, providing data that is representative of diverse gender and racial experiences, and reshaping the teams developing AI to make them more diverse and inclusive.

According to the Global Gender Gap Report of 2023, there are only 30 per cent women currently working in AI .  

“When technology is developed with just one perspective, it’s like looking at the world half-blind,” concurred Mahfouz. She is currently working on a project to create an AI-powered platform that would connect Afghan women with each other. 

“More women researchers are needed in the field. The unique lived experiences of women can profoundly shape the theoretical foundations of technology. It can also open new applications of the technology,” she added. 

“To prevent gender bias in AI, we must first address gender bias in our society,” said Doğuç from Turkey.

There is a critical need for drawing upon diverse fields of expertise when developing AI, including gender expertise, so that machine learning systems can serve us better and support the drive for a more equal and sustainable world.

In a rapidly advancing AI industry, the lack of gender perspectives, data, and decision-making can perpetuate profound inequality for years to come.

The AI field needs more women, and that requires enabling and increasing girls’ and women’s access to and leadership in STEM and ICT education and careers.

The World Economic Forum reported in 2023 that women accounted for just 29 per cent of all science, technology, engineering and math (STEM) workers. Although more women are graduating and entering STEM jobs today than ever before, they are concentrated in entry level jobs and less likely to hold leadership positions.

How can AI governance help accelerate progress towards gender equality?

International cooperation on digital technology has focused on technical and infrastructural issues and the digital economy, often at the expense of how technological developments were affecting society and generating disruption across all its layers – especially for the most vulnerable and historically excluded. There is a global governance deficit in addressing the challenges and risks of AI and harnessing its potential to leave no one behind.

“Right now, there is no mechanism to constrain developers from releasing AI systems before they are ready and safe. There’s a need for a global multistakeholder governance model that prevents and redresses when AI systems exhibit gender or racial bias, reinforce harmful stereotypes, or does not meet privacy and security standards,” said Helene Molinier, UN Women’s Advisor on Digital Gender Equality Cooperation in a recent interview with Devex.

In the current AI architecture, benefits and risks are not equitably distributed, with power concentrated in the hands of a few corporations, States and individuals, who control talent, data and computer resources. There is also no mechanism to look at broader considerations, like new forms of social vulnerability generated by AI, the disruption of industries and labour markets, the propensity for emerging technology to be used as a tool of oppression, the sustainability of the AI supply chain, or the impact of AI on future generations.

In 2024, the negotiation of the Global Digital Compact (GDC) offers a unique opportunity to build political momentum and place gender perspectives on digital technology at the core of a new digital governance framework. Without it, we face the risk of overlaying AI onto existing gender gaps, causing gender-based discrimination and harm to be left unchanged – and even amplified and perpetuated by AI systems.

UN Women position paper on the GDC provide concrete recommendations to harness the speed, scale, and scope of digital transformation for the empowerment of women and girls in all their diversity, and to trigger transformations that set countries on paths to an equitable digital future for all.

• Science and technology for development
• Innovation and technology

Related content

UN Women statement for the International Girls in ICT Day 2024

UN Women statement for the International Day for Women and Girls in Science

Creating safe digital spaces free of trolls, doxing, and hate speech