problem solving of inequalities

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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

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Solving Inequality Word Questions

(You might like to read Introduction to Inequalities and Solving Inequalities first.)

In Algebra we have "inequality" questions like:

Sam and Alex play in the same soccer team. Last Saturday Alex scored 3 more goals than Sam, but together they scored less than 9 goals. What are the possible number of goals Alex scored?

How do we solve them?

The trick is to break the solution into two parts:

Turn the English into Algebra.

Then use Algebra to solve.

Turning English into Algebra

To turn the English into Algebra it helps to:

• Read the whole thing first
• Do a sketch if needed
• Assign letters for the values
• Find or work out formulas

We should also write down what is actually being asked for , so we know where we are going and when we have arrived!

The best way to learn this is by example, so let's try our first example:

Assign Letters:

• the number of goals Alex scored: A
• the number of goals Sam scored: S

We know that Alex scored 3 more goals than Sam did, so: A = S + 3

And we know that together they scored less than 9 goals: S + A < 9

We are being asked for how many goals Alex might have scored: A

Sam scored less than 3 goals, which means that Sam could have scored 0, 1 or 2 goals.

Alex scored 3 more goals than Sam did, so Alex could have scored 3, 4, or 5 goals .

• When S = 0, then A = 3 and S + A = 3, and 3 < 9 is correct
• When S = 1, then A = 4 and S + A = 5, and 5 < 9 is correct
• When S = 2, then A = 5 and S + A = 7, and 7 < 9 is correct
• (But when S = 3, then A = 6 and S + A = 9, and 9 < 9 is incorrect)

Lots More Examples!

Example: Of 8 pups, there are more girls than boys. How many girl pups could there be?

• the number of girls: g
• the number of boys: b

We know that there are 8 pups, so: g + b = 8, which can be rearranged to

We also know there are more girls than boys, so:

We are being asked for the number of girl pups: g

So there could be 5, 6, 7 or 8 girl pups.

Could there be 8 girl pups? Then there would be no boys at all, and the question isn't clear on that point (sometimes questions are like that).

• When g = 8, then b = 0 and g > b is correct (but is b = 0 allowed?)
• When g = 7, then b = 1 and g > b is correct
• When g = 6, then b = 2 and g > b is correct
• When g = 5, then b = 3 and g > b is correct
• (But if g = 4, then b = 4 and g > b is incorrect)

A speedy example:

Example: Joe enters a race where he has to cycle and run. He cycles a distance of 25 km, and then runs for 20 km. His average running speed is half of his average cycling speed. Joe completes the race in less than 2½ hours, what can we say about his average speeds?

• Average running speed: s
• So average cycling speed: 2s
• Speed = Distance Time
• Which can be rearranged to: Time = Distance Speed

We are being asked for his average speeds: s and 2s

The race is divided into two parts:

• Distance = 25 km
• Average speed = 2s km/h
• So Time = Distance Average Speed = 25 2s hours
• Distance = 20 km
• Average speed = s km/h
• So Time = Distance Average Speed = 20 s hours

Joe completes the race in less than 2½ hours

• The total time < 2½
• 25 2s + 20 s < 2½

So his average speed running is greater than 13 km/h and his average speed cycling is greater than 26 km/h

In this example we get to use two inequalities at once:

Example: The velocity v m/s of a ball thrown directly up in the air is given by v = 20 − 10t , where t is the time in seconds. At what times will the velocity be between 10 m/s and 15 m/s?

• velocity in m/s: v
• the time in seconds: t
• v = 20 − 10t

We are being asked for the time t when v is between 5 and 15 m/s:

So the velocity is between 10 m/s and 15 m/s between 0.5 and 1 second after.

And a reasonably hard example to finish with:

Example: A rectangular room fits at least 7 tables that each have 1 square meter of surface area. The perimeter of the room is 16 m. What could the width and length of the room be?

Make a sketch: we don't know the size of the tables, only their area, they may fit perfectly or not!

• the length of the room: L
• the width of the room: W

The formula for the perimeter is 2(W + L) , and we know it is 16 m

• 2(W + L) = 16
• L = 8 − W

We also know the area of a rectangle is the width times the length: Area = W × L

And the area must be greater than or equal to 7:

• W × L ≥ 7

We are being asked for the possible values of W and L

Let's solve:

So the width must be between 1 m and 7 m (inclusive) and the length is 8−width .

• Say W = 1, then L = 8−1 = 7, and A = 1 x 7 = 7 m 2 (fits exactly 7 tables)
• Say W = 0.9 (less than 1), then L = 7.1, and A = 0.9 x 7.1 = 6.39 m 2 (7 won't fit)
• Say W = 1.1 (just above 1), then L = 6.9, and A = 1.1 x 6.9 = 7.59 m 2 (7 fit easily)
• Likewise for W around 7 m

Inequalities

In Mathematics, equations are not always about being balanced on both sides with an 'equal to' symbol. Sometimes it can be about 'not an equal to' relationship like something is greater than the other or less than. In mathematics, inequality refers to a relationship that makes a non-equal comparison between two numbers or other mathematical expressions. These mathematical expressions come under algebra and are called inequalities.

Let us learn the rules of inequalities, and how to solve and graph them.

What is an Inequality?

Inequalities  are the mathematical expressions in which both sides are not equal. In inequality, unlike in equations, we compare two values. The equal sign in between is replaced by less than (or less than or equal to), greater than (or greater than or equal to), or not equal to sign.

Olivia is selected in the 12U Softball. How old is Olivia? You don't know the age of Olivia, because it doesn't say "equals". But you do know her age should be less than or equal to 12, so it can be written as Olivia's Age ≤ 12. This is a practical scenario related to inequalities.

Inequality Meaning

The meaning of inequality is to say that two things are NOT equal. One of the things may be less than, greater than, less than or equal to, or greater than or equal to the other things.

• p ≠ q means that p is not equal to q
• p < q means that p is less than q
• p > q means that p is greater than q
• p ≤ q means that p is less than or equal to q
• p ≥ q means that p is greater than or equal to q

There are different types of inequalities. Some of the important inequalities are:

• Polynomial inequalities
• Absolute value inequalities
• Rational inequalities

Rules of Inequalities

The rules of inequalities are special. Here are some listed with inequalities examples.

Inequalities Rule 1

When inequalities are linked up you can jump over the middle inequality.

• If, p < q and q < d, then p < d
• If, p > q and q > d, then p > d

Example: If Oggy is older than Mia and Mia is older than Cherry, then Oggy must be older than Cherry.

Inequalities Rule 2

Swapping of numbers p and q results in:

• If, p > q, then q < p
• If, p < q, then q > p

Example: Oggy is older than Mia, so Mia is younger than Oggy.

Inequalities Rule 3

Adding the number d to both sides of inequality: If p < q, then p + d < q + d

Example: Oggy has less money than Mia. If both Oggy and Mia get $5 more, then Oggy will still have less money than Mia.

• If p < q, then p − d < q − d
• If p > q, then p + d > q + d, and
• If p > q, then p − d > q − d

So, the addition and subtraction of the same value to both p and q will not change the inequality.

Inequalities Rule 4

If you multiply numbers p and q by a positive number , there is no change in inequality. If you multiply both p and q by a negative number , the inequality swaps: p<q becomes q<p after multiplying by (-2)

Here are the rules:

• If p < q, and d is positive, then pd < qd
• If p < q, and d is negative, then pd > qd (inequality swaps)

Positive case example: Oggy's score of 5 is lower than Mia's score of 9 (p < q). If Oggy and Mia double their scores '×2', Oggy's score will still be lower than Mia's score, 2p < 2q. If the scores turn minuses, then scores will be −p > −q.

Inequalities Rule 5

Putting minuses in front of p and q changes the direction of the inequality.

• If p < q then −p > −q
• If p > q, then −p < −q
• It is the same as multiplying by (-1) and changes direction.

Inequalities Rule 6

Taking the reciprocal 1/value of both p and q changes the direction of the inequality. When p and q are both positive or both negative:

• If, p < q, then 1/p > 1/q
• If p > q, then 1/p < 1/q

Inequalities Rule 7

A square of a number is always greater than or equal to zero p 2  ≥ 0.

Example: (4) 2 = 16, (−4) 2 = 16, (0) 2 = 0

Inequalities Rule 8

Taking a square root will not change the inequality. If p ≤ q, then √p ≤ √q (for p, q ≥ 0).

Example: p=2, q=7 2 ≤ 7, then √2 ≤ √7

The rules of inequalities are summarized in the following table.

Solving Inequalities

Here are the steps for  solving inequalities :

• Step - 1: Write the inequality as an equation.
• Step - 2: Solve the equation for one or more values.
• Step - 3: Represent all the values on the number line.
• Step - 4: Also, represent all excluded values on the number line using open circles.
• Step - 5: Identify the intervals.
• Step - 6: Take a random number from each interval, substitute it in the inequality and check whether the inequality is satisfied.
• Step - 7: Intervals that are satisfied are the solutions.

But for solving simple inequalities (linear), we usually apply algebraic operations like addition , subtraction , multiplication , and division . Consider the following example:

2x + 3 > 3x + 4

Subtracting 3x and 3 from both sides,

2x - 3x > 4 - 3

Multiplying both sides by -1,

Notice that we have changed the ">" symbol into "<" symbol. Why? This is because we have multiplied both sides of the inequality by a negative number. The process of solving inequalities mentioned above works for a simple linear inequality. But to solve any other complex inequality, we have to use the following process.

Let us use this procedure to solve inequalities of different types.

Graphing Inequalities

While graphing inequalities , we have to keep the following things in mind.

• If the endpoint is included (i.e., in case of ≤ or ≥) use a closed circle.
• If the endpoint is NOT included (i.e., in case of < or >), use an open circle.
• Use open circle at either ∞ or -∞.
• Draw a line from the endpoint that extends to the right side if the variable is greater than the number.
• Draw a line from the endpoint that extends to the left side if the variable is lesser than the number.

Writing Inequalities in Interval Notation

While writing the solution of an inequality in the interval notation , we have to keep the following things in mind.

• If the endpoint is included (i.e., in case of ≤ or ≥) use the closed brackets '[' or ']'
• If the endpoint is not included (i.e., in case of < or >), use the open brackets '(' or ')'
• Use always open bracket at either ∞ or -∞.

Here are some examples to understand the same:

Graphing Inequalities with Two Variables

For graphing  inequalities with two variables , you will have to plot the "equals" line and then, shade the appropriate area. There are three steps:

• Write the equation such as "y" is on the left and everything else on the right.
• Plot the "y=" line (draw a solid line for y≤ or y≥, and a dashed line for y< or y>)
• Shade the region above the line for a "greater than" (y> or y≥) or below the line for a "less than" (y< or y≤).

Let us try some example: This is a graph of a linear inequality: y ≤ x + 4

You can see, y = x + 4 line and the shaded area (in yellow) is where y is less than or equal to x + 4. Let us now see how to solve different types of inequalities and how to graph the solution in each case.

Solving Polynomial Inequalities

The polynomial inequalities are inequalities that can be expressed as a polynomial on one side and 0 on the other side of the inequality. There are different types of polynomial inequalities but the important ones are:

• Linear Inequalities
• Quadratic Inequalities

Solving Linear Inequalities

A linear inequality is an inequality that can be expressed with a linear expression on one side and a 0 on the other side. Solving linear inequalities is as same as solving linear equations , but just the rules of solving inequalities (that was explained before) should be taken care of. Let us see some examples.

Solving One Step Inequalities

Consider an inequality 2x < 6 (which is a linear inequality with one variable ). To solve this, just one step is sufficient which is dividing both sides by 2. Then we get x < 3. Therefore, the solution of the inequality is x < 3 (or) (-∞, 3).

Solving Two Step Inequalities

Consider an inequality -2x + 3 > 6. To solve this, we need two steps . The first step is subtracting 3 from both sides, which gives -2x > 3. Then we need to divide both sides by -2 and it results in x < -3/2 (note that we have changed the sign of the inequality). So the solution of the inequality is x < -3/2 (or) (-∞, -3/2).

Solving Compound Inequalities

Compound inequalities refer to the set of inequalities with either "and" or "or" in between them. For solving inequalities, in this case, just solve each inequality independently and then find the final solution according to the following rules:

• The final solution is the intersection of the solutions of the independent inequalities if we have "and" between them.
• The final solution is the union of the solutions of the independent inequalities if we have "or" between them.

Example: Solve the compound inequality 2x + 3 < -5 and x + 6 < 3.

By first inequality: 2x + 3 < -5 2x < -8 x < -4

By second inequality, x + 6 < 3 x < -3

Since we have "and" between them, we have to find the intersection of the sets x < -4 and x < -3. A number line may be helpful in this case. Then the final solution is:

x < -3 (or) (-∞, -3).

Solving Quadratic Inequalities

A quadratic inequality involves a quadratic expression in it. Here is the process of solving quadratic inequalities . The process is explained with an example where we are going to solve the inequality x 2 - 4x - 5 ≥ 0.

• Step 1: Write the inequality as equation. x 2 - 4x - 5 = 0
• Step 2: Solve the equation. Here we can use any process of solving quadratic equations . Then (x - 5) (x + 1) = 0 x = 5; x = -1.

• Step 5: The inequalities with "true" from the above table are solutions. Therefore, the solutions of the quadratic inequality x 2 - 4x - 5 ≥ 0 is (-∞, -1] U [5, ∞).

We can use the same process for solving cubic inequalities, biquadratic inequalities, etc.

Solving Absolute Value Inequalities

An absolute value inequality includes an algebraic expression inside the absolute value sign. Here is the process of solving absolute value inequalities where the process is explained with an example of solving an absolute value inequality |x + 3| ≤ 2. If you want to learn different methods of solving absolute value inequalities, click here .

• Step 1: Consider the absolute value inequality as equation. |x + 3| = 2
• Step 2: Solve the equation. x + 3 = ±2 x + 3 = 2; x + 3 = -2 x = -1; x = -5

• Step 5: The intervals that satisfied the inequality are the solution intervals. Therefore, the solution of the absolute value inequality |x + 3| ≤ 2 is [-5, -1].

Solving Rational Inequalities

Rational inequalities are inequalities that involve rational expressions (fractions with variables). To solve the rational inequalities (inequalities with fractions), we just use the same procedure as other inequalities but we have to take care of the excluded points . For example, while solving the rational inequality (x + 2) / (x - 2) < 3, we should note that the rational expression (x + 2) / (x - 2) is NOT defined at x = 2 (set the denominator x - 2 = 0 ⇒x = 2). Let us solve this inequality step by step.

• Step 1: Consider the inequality as the equation. (x + 2) / (x - 2) = 3
• Step 2: Solve the equation. x + 2 = 3(x - 2) x + 2 = 3x - 6 2x = 8 x = 4

• Step 5: The intervals that have come up with "true" in Step 4 are the solutions. Therefore, the solution of the rational inequality (x + 2) / (x - 2) < 3 is (-∞, 2) U (4, ∞).

Important Notes on Inequalities:

Here are the notes about inequalities:

• If we have strictly less than or strictly greater than symbol, then we never get any closed interval in the solution.
• We always get open intervals at ∞ or -∞ symbols because they are NOT numbers to include.
• Write open intervals always at excluded values when solving rational inequalities.
• Excluded values should be taken care of only in case of rational inequalities.

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Inequalities Examples

Example 1: Using the techniques of solving inequalities, solve: -19 < 3x + 2 ≤ 17 and write the answer in the interval notation.

Given that -19 < 3x + 2 ≤ 17.

This is a compound inequality.

Subtracting 2 from all sides,

-21 < 3x ≤ 15

Dividing all sides by 3,

-7 < x ≤ 5

Answer: The solution is (-7, 5].

Example 2: While solving inequalities, explain why each of the following statements is incorrect. Also, correct them. a) 2x < 5 ⇒ x > 5/2 b) x > 3 ⇒ x ∈ [3, ∞) c) -x > -7 ⇒ x > 7.

a) 2x < 5. Here, when we divide both sides by 2, which is a positive number, the sign does not change. So the correct inequality is x < 5/2.

b) x > 3. It does not include an equal to symbol. So 3 should NOT be included in the interval. So the correct interval is (3, ∞).

c) -x > -7. When we divide both sides by -1, a negative number, the sign should change. So the correct inequality is x < 7.

Answer: The corrected ones are a) x < 5/2; b) x ∈ (3, ∞); c) x < 7.

Example 3: Solve the inequality x 2 - 7x + 10 < 0.

First, solve the equation x 2 - 7x + 10 = 0.

(x - 2) (x - 5) = 0.

x = 2, x = 5.

If we represent these numbers on the number line, we get the following intervals: (-∞, 2), (2, 5), and (5, ∞).

Let us take some random numbers from each interval to test the given quadratic inequality.

Therefore, the only interval that satisfies the inequality is (2, 5).

Answer: The solution is (2, 5).

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Practice Questions on Inequalities

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FAQs on Inequalities

What are inequalities in math.

When two or more algebraic expressions are compared using the symbols <, > ≤, or ≥, then they form an inequality. They   are the mathematical expressions in which both sides are not equal.

How Do you Solve Inequalities On A Number Line?

To plot an inequality in math, such as x>3, on a number line,

• Step 1: Draw a circle over the number (e.g., 3).
• Step 2: Check if the sign includes equal to (≥ or ≤) or not. If equal to sign is there along with > or <, then fill in the circle otherwise leave the circle unfilled.
• Step 3: On the number line, extend the line from 3(after encircling it) to show it is greater than or equal to 3.

How to Calculate Inequalities in Math?

To  calculate inequalities :

• just make it an equation
• mark the zeros  on the number line to get intervals
• test the intervals by taking any one number from it against the inequality.

Explain the Process of Solving Inequalities Graphically.

Solving inequalities graphically is possible when we have a system of two inequalities in two variables. In this case, we consider both inequalities as two linear equations and graph them. Then we get two lines. Shade the upper/lower portion of each of the lines that satisfies the inequality. The common portion of both shaded regions is the solution region.

What is the Difference Between Equations and Inequalities?

Here are the differences between equations and inequalities.

What Happens When you Square An Inequality?

A square of a number is always greater than or equal to zero p 2  ≥ 0. Example: (4) 2 = 16, (−4) 2 = 16, (0) 2 = 0

What are the Steps to Calculate Inequalities with Fractions?

Calculating inequalities with fractions is just like solving any other inequality. One easy way of solving such inequalities is to multiply every term on both sides by the LCD of all denominators so that all fractions become integers . For example, to solve (1/2) x + 1 > (3/4) x + 2, multiply both sides by 4. Then we get 2x + 4 > 3x + 8 ⇒ -x > 4 ⇒ x < -4.

What are the Steps for Solving Inequalities with Variables on Both Sides?

When an inequality has a variable on both sides, we have to try to isolate the variable. But in this process, flip the inequality sign whenever we are dividing or multiplying both sides by a negative number. Here is an example. 3x - 7 < 5x - 11 ⇒ -2x < -4 ⇒ x > 2.

How Do you Find the Range of Inequality?

You can find the range of values of x, by solving the inequality by considering it as a normal linear equation.

What Are the 5 Inequality Symbols?

The 5 inequality symbols are less than (<), greater than (>), less than or equal (≤), greater than or equal (≥), and the not equal symbol (≠).

How Do you Tell If It's An Inequality?

Equations and inequalities are mathematical sentences formed by relating two expressions to each other. In an equation, the two expressions are supposed to be equal and shown by the symbol =. Whereas in inequality, the two expressions are not necessarily equal and are indicated by the symbols: >, <, ≤ or ≥.

How to Graph the Solution After Solving Inequalities?

After solving inequalities, we can graph the solution keeping the following things in mind.

• Use an open circle at the number if it is not included and use a closed circle if it is included.
• Draw a line to the right side of the number in case of '>' and to the left side of the number in case of '<'.

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Equations and Inequalities Involving Signed Numbers

In chapter 2 we established rules for solving equations using the numbers of arithmetic. Now that we have learned the operations on signed numbers, we will use those same rules to solve equations that involve negative numbers. We will also study techniques for solving and graphing inequalities having one unknown.

SOLVING EQUATIONS INVOLVING SIGNED NUMBERS

Upon completing this section you should be able to solve equations involving signed numbers.

Example 1 Solve for x and check: x + 5 = 3

Using the same procedures learned in chapter 2, we subtract 5 from each side of the equation obtaining

Example 2 Solve for x and check: - 3x = 12

Dividing each side by -3, we obtain

LITERAL EQUATIONS

• Identify a literal equation.
• Apply previously learned rules to solve literal equations.

An equation having more than one letter is sometimes called a literal equation . It is occasionally necessary to solve such an equation for one of the letters in terms of the others. The step-by-step procedure discussed and used in chapter 2 is still valid after any grouping symbols are removed.

Example 1 Solve for c: 3(x + c) - 4y = 2x - 5c

First remove parentheses.

At this point we note that since we are solving for c, we want to obtain c on one side and all other terms on the other side of the equation. Thus we obtain

Sometimes the form of an answer can be changed. In this example we could multiply both numerator and denominator of the answer by (- l) (this does not change the value of the answer) and obtain

The advantage of this last expression over the first is that there are not so many negative signs in the answer.

The most commonly used literal expressions are formulas from geometry, physics, business, electronics, and so forth.

Notice in this example that r was left on the right side and thus the computation was simpler. We can rewrite the answer another way if we wish.

GRAPHING INEQUALITIES

• Use the inequality symbol to represent the relative positions of two numbers on the number line.
• Graph inequalities on the number line.

The symbols are inequality symbols or order relations and are used to show the relative sizes of the values of two numbers. We usually read the symbol as "greater than." For instance, a > b is read as "a is greater than b." Notice that we have stated that we usually read a < b as a is less than b. But this is only because we read from left to right. In other words, "a is less than b" is the same as saying "b is greater than a." Actually then, we have one symbol that is written two ways only for convenience of reading. One way to remember the meaning of the symbol is that the pointed end is toward the lesser of the two numbers.

In simpler words this definition states that a is less than b if we must add something to a to get b. Of course, the "something" must be positive.

If you think of the number line, you know that adding a positive number is equivalent to moving to the right on the number line. This gives rise to the following alternative definition, which may be easier to visualize.

Example 1 3 < 6, because 3 is to the left of 6 on the number line.

Example 2 - 4 < 0, because -4 is to the left of 0 on the number line.

Example 3 4 > - 2, because 4 is to the right of -2 on the number line.

Example 4 - 6 < - 2, because -6 is to the left of -2 on the number line.

The mathematical statement x < 3, read as "x is less than 3," indicates that the variable x can be any number less than (or to the left of) 3. Remember, we are considering the real numbers and not just integers, so do not think of the values of x for x < 3 as only 2, 1,0, - 1, and so on.

As a matter of fact, to name the number x that is the largest number less than 3 is an impossible task. It can be indicated on the number line, however. To do this we need a symbol to represent the meaning of a statement such as x < 3.

The symbols ( and ) used on the number line indicate that the endpoint is not included in the set.

Example 5 Graph x < 3 on the number line.

Note that the graph has an arrow indicating that the line continues without end to the left.

Example 6 Graph x > 4 on the number line.

Example 7 Graph x > -5 on the number line.

Example 8 Make a number line graph showing that x > - 1 and x < 5. (The word "and" means that both conditions must apply.)

Example 9 Graph - 3 < x < 3.

Example 10 x >; 4 indicates the number 4 and all real numbers to the right of 4 on the number line.

The symbols [ and ] used on the number line indicate that the endpoint is included in the set.

Example 13 Write an algebraic statement represented by the following graph.

Example 14 Write an algebraic statement for the following graph.

Example 15 Write an algebraic statement for the following graph.

SOLVING INEQUALITIES

Upon completing this section you should be able to solve inequalities involving one unknown.

The solutions for inequalities generally involve the same basic rules as equations. There is one exception, which we will soon discover. The first rule, however, is similar to that used in solving equations.

If the same quantity is added to each side of an inequality , the results are unequal in the same order.

Example 1 If 5 < 8, then 5 + 2 < 8 + 2.

Example 2 If 7 < 10, then 7 - 3 < 10 - 3.

We can use this rule to solve certain inequalities.

Example 3 Solve for x: x + 6 < 10

If we add -6 to each side, we obtain

Graphing this solution on the number line, we have

We will now use the addition rule to illustrate an important concept concerning multiplication or division of inequalities.

Suppose x > a.

Now add - x to both sides by the addition rule.

Now add -a to both sides.

The last statement, - a > -x, can be rewritten as - x < -a. Therefore we can say, "If x > a, then - x < -a. This translates into the following rule:

If an inequality is multiplied or divided by a negative number, the results will be unequal in the opposite order.

Example 5 Solve for x and graph the solution: -2x>6

To obtain x on the left side we must divide each term by - 2. Notice that since we are dividing by a negative number, we must change the direction of the inequality.

Take special note of this fact. Each time you divide or multiply by a negative number, you must change the direction of the inequality symbol. This is the only difference between solving equations and solving inequalities.

Once we have removed parentheses and have only individual terms in an expression, the procedure for finding a solution is almost like that in chapter 2.

Let us now review the step-by-step method from chapter 2 and note the difference when solving inequalities.

First Eliminate fractions by multiplying all terms by the least common denominator of all fractions. (No change when we are multiplying by a positive number.) Second Simplify by combining like terms on each side of the inequality. (No change) Third Add or subtract quantities to obtain the unknown on one side and the numbers on the other. (No change) Fourth Divide each term of the inequality by the coefficient of the unknown. If the coefficient is positive, the inequality will remain the same. If the coefficient is negative, the inequality will be reversed. (This is the important difference between equations and inequalities.)

• A literal equation is an equation involving more than one letter.
• The symbols are inequality symbols or order relations .
• a a is to the left of b on the real number line.
• To solve a literal equation for one letter in terms of the others follow the same steps as in chapter 2.
• To solve an inequality use the following steps: Step 1 Eliminate fractions by multiplying all terms by the least common denominator of all fractions. Step 2 Simplify by combining like terms on each side of the inequality. Step 3 Add or subtract quantities to obtain the unknown on one side and the numbers on the other. Step 4 Divide each term of the inequality by the coefficient of the unknown. If the coefficient is positive, the inequality will remain the same. If the coefficient is negative, the inequality will be reversed. Step 5 Check your answer.

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• What are the 4 inequalities?
• There are four types of inequalities: greater than, less than, greater than or equal to, and less than or equal to.
• What is a inequality in math?
• In math, inequality represents the relative size or order of two values.
• How do you solve inequalities?
• To solve inequalities, isolate the variable on one side of the inequality, If you multiply or divide both sides by a negative number, flip the direction of the inequality.
• What are the 2 rules of inequalities?
• The two rules of inequalities are: If the same quantity is added to or subtracted from both sides of an inequality, the inequality remains true. If both sides of an inequality are multiplied or divided by the same positive quantity, the inequality remains true. If we multiply or divide both sides of an inequality by the same negative number, we must flip the direction of the inequality to maintain its truth.

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2.2: Solving Inequalities

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• Page ID 62899

• Amy Givler Chapman, Meagan Herald, Jessica Libertini
• Virginia Military Institute

In this section, we will look at solving inequalities. You will often work with inequalities in calculus, particularly when you work with derivatives. A derivative is a function that tells you how quickly a related function is changing. A positive derivative tells you the function is increasing and a negative derivative tells you the function is decreasing. This means you will need to be able to identify when the derivative is greater than zero and when it is less than zero.

When solving inequalities, mathematicians express their answers using interval notation, a special way of expressing an interval of numbers. The intervals will tell us when the inequality is a true statement, i.e., they tell us all the input values that make the inequality valid. You will also hear mathematicians use the phrase “the inequality holds for...’; this is another way of saying that these are the inputs that make the inequality true. Let’s familiarize ourselves with interval notation before we look at inequalities.

2.2.1 Interval Notation

Before we get to solving inequalities, we’ll discuss interval notation. Interval notation provides us with a way to describe ranges of numbers concisely. Unlike order of operations, with interval notation parentheses and brackets have different meaning. For example, \([1,4.5]\) is the range of numbers between 1 and 4.5, including those endpoints. For example, 1, 2, \(\pi\) , and 4.5 are all included in that interval, but -1.2, 85, and 4.5000001 are not. However, if we look at \((1,4.5)\) , 2 and \(\pi\) are still in this interval but 1 and 4.5 are not. Brackets tell us we include the endpoint and parentheses tell us that we don’t.

With interval notation, we can mix parentheses and brackets if we need to include one endpoint but not the other. For example, \([1,4.5)\) contains 1 but not 4.5 and \((1,4.5]\) contains 4.5 but not 1.

Example \(\PageIndex{1}\): Interval Notation

Determine if each of the following numbers is included in the interval \([-5, 27)\) .

\[\begin{array}{lll}{\text{1. }2}&{\qquad}&{\text{5. }27} \\ {\text{2. }\pi}&{\qquad}&{\text{6. }32} \\ {\text{3. }-5}&{\qquad}&{\text{7. }-5.000001} \\ {\text{4. }-8}&{\qquad}&{\text{8. }-4.999999}\end{array}\nonumber\]

For each of these, we need to determine if the number is between the two numbers given in the interval.

\(2\) is bigger than \(-5\) and smaller than \(27\) so it is in the interval.

\(\pi\) is bigger than \(-5\) and smaller than \(27\) so it is in the interval.

\(-5\) is one of our endpoints, so we need to see if it has a bracket or a parenthesis on that end. It has a bracket, so it is included in the interval.

\(-8\) is smaller than \(-5\) , so it is not included in the interval.

\(27\) is one of our endpoints, so we need to see if it has a bracket or a parenthesis on that end. It has a parenthesis, so it is not included in the interval.

\(32\) is bigger than \(27\) , so it is not included in the interval.

\(-5.000001\) is smaller than \(-5\) , so it is not included in the interval.

\(-4.999999\) is bigger than \(-5\) and smaller than \(27\) so it is in the interval.

\[2,\: \pi,\: -5,\text{ and }-4.999999\text{ are in the interval}\]

\[-8,\: 27,\: 32,\text{ and }-5.000001\text{ are not in the interval}\]

We can also use interval notation to express ranges that don’t have an upper bound. For example, if we wanted to use interval notation to write the range for all positive numbers, we would write \((0,\infty)\) . We know there’s no limit to how big a positive number can get, so we use \(\infty\) to indicate that we are just looking at numbers bigger than \(0\) . Similarly, we can write \((-\infty,0)\) to express the range for all negative numbers. Note that for both of these we use a parenthesis with the infinity symbol and not a bracket since infinity isn’t a number.

Additionally, we can use interval notation to express more complicated ranges of numbers. We can combine ranges using \(\cup\) , the shorthand mathematical way of writing “or”. For example, \([1,3]\cup(4,\infty)\) means the range of values between 1 and 3, including the endpoints, as well as any numbers bigger than 4. So 2, 4.1, and 20 are all in this interval, but -2, 3.5, and 4 are not. We can also use notation to limit ranges using \(\cap\) , the shorthand mathematical way of writing “and also.” For example, if we have two ranges, say \((1,4]\) and \([2,8)\) , and are only interested in the numbers that are in both ranges, we can write \((1,4] \cap [2,8)\) . We can use this symbol to help show our work, but for our final answer we should always simplify so that we don’t need to use the \(\cap\) symbol (it’s fine, and quite common, to use the \(\cup\) symbol as part of your final answer). We said that \(\cap\) means we only want the numbers that are in both intervals; the smallest number contained by both intervals is 2 and the largest is 4, so we can write \((1,4] \cap [2,8) = [2,4]\) instead.

2.2.2 Interval Notation and Inequalities 

Interval notation also gives us another way of expressing an inequality. For example, \(x\geq 2\) can be written as \(x \in [2,\infty)\) . Here the \(\in\) symbol is read as the word “in”. We would read this out loud by saying “x is greater than or equal to 2” is the same as “x is in the range from 2, inclusive, to infinity.” The statement \(x \in (2,\infty)\) is a bit different; it’s the same as \(x > 2\) since we don’t want to include 2 as part of our range. Here, we would read the range as “x in 2, exclusive, to infinity.” The symbols we learned earlier, \(\cup\) and \(\cap\) are read as “union” and “intersect,” respectively.

When working with inequalities, we will start all inequality problems by turning them into equality problems. This will allow us to use some techniques we’ve already seen when we discussed factoring and roots of a function. The solution(s) to the equality problem will tell us the “break points,” the input values where the inequality may switch from being true to being false. We’ll test values on both sides of each break point to see where the inequality is true. We will work carefully to make sure we find all the break points because we don’t want to miss any places where the inequality could switch from true to false. Let’s look at a few straightforward examples before we move onto the more complicated inequalities.

Example \(\PageIndex{2}\): Polynomial Inequality

Solve \(x^2 -6x +8 >0\) .

Our first step is to convert this into an equality statement by changing the \(>\) symbol to an \(=\) symbol: \[x^2-6x+8 =0\] Now, we can use any solution method we learned for finding the roots of a quadratic function to solve. Here we have a quadratic that factors nicely, so we will take that approach, but you could use the quadratic formula if you prefer. \[\begin{align}\begin{aligned}\begin{split} x^2 -6x+8 & = 0 \\ (x-2)(x-4) & = 0 \\ x & =2,4 \end{split}\end{aligned}\end{align}\] This tells us that the break points are \(x=2\) and \(x=4\) . These are the only places that the inequality could change from being true to being false for this type of function. We’ll test values on each side of both break points; this means we need to test a value that is less than 2, a value between 2 and 4, and a value bigger than 4. We like to work from left to right, so we will start with testing a value less than 2. We can pick any number that is less than 2, but we will use 0 because it is easy to work with. If we substitute in \(x=0\) we get: \[x^2-6x + 8 = (0)^2-6(0)+8 = 8 >0\] We get 8, which is bigger than 0, so the inequality is true for all values less than 2. Next, we need to test a value between 2 and 4; 3 seems like the easiest option. \[x^2-6x + 8 = (3)^2-6(3)+8 = 9-18+8 = -1 <0\] We get a negative number, so the inequality is false for everything between 2 and 4. Now, we need to test something bigger than 4. We’ll use 5, but you can pick any number, as long as it’s bigger than 4. \[x^2-6x + 8 = (5)^2-6(5)+8 = 25 - 30 + 8 = 3 >0\] The result is positive, so the inequality is true. Now, we have that the inequality is true for numbers less than 2 and numbers greater than 4. It is not true for \(x=2\) or \(x=4\) since both of these make the left side 0 and we want the left side to be bigger than 0, not equal to it. In interval notation, we have:

\[x\in (-\infty,2)\cup(4,\infty)\]

In this example, we have a strict inequality . We say it is strict because it is \(>\) and not \(\geq\) . Similarly, we would say an inequality with \(<\) is strict. With strict inequalities, our final answer will not include the break points, so we will use parentheses at these break points because we do not want to include these points.

Many people will use a number line when working with inequalities. When using a number line, you would mark each break point, and then shade or otherwise mark the intervals where the inequality is true. For the previous problem, this would look like the following:

Figure \(\PageIndex{1}\)

Since we have a strict inequality (meaning we have \(>\) or \(<\) so that we are strictly greater than or strictly less than), we use open circles to mark our break points. This reminds us that we do not include these points in our intervals. Some people will use check-marks and X’s instead, with a check-marks indicating where the inequality holds and a X where it doesn’t. This would look like the following:

Figure \(\PageIndex{2}\)

These number lines become quite useful if you have a lot of break points. They make it very clear so that you can be sure to test a point in each interval. We can also use a table to summarize results, rather than using a number line. The table has a few key advantages: it clearly summarizes your work making your thought process easier to follow and will help eliminate careless errors from your work. A table for the previous example might look like:

Table \(\PageIndex{1}\)

Any of these methods are appropriate for clearly showing your work; the one you choose is a matter of personal preference.

2.2.3: Incorporating Undefined Points

We noted earlier that our inequality can change from true to false at our break points, the points where the equality statement is true. The inequality can also change from true to false at places where the function is undefined. For example, we know that the function \(f(x) = \frac{1}{x}\) is positive when \(x\) is positive and negative when \(x\) is negative. This means that the inequality \(\frac{1}{x}>0\) holds, or is true, only for \(x \in (0,\infty)\) . However, there are no places where \(f(x)=0\) . Since \(f(x)\) is undefined at \(x=0\) , it introduces a different type of break point; one where the graph of the function “breaks” because it cannot be graphed where it is undefined. Let’s take a look at an example where we have to incorporate undefined points by including additional break points.

Example \(\PageIndex{3}\): Solving a Rational Inequality

Solve \(\displaystyle \frac{x-5}{x^2-4} \geq 0\) .

We’ll start by turning the inequality into an equality statement and solving for \(x\) . To help solve for \(x\) , we will get rid of the fraction by multiplying both sides by the full denominator; this will allow us to cancel the denominator on the left side. \[\begin{align}\begin{aligned}\begin{split} \frac{x-5}{x^2-4} &= 0 \\ (x^2-4) \Bigg(\frac{x-5}{x^2-4} \Bigg) & = (x^2-4)(0) \\ x-5 &= 0 \\ x &= 5 \end{split}\end{aligned}\end{align}\] This gives us a break point at \(x=5\) . Next, we will need to see if the function is undefined at any points. Since it is a rational function (a fraction with a polynomial in the numerator and a polynomial in the denominator), we know it is undefined anywhere that the denominator equals zero. Let’s look for these points: \[\begin{align}\begin{aligned}\begin{split} x^2 - 4 &= 0 \\ (x-2)(x+2) & = 0 \\ x &= 2, -2 \end{split}\end{aligned}\end{align}\] We see that \(\frac{x-5}{x^2-4}\) is undefined for \(x=2\) and \(x=-2\) . This gives us two additional break points. That means we have three break points: \(x=5\) , \(x=2\) , and \(x=-2\) . Let’s mark these on a number line. Since this is not a strict inequality, we will use closed circles to mark the break point at \(x=5\) . However, we still need to use open circles at \(x=2\) and \(x=-2\) because the function is undefined at these points and we will not include them in our intervals.

Figure \(\PageIndex{3}\)

Now, we need to check values in each interval. First, we’ll check something less than \(-2\) ; we’ll use \(x=-3\) . Substituting, gives \(\frac{(-3)-5}{(-3)^2 -4} = \frac{-8}{5}\) . This is less than 0; this means we can place an x-mark over this interval:

Figure \(\PageIndex{4}\)

Now, to check something between \(-2\) and \(2\) . We’ll use \(x=0\) . Substituting gives \(\frac{(0)-5}{(0)^2-4} = \frac{-5}{-4} =\frac{5}{4} >0\) . This means we can place a check-mark over this interval:

Figure \(\PageIndex{5}\)

Now, we need to check a value between \(2\) and \(5\) . We’ll use \(x=3\) . We get \(\frac{(3)-5}{(3)^2-4}=\frac{-2}{5} <0\) , so this interval gets an x-mark.

Figure \(\PageIndex{6}\)

Lastly, we need to check a value greater than \(5\) . We’ll use \(x=6\) . This gives \(\frac{(6)-5}{(6)^2-4} = \frac{1}{32} >0\) , so this interval gets a check-mark.

Figure \(\PageIndex{7}\)

We now have a mark over every interval, so we can determine our final answer. We can see that the inequality is true for \(x \in (-2,2) \cup [5,\infty)\) . Note that we included \(x=5\) since it has a closed circle and excluded \(x=-2\) and \(x=2\) since they have open circles. Our final answer is

\[\frac{x-5}{x^{2}-4}\geq 0\text{ holds for }x\in (-2,2)\cup [5,\:\infty )\]

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Mathematics > Optimization and Control

Title: popov mirror-prox for solving variational inequalities.

Abstract: We consider the mirror-prox algorithm for solving monotone Variational Inequality (VI) problems. As the mirror-prox algorithm is not practically implementable, except in special instances of VIs (such as affine VIs), we consider its implementation with Popov method updates. We provide convergence rate analysis of our proposed method for a monotone VI with a Lipschitz continuous mapping. We establish a convergence rate of $O(1/t)$, in terms of the number $t$ of iterations, for the dual gap function. Simulations on a two player matrix game corroborate our findings.

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