equations and inequalities homework 2

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• Order of operations
• Evaluating expressions
• Simplifying algebraic expressions
• Multi-step equations
• Work word problems
• Distance-rate-time word problems
• Mixture word problems
• Absolute value equations
• Multi-step inequalities
• Compound inequalities
• Absolute value inequalities
• Discrete relations
• Continuous relations
• Evaluating and graphing functions
• Review of linear equations
• Graphing absolute value functions
• Graphing linear inequalities
• Direct and inverse variation
• Systems of two linear inequalities
• Systems of two equations
• Systems of two equations, word problems
• Points in three dimensions
• Systems of three equations, elimination
• Systems of three equations, substitution
• Basic matrix operations
• Matrix multiplication
• All matrix operations combined
• Matrix inverses
• Geometric transformations with matrices
• Operations with complex numbers
• Properties of complex numbers
• Rationalizing imaginary denominators
• Properties of parabolas
• Vertex form
• Graphing quadratic inequalities
• Factoring quadratic expressions
• Solving quadratic equations w/ square roots
• Solving quadratic equations by factoring
• Completing the square
• Solving equations by completing the square
• Solving equations with the quadratic formula
• The discriminant
• Naming and simple operations
• Factoring a sum/difference of cubes
• Factoring by grouping
• Factoring quadratic form
• Factoring using all techniques
• Factors and Zeros
• The Remainder Theorem
• Irrational and Imaginary Root Theorems
• Descartes' Rule of Signs
• More on factors, zeros, and dividing
• The Rational Root Theorem
• Polynomial equations
• Basic shape of graphs of polynomials
• Graphing polynomial functions
• The Binomial Theorem
• Evaluating functions
• Function operations
• Inverse functions
• Simplifying radicals
• Operations with radical expressions
• Dividing radical expressions
• Radicals and rational exponents
• Simplifying rational exponents
• Square root equations
• Rational exponent equations
• Graphing radicals
• Graphing & properties of parabolas
• Equations of parabolas
• Graphing & properties of circles
• Equations of circles
• Graphing & properties of ellipses
• Equations of ellipses
• Graphing & properties of hyperbolas
• Equations of hyperbolas
• Classifying conic sections
• Eccentricity
• Systems of quadratic equations
• Graphing simple rational functions
• Graphing general rational functions
• Simplifying rational expressions
• Multiplying / dividing rational expressions
• Adding / subtracting rational expressions
• Complex fractions
• Solving rational equations
• The meaning of logarithms
• Properties of logarithms
• The change of base formula
• Writing logs in terms of others
• Logarithmic equations
• Inverse functions and logarithms
• Exponential equations not requiring logarithms
• Exponential equations requiring logarithms
• Graphing logarithms
• Graphing exponential functions
• Discrete exponential growth and decay word problems
• Continuous exponential growth and decay word problems
• General sequences
• Arithmetic sequences
• Geometric sequences
• Comparing Arithmetic/Geometric Sequences
• General series
• Arithmetic series
• Arithmetic/Geometric Means w/ Sequences
• Finite geometric series
• Infinite geometric series
• Right triangle trig: Evaluating ratios
• Right triangle trig: Missing sides/angles
• Angles and angle measure
• Co-terminal angles and reference angles
• Arc length and sector area
• Trig ratios of general angles
• Exact trig ratios of important angles
• The Law of Sines
• The Law of Cosines
• Graphing trig functions
• Translating trig functions
• Angle Sum/Difference Identities
• Double-/Half-Angle Identities
• Sample spaces and The Counting Principle
• Independent and dependent events
• Mutualy exclusive events
• Permutations
• Combinations
• Permutations vs combinations
• Probability using permutations and combinations

5.2 Solving Systems of Equations by Substitution

Learning objectives.

By the end of this section, you will be able to:

• Solve a system of equations by substitution
• Solve applications of systems of equations by substitution

Be Prepared 5.4

Before you get started, take this readiness quiz.

Simplify −5 ( 3 − x ) −5 ( 3 − x ) . If you missed this problem, review Example 1.136 .

Be Prepared 5.5

Simplify 4 − 2 ( n + 5 ) 4 − 2 ( n + 5 ) . If you missed this problem, review Example 1.123 .

Be Prepared 5.6

Solve for y y : 8 y − 8 = 32 − 2 y 8 y − 8 = 32 − 2 y If you missed this problem, review Example 2.34 .

Be Prepared 5.7

Solve for x x : 3 x − 9 y = −3 3 x − 9 y = −3 If you missed this problem, review Example 2.65 .

Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.

In this section, we will solve systems of linear equations by the substitution method.

Solve a System of Equations by Substitution

We will use the same system we used first for graphing.

We will first solve one of the equations for either x or y . We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

We’ll fill in all these steps now in Example 5.13 .

Example 5.13

How to solve a system of equations by substitution.

Solve the system by substitution. { 2 x + y = 7 x − 2 y = 6 { 2 x + y = 7 x − 2 y = 6

Try It 5.25

Solve the system by substitution. { −2 x + y = −11 x + 3 y = 9 { −2 x + y = −11 x + 3 y = 9

Try It 5.26

Solve the system by substitution. { x + 3 y = 10 4 x + y = 18 { x + 3 y = 10 4 x + y = 18

Solve a system of equations by substitution.

• Step 1. Solve one of the equations for either variable.
• Step 2. Substitute the expression from Step 1 into the other equation.
• Step 3. Solve the resulting equation.
• Step 4. Substitute the solution in Step 3 into one of the original equations to find the other variable.
• Step 5. Write the solution as an ordered pair.
• Step 6. Check that the ordered pair is a solution to both original equations.

If one of the equations in the system is given in slope–intercept form, Step 1 is already done! We’ll see this in Example 5.14 .

Example 5.14

Solve the system by substitution.

{ x + y = −1 y = x + 5 { x + y = −1 y = x + 5

The second equation is already solved for y . We will substitute the expression in place of y in the first equation.

Try It 5.27

Solve the system by substitution. { x + y = 6 y = 3 x − 2 { x + y = 6 y = 3 x − 2

Try It 5.28

Solve the system by substitution. { 2 x − y = 1 y = −3 x − 6 { 2 x − y = 1 y = −3 x − 6

If the equations are given in standard form, we’ll need to start by solving for one of the variables. In this next example, we’ll solve the first equation for y .

Example 5.15

Solve the system by substitution. { 3 x + y = 5 2 x + 4 y = −10 { 3 x + y = 5 2 x + 4 y = −10

We need to solve one equation for one variable. Then we will substitute that expression into the other equation.

Try It 5.29

Solve the system by substitution. { 4 x + y = 2 3 x + 2 y = −1 { 4 x + y = 2 3 x + 2 y = −1

Try It 5.30

Solve the system by substitution. { − x + y = 4 4 x − y = 2 { − x + y = 4 4 x − y = 2

In Example 5.15 it was easiest to solve for y in the first equation because it had a coefficient of 1. In Example 5.16 it will be easier to solve for x .

Example 5.16

Solve the system by substitution. { x − 2 y = −2 3 x + 2 y = 34 { x − 2 y = −2 3 x + 2 y = 34

We will solve the first equation for x x and then substitute the expression into the second equation.

Try It 5.31

Solve the system by substitution. { x − 5 y = 13 4 x − 3 y = 1 { x − 5 y = 13 4 x − 3 y = 1

Try It 5.32

Solve the system by substitution. { x − 6 y = −6 2 x − 4 y = 4 { x − 6 y = −6 2 x − 4 y = 4

When both equations are already solved for the same variable, it is easy to substitute!

Example 5.17

Solve the system by substitution. { y = −2 x + 5 y = 1 2 x { y = −2 x + 5 y = 1 2 x

Since both equations are solved for y , we can substitute one into the other.

Try It 5.33

Solve the system by substitution. { y = 3 x − 16 y = 1 3 x { y = 3 x − 16 y = 1 3 x

Try It 5.34

Solve the system by substitution. { y = − x + 10 y = 1 4 x { y = − x + 10 y = 1 4 x

Be very careful with the signs in the next example.

Example 5.18

Solve the system by substitution. { 4 x + 2 y = 4 6 x − y = 8 { 4 x + 2 y = 4 6 x − y = 8

We need to solve one equation for one variable. We will solve the first equation for y .

Try It 5.35

Solve the system by substitution. { x − 4 y = −4 −3 x + 4 y = 0 { x − 4 y = −4 −3 x + 4 y = 0

Try It 5.36

Solve the system by substitution. { 4 x − y = 0 2 x − 3 y = 5 { 4 x − y = 0 2 x − 3 y = 5

In Example 5.19 , it will take a little more work to solve one equation for x or y .

Example 5.19

Solve the system by substitution. { 4 x − 3 y = 6 15 y − 20 x = −30 { 4 x − 3 y = 6 15 y − 20 x = −30

We need to solve one equation for one variable. We will solve the first equation for x .

Since 0 = 0 is a true statement, the system is consistent. The equations are dependent. The graphs of these two equations would give the same line. The system has infinitely many solutions.

Try It 5.37

Solve the system by substitution. { 2 x − 3 y = 12 −12 y + 8 x = 48 { 2 x − 3 y = 12 −12 y + 8 x = 48

Try It 5.38

Solve the system by substitution. { 5 x + 2 y = 12 −4 y − 10 x = −24 { 5 x + 2 y = 12 −4 y − 10 x = −24

Look back at the equations in Example 5.19 . Is there any way to recognize that they are the same line?

Let’s see what happens in the next example.

Example 5.20

Solve the system by substitution. { 5 x − 2 y = −10 y = 5 2 x { 5 x − 2 y = −10 y = 5 2 x

The second equation is already solved for y , so we can substitute for y in the first equation.

Since 0 = −10 is a false statement the equations are inconsistent. The graphs of the two equation would be parallel lines. The system has no solutions.

Try It 5.39

Solve the system by substitution. { 3 x + 2 y = 9 y = − 3 2 x + 1 { 3 x + 2 y = 9 y = − 3 2 x + 1

Try It 5.40

Solve the system by substitution. { 5 x − 3 y = 2 y = 5 3 x − 4 { 5 x − 3 y = 2 y = 5 3 x − 4

Solve Applications of Systems of Equations by Substitution

We’ll copy here the problem solving strategy we used in the Solving Systems of Equations by Graphing section for solving systems of equations. Now that we know how to solve systems by substitution, that’s what we’ll do in Step 5.

How to use a problem solving strategy for systems of linear equations.

• Step 1. Read the problem. Make sure all the words and ideas are understood.
• Step 2. Identify what we are looking for.
• Step 3. Name what we are looking for. Choose variables to represent those quantities.
• Step 4. Translate into a system of equations.
• Step 5. Solve the system of equations using good algebra techniques.
• Step 6. Check the answer in the problem and make sure it makes sense.
• Step 7. Answer the question with a complete sentence.

Some people find setting up word problems with two variables easier than setting them up with just one variable. Choosing the variable names is easier when all you need to do is write down two letters. Think about this in the next example—how would you have done it with just one variable?

Example 5.21

The sum of two numbers is zero. One number is nine less than the other. Find the numbers.

Try It 5.41

The sum of two numbers is 10. One number is 4 less than the other. Find the numbers.

Try It 5.42

The sum of two number is −6. One number is 10 less than the other. Find the numbers.

In the Example 5.22 , we’ll use the formula for the perimeter of a rectangle, P = 2 L + 2 W .

Example 5.22

The perimeter of a rectangle is 88. The length is five more than twice the width. Find the length and the width.

Try It 5.43

The perimeter of a rectangle is 40. The length is 4 more than the width. Find the length and width of the rectangle.

Try It 5.44

The perimeter of a rectangle is 58. The length is 5 more than three times the width. Find the length and width of the rectangle.

For Example 5.23 we need to remember that the sum of the measures of the angles of a triangle is 180 degrees and that a right triangle has one 90 degree angle.

Example 5.23

The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles.

We will draw and label a figure.

Try It 5.45

The measure of one of the small angles of a right triangle is 2 more than 3 times the measure of the other small angle. Find the measure of both angles.

Try It 5.46

The measure of one of the small angles of a right triangle is 18 less than twice the measure of the other small angle. Find the measure of both angles.

Example 5.24

Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $25,000 plus $15 for each training session. Option B would pay her $10,000 + $40 for each training session. How many training sessions would make the salary options equal?

Try It 5.47

Geraldine has been offered positions by two insurance companies. The first company pays a salary of $12,000 plus a commission of $100 for each policy sold. The second pays a salary of $20,000 plus a commission of $50 for each policy sold. How many policies would need to be sold to make the total pay the same?

Try It 5.48

Kenneth currently sells suits for company A at a salary of $22,000 plus a $10 commission for each suit sold. Company B offers him a position with a salary of $28,000 plus a $4 commission for each suit sold. How many suits would Kenneth need to sell for the options to be equal?

Access these online resources for additional instruction and practice with solving systems of equations by substitution.

• Instructional Video-Solve Linear Systems by Substitution
• Instructional Video-Solve by Substitution

Section 5.2 Exercises

Practice makes perfect.

In the following exercises, solve the systems of equations by substitution.

{ 2 x + y = −4 3 x − 2 y = −6 { 2 x + y = −4 3 x − 2 y = −6

{ 2 x + y = −2 3 x − y = 7 { 2 x + y = −2 3 x − y = 7

{ x − 2 y = −5 2 x − 3 y = −4 { x − 2 y = −5 2 x − 3 y = −4

{ x − 3 y = −9 2 x + 5 y = 4 { x − 3 y = −9 2 x + 5 y = 4

{ 5 x − 2 y = −6 y = 3 x + 3 { 5 x − 2 y = −6 y = 3 x + 3

{ −2 x + 2 y = 6 y = −3 x + 1 { −2 x + 2 y = 6 y = −3 x + 1

{ 2 x + 3 y = 3 y = − x + 3 { 2 x + 3 y = 3 y = − x + 3

{ 2 x + 5 y = −14 y = −2 x + 2 { 2 x + 5 y = −14 y = −2 x + 2

{ 2 x + 5 y = 1 y = 1 3 x − 2 { 2 x + 5 y = 1 y = 1 3 x − 2

{ 3 x + 4 y = 1 y = − 2 5 x + 2 { 3 x + 4 y = 1 y = − 2 5 x + 2

{ 3 x − 2 y = 6 y = 2 3 x + 2 { 3 x − 2 y = 6 y = 2 3 x + 2

{ −3 x − 5 y = 3 y = 1 2 x − 5 { −3 x − 5 y = 3 y = 1 2 x − 5

{ 2 x + y = 10 − x + y = −5 { 2 x + y = 10 − x + y = −5

{ −2 x + y = 10 − x + 2 y = 16 { −2 x + y = 10 − x + 2 y = 16

{ 3 x + y = 1 −4 x + y = 15 { 3 x + y = 1 −4 x + y = 15

{ x + y = 0 2 x + 3 y = −4 { x + y = 0 2 x + 3 y = −4

{ x + 3 y = 1 3 x + 5 y = −5 { x + 3 y = 1 3 x + 5 y = −5

{ x + 2 y = −1 2 x + 3 y = 1 { x + 2 y = −1 2 x + 3 y = 1

{ 2 x + y = 5 x − 2 y = −15 { 2 x + y = 5 x − 2 y = −15

{ 4 x + y = 10 x − 2 y = −20 { 4 x + y = 10 x − 2 y = −20

{ y = −2 x − 1 y = − 1 3 x + 4 { y = −2 x − 1 y = − 1 3 x + 4

{ y = x − 6 y = − 3 2 x + 4 { y = x − 6 y = − 3 2 x + 4

{ y = 2 x − 8 y = 3 5 x + 6 { y = 2 x − 8 y = 3 5 x + 6

{ y = − x − 1 y = x + 7 { y = − x − 1 y = x + 7

{ 4 x + 2 y = 8 8 x − y = 1 { 4 x + 2 y = 8 8 x − y = 1

{ − x − 12 y = −1 2 x − 8 y = −6 { − x − 12 y = −1 2 x − 8 y = −6

{ 15 x + 2 y = 6 −5 x + 2 y = −4 { 15 x + 2 y = 6 −5 x + 2 y = −4

{ 2 x − 15 y = 7 12 x + 2 y = −4 { 2 x − 15 y = 7 12 x + 2 y = −4

{ y = 3 x 6 x − 2 y = 0 { y = 3 x 6 x − 2 y = 0

{ x = 2 y 4 x − 8 y = 0 { x = 2 y 4 x − 8 y = 0

{ 2 x + 16 y = 8 − x − 8 y = −4 { 2 x + 16 y = 8 − x − 8 y = −4

{ 15 x + 4 y = 6 −30 x − 8 y = −12 { 15 x + 4 y = 6 −30 x − 8 y = −12

{ y = −4 x 4 x + y = 1 { y = −4 x 4 x + y = 1

{ y = − 1 4 x x + 4 y = 8 { y = − 1 4 x x + 4 y = 8

{ y = 7 8 x + 4 −7 x + 8 y = 6 { y = 7 8 x + 4 −7 x + 8 y = 6

{ y = − 2 3 x + 5 2 x + 3 y = 11 { y = − 2 3 x + 5 2 x + 3 y = 11

In the following exercises, translate to a system of equations and solve.

The sum of two numbers is 15. One number is 3 less than the other. Find the numbers.

The sum of two numbers is 30. One number is 4 less than the other. Find the numbers.

The sum of two numbers is −26. One number is 12 less than the other. Find the numbers.

The perimeter of a rectangle is 50. The length is 5 more than the width. Find the length and width.

The perimeter of a rectangle is 60. The length is 10 more than the width. Find the length and width.

The perimeter of a rectangle is 58. The length is 5 more than three times the width. Find the length and width.

The perimeter of a rectangle is 84. The length is 10 more than three times the width. Find the length and width.

The measure of one of the small angles of a right triangle is 14 more than 3 times the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 26 more than 3 times the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 15 less than twice the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 45 less than twice the measure of the other small angle. Find the measure of both angles.

Maxim has been offered positions by two car dealers. The first company pays a salary of $10,000 plus a commission of $1,000 for each car sold. The second pays a salary of $20,000 plus a commission of $500 for each car sold. How many cars would need to be sold to make the total pay the same?

Jackie has been offered positions by two cable companies. The first company pays a salary of $ 14,000 plus a commission of $100 for each cable package sold. The second pays a salary of $20,000 plus a commission of $25 for each cable package sold. How many cable packages would need to be sold to make the total pay the same?

Amara currently sells televisions for company A at a salary of $17,000 plus a $100 commission for each television she sells. Company B offers her a position with a salary of $29,000 plus a $20 commission for each television she sells. How many televisions would Amara need to sell for the options to be equal?

Mitchell currently sells stoves for company A at a salary of $12,000 plus a $150 commission for each stove he sells. Company B offers him a position with a salary of $24,000 plus a $50 commission for each stove he sells. How many stoves would Mitchell need to sell for the options to be equal?

Everyday Math

When Gloria spent 15 minutes on the elliptical trainer and then did circuit training for 30 minutes, her fitness app says she burned 435 calories. When she spent 30 minutes on the elliptical trainer and 40 minutes circuit training she burned 690 calories. Solve the system { 15 e + 30 c = 435 30 e + 40 c = 690 { 15 e + 30 c = 435 30 e + 40 c = 690 for e e , the number of calories she burns for each minute on the elliptical trainer, and c c , the number of calories she burns for each minute of circuit training.

Stephanie left Riverside, California, driving her motorhome north on Interstate 15 towards Salt Lake City at a speed of 56 miles per hour. Half an hour later, Tina left Riverside in her car on the same route as Stephanie, driving 70 miles per hour. Solve the system { 56 s = 70 t s = t + 1 2 { 56 s = 70 t s = t + 1 2 .

• ⓐ for t t to find out how long it will take Tina to catch up to Stephanie.
• ⓑ what is the value of s s , the number of hours Stephanie will have driven before Tina catches up to her?

Writing Exercises

Solve the system of equations { x + y = 10 x − y = 6 { x + y = 10 x − y = 6

ⓐ by graphing. ⓑ by substitution. ⓒ Which method do you prefer? Why?

Solve the system of equations { 3 x + y = 12 x = y − 8 { 3 x + y = 12 x = y − 8 by substitution and explain all your steps in words.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

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Equations and Inequalities Homework (Algebra 2 - Unit 1)

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Equations and Inequalities Homework Bundle:

This resource is a bundled set of homework practice sets and daily content quizzes for Unit 1: EXPRESSIONS, EQUATIONS & INEQUALITIES designed for Algebra 2 Honors students. The file includes 10 pages of homework assignments and two different forms of a daily content quiz that you can use as a homework check, group work, or exit tickets.

The unit includes the following topics:

1) Properties of Real Numbers

2) Algebraic Expressions

3) Solving Equations

4) Solving Inequalities

5) Absolute Value Equations and Inequalities

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10.2.1: Solving One-Step Inequalities

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Learning Objectives

• Represent inequalities on a number line.
• Use the addition property of inequality to isolate variables and solve algebraic inequalities, and express their solutions graphically.
• Use the multiplication property of inequality to isolate variables and solve algebraic inequalities, and express their solutions graphically.

Introduction

Sometimes there is a range of possible values to describe a situation. When you see a sign that says “Speed Limit 25,” you know that it doesn’t mean that you have to drive exactly at a speed of 25 miles per hour (mph). This sign means that you are not supposed to go faster than 25 mph, but there are many legal speeds you could drive, such as 22 mph, 24.5 mph or 19 mph. In a situation like this, which has more than one acceptable value, inequalities are used to represent the situation rather than equations .

What is an Inequality?

An inequality is a mathematical statement that compares two expressions using an inequality sign. In an inequality, one expression of the inequality can be greater or less than the other expression. Special symbols are used in these statements. The box below shows the symbol, meaning, and an example for each inequality sign.

Inequality Signs

\(\ x \neq y \quad x \text{ is} {\bf\text { not equal }}\text{to } y\).

Example : The number of days in a week is not equal to 9.

\(\ x>y \quad x {\bf\text { is greater than }} y . \text { Example: } 6>3\)

Example : The number of days in a month is greater than the number of days in a week.

\(\ x<y \quad x {\bf\text { is less than }} y\)

Example : The number of days in a week is less than the number of days in a year.

\(\ x \geq y \quad x {\bf\text { is greater than or equal to }} y\)

Example : 31 is greater than or equal to the number of days in a month.

\(\ x \leq y \quad x {\bf\text { is less than or equal to }} y\)

Example : The speed of a car driving legally in a 25 mph zone is less than or equal to 25 mph.

The important thing about inequalities is that there can be multiple solutions. For example, the inequality “31 ≥ the number of days in a month” is a true statement for every month of the year—no month has more than 31 days. It holds true for January, which has 31 days (\(\ 31 \geq 31\)); September, which has 30 days (≥ 30); and February, which has either 28 or 29 days depending upon the year (\(\ 31 \geq 28 \text { and } 31 \geq 29\)).

The inequality \(\ x>y\) can also be written as \(\ y<x\). The sides of any inequality can be switched as long as the inequality symbol between them is also reversed.

Representing Inequalities on a Number Line

Inequalities can be graphed on a number line. Below are three examples of inequalities and their graphs.

Each of these graphs begins with a circle—either an open or closed (shaded) circle. This point is often called the end point of the solution. A closed, or shaded, circle is used to represent the inequalities greater than or equal to (≥) or less than or equal to (≤) . The point is part of the solution. An open circle is used for greater than (>) or less than (<). The point is not part of the solution.

The graph then extends endlessly in one direction. This is shown by a line with an arrow at the end. For example, notice that for the graph of \(\ x \geq-3\) shown above, the end point is -3, represented with a closed circle since the inequality is greater than or equal to -3. The blue line is drawn to the right on the number line because the values in this area are greater than -3. The arrow at the end indicates that the solutions continue infinitely.

Solving Inequalities Using Addition & Subtraction Properties

You can solve most inequalities using the same methods as those for solving equations. Inverse operations can be used to solve inequalities. This is because when you add or subtract the same value from both sides of an inequality, you have maintained the inequality. These properties are outlined in the blue box below.

Addition and Subtraction Properties of Inequality

\(\ \text { If } a>b, \text { then } a+c>b+c\)

\(\ \text { If } a>b, \text { then } a-c>b-c\)

Because inequalities have multiple possible solutions, representing the solutions graphically provides a helpful visual of the situation. The example below shows the steps to solve and graph an inequality.

Solve for \(\ x\).

\(\ x+3<5\)

\(\ x<2\)

The graph of the inequality \(\ x<2\) is shown below.

Just as you can check the solution to an equation, you can check a solution to an inequality. First, you check the end point by substituting it in the related equation. Then you check to see if the inequality is correct by substituting any other solution to see if it is one of the solutions. Because there are multiple solutions, it is a good practice to check more than one of the possible solutions. This can also help you check that your graph is correct.

The example below shows how you could check that \(\ x<2\) is the solution to \(\ x+3<5\).

Check that \(\ x<2\) is the solution to \(\ x+3<5\).

\(\ x<2\) is the solution to \(\ x+3<5\)

The following examples show additional inequality problems. The graph of the solution to the inequality is also shown. Remember to check the solution. This is a good habit to build!

Advanced Example

\(\ \frac{15}{2}+x>-\frac{37}{4}\)

\(\ x>-\frac{67}{4}\)

\(\ x-10 \leq-12\)

\(\ x \leq-2\)

The graph of this solution in shown below. Notice that a closed circle is used because the inequality is “less than or equal to” (≤). The blue arrow is drawn to the left of the point -2 because these are the values that are less than -2.

Check that \(\ x \leq-2\) is the solution to \(\ x-10 \leq-12\).

\(\ x \leq-2\) is the solution to \(\ x-10 \leq 12\).

Solve for \(\ a\).

\(\ a-17>-17\)

\(\ a>0\)

The graph of this solution in shown below. Notice that an open circle is used because the inequality is “greater than” (>). The arrow is drawn to the right of 0 because these are the values that are greater than 0.

Check that \(\ a>0\) is the solution to

\(\ a-17>-17\).

\(\ a>0\) is the solution to \(\ a-17>-17\).

Advanced Question

Solve for \(\ x\): \(\ 0.5 x \leq 7-0.5 x\).

• \(\ x \leq 0\)
• \(\ x>35\)
• \(\ x \leq 7\)
• \(\ x \geq 5\)
• Incorrect. To find the value of \(\ x\), try adding \(\ 0.5 x\) to both sides. The correct answer is \(\ x \leq 7\).
• Correct. When you add \(\ 0.5 x\) to both sides it creates \(\ 1 x\), so \(\ x \leq 7\).

Solving Inequalities Involving Multiplication

Solving an inequality with a variable that has a coefficient other than 1 usually involves multiplication or division. The steps are like solving one-step equations involving multiplication or division EXCEPT for the inequality sign. Let’s look at what happens to the inequality when you multiply or divide each side by the same number.

When you multiply by a negative number, “reverse” the inequality sign.

Whenever you multiply or divide both sides of an inequality by a negative number, the inequality sign must be reversed in order to keep a true statement.

These rules are summarized in the box below.

Multiplication and Division Properties of Inequality

\(\ \begin{aligned} \text { If } a>b \text { , then } a c>b c \text { , if } c>0\\ \text { If } a>b \text { , then } a c<b c \text { , if } c<0\\ \\ \text { If } a>b \text { , then } \frac{a}{c}>\frac{b}{c} \text { , if } c>0\\ \text { If } a>b \text { , then } \frac{a}{c}<\frac{b}{c} \text { , if } c<0 \end{aligned}\)

Keep in mind that you only change the sign when you are multiplying and dividing by a negative number. If you add or subtract a negative number, the inequality stays the same.

\(\ -\frac{1}{3}>-12 x\)

\(\ x>\frac{1}{36}\)

\(\ 3 x>12\)

\(\ x>4\)

The graph of this solution is shown below.

There was no need to make any changes to the inequality sign because both sides of the inequality were divided by positive 3. In the next example, there is division by a negative number, so there is an additional step in the solution!

\(\ -2 x>6\)

\(\ x<-3\)

Because both sides of the inequality were divided by a negative number, -2, the inequality symbol was switched from > to <. The graph of this solution is shown below.

Solve for \(\ y\): \(\ -10 y \geq 150\)

• \(\ y=-15\)
• \(\ y \geq-15\)
• \(\ y \leq-15\)
• \(\ y \geq 15\)
• Incorrect. While -15 is a solution to the inequality, it is not the only solution. The solution must include an inequality sign. The correct answer is \(\ y \leq-15\).
• Incorrect. This solution does not satisfy the inequality. For example \(\ y=0\), which is a value greater than -15, results in an untrue statement. 0 is not greater than When dividing by a negative number, you must change the inequality symbol. The correct answer is \(\ y \leq-15\).
• Correct. Dividing both sides by -10 leaves \(\ y\) isolated on the left side of the inequality and -15 on the right. Since you divided by a negative number, the ≥ must be switched to ≤.
• Incorrect. Divide by -10, not 10, to isolate the variable. The correct answer is \(\ y \leq-15\).

Solve for \(\ a\): \(\ -\frac{a}{5}<\frac{35}{8}\)

• \(\ a>-\frac{175}{8}\)
• \(\ a<-\frac{175}{8}\)
• \(\ a>-\frac{7}{8}\)
• \(\ a<-\frac{7}{8}\)
• Correct. By multiplying both sides by -5 and flipping the inequality sign from < to >, you found that \(\ a>-\frac{175}{8}\).
• Incorrect. You correctly multiplied by -5, but remember that the inequality sign flips when you multiply by a negative number. The correct response is: \(\ a>-\frac{175}{8}\).
• Incorrect. It looks like you divided both sides by -5. While you remembered to flip the inequality sign correctly, division is not the correct operation here. The correct response is: \(\ a>-\frac{175}{8}\).
• Incorrect. It looks like you divided both sides by -5. Division is not the correct operation here, and remember to flip the inequality sign when you multiply or divide by a negative number. The correct response is: \(\ a>-\frac{175}{8}\).

Solving inequalities is very similar to solving equations, except you have to reverse the inequality symbols when you multiply or divide both sides of an inequality by a negative number. Since inequalities can have multiple solutions, it is customary to represent the solution to an inequality graphically as well as algebraically. Because there is usually more than one solution to an inequality, when you check your answer you should check the end point and one other value to check the direction of the inequality.

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Unit 10: One-step and two-step equations & inequalities

Combining like terms.

• Intro to combining like terms (Opens a modal)
• Combining like terms with negative coefficients & distribution (Opens a modal)
• Combining like terms with negative coefficients (Opens a modal)
• Combining like terms with rational coefficients (Opens a modal)
• Combining like terms with negative coefficients Get 5 of 7 questions to level up!
• Combining like terms with negative coefficients & distribution Get 3 of 4 questions to level up!
• Combining like terms with rational coefficients Get 3 of 4 questions to level up!

The distributive property & equivalent expressions

• The distributive property with variables (Opens a modal)
• Factoring with the distributive property (Opens a modal)
• Equivalent expressions: negative numbers & distribution (Opens a modal)
• Distributive property with variables (negative numbers) Get 3 of 4 questions to level up!
• Equivalent expressions: negative numbers & distribution Get 5 of 7 questions to level up!

Interpreting linear expressions

• Interpreting linear expressions: diamonds (Opens a modal)
• Interpreting linear expressions: flowers (Opens a modal)
• Writing expressions word problems (Opens a modal)
• Interpreting linear expressions Get 3 of 4 questions to level up!
• Writing expressions word problems Get 5 of 7 questions to level up!

Two-step equations intro

• Same thing to both sides of equations (Opens a modal)
• Intro to two-step equations (Opens a modal)
• Two-step equations intuition (Opens a modal)
• Worked example: two-step equations (Opens a modal)
• Two-step equations Get 5 of 7 questions to level up!

Two-step equations with decimals and fractions

• Two-step equations with decimals and fractions (Opens a modal)
• Find the mistake: two-step equations (Opens a modal)
• Two-step equations review (Opens a modal)
• Two-step equations with decimals and fractions Get 5 of 7 questions to level up!
• Find the mistake: two-step equations Get 3 of 4 questions to level up!

Two-step equation word problems

• Equation word problem: super yoga (1 of 2) (Opens a modal)
• Equation word problem: super yoga (2 of 2) (Opens a modal)
• Two-step equation word problem: computers (Opens a modal)
• Two-step equation word problem: garden (Opens a modal)
• Two-step equation word problem: oranges (Opens a modal)
• Interpret two-step equation word problems Get 3 of 4 questions to level up!
• Two-step equations word problems Get 3 of 4 questions to level up!

One-step inequalities

• Testing solutions to inequalities (Opens a modal)
• One-step inequalities examples (Opens a modal)
• One-step inequalities: -5c ≤ 15 (Opens a modal)
• One-step inequality word problem (Opens a modal)
• One-step inequalities review (Opens a modal)
• Testing solutions to inequalities Get 3 of 4 questions to level up!
• One-step inequalities Get 5 of 7 questions to level up!

Two-step inequalities

• Two-step inequalities (Opens a modal)
• Two-step inequality word problem: apples (Opens a modal)
• Two-step inequality word problem: R&B (Opens a modal)
• Two-step inequalities Get 5 of 7 questions to level up!
• Two-step inequality word problems Get 3 of 4 questions to level up!