Unit 3 equations and inequalities answer key

Q: How do you solve equations and inequalities?

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.

Q: How do you solve algebraic equations?

A General Rule for Solving Equations.

If you're seeing this message, it means we're having trouble loading external resources on our website.

Table of Contents Show

Introduction
Simultaneous Equations
Solving Simultaneous Linear Equations by Elimination
Solving Simultaneous Linear Equations by Substitution
Inequalities
Solving Algebraic Inequalities
How do you solve equations and inequalities?
How do you solve algebraic equations?

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Equations and Inequalities (Pre-Algebra Curriculum - Unit 3) | All Things Algebra®

DISTANCE LEARNING UPDATE: This unit now contains a Google document with:

(1) Links to instructional videos.
Videos are created by fellow teachers for their students using the guided notes from the unit. Please watch through first before sharing with your students. They might not always be perfect but we are trying our best to help out where we can in a very limited time.

(2) A link to a Google Slides version of the unit.
Each page is set to the background in Google Slides. There are no text boxes; this is the PDF in Google Slides. We are unable to do text boxes at this time but hope this saves you a step if you wish to use it in Slides instead!

-------------------------------------------------------------------------------------------------------------------------

This bundle includes notes, homework assignments, four quizzes, a study guide, and a unit test that cover the following topics:

• One-Step Equations

• Rational Equations

• Two-Step Equations

• Solving Equations by Square Roots

• Multi-Step Equations (Variables on One Side)

• Multi-Step Equations (Variables on Both Sides)

• Special Solutions: No Solution and Infinite Solution

• Solving Equations by Clearing Fractions

• Translating and Solving Equations

• Equation Word Problems

• Representing and Graphing Inequalities

• One- and Two-Step Inequalities

• Multi-Step Inequalities

• Translating and Solving Inequalities

• Inequality Word Problems

NEW: Assessments are now EDITABLE! Now you can easily make multiple versions or customize to fit your needs! PowerPoint and Equation Editor (usually built in to PowerPoint) are required to edit these files. There is a folder titled "Editable Assessments" when you download. This is where you will find editable versions of each quiz and the unit test. If your Equation Editor is incompatible with mine, simply delete my equation and insert your own.

This resource is included in the following bundle(s):

Pre-Algebra Curriculum

More Pre-Algebra Units:

Unit 1 – The Real Numbers

Unit 2 – Algebraic Expressions

Unit 4 – Ratios, Proportions, and Percents

Unit 5 – Functions and Linear Representations

Unit 6 – Systems of Equations

Unit 7 – Geometry

Unit 8 – Measurement: Area and Volume

Unit 9 – Probability and Statistics

LICENSING TERMS: This purchase includes a license for one teacher only for personal use in their classroom. Licenses are non-transferable, meaning they can not be passed from one teacher to another. No part of this resource is to be shared with colleagues or used by an entire grade level, school, or district without purchasing the proper number of licenses. If you are a coach, principal, or district interested in transferable licenses to accommodate yearly staff changes, please contact me for a quote at .

COPYRIGHT TERMS: This resource may not be uploaded to the internet in any form, including classroom/personal websites or network drives, unless the site is password protected and can only be accessed by students.

Introduction

This section will cover how to:

solve simultaneous linear equations by elimination
solve simultaneous linear equations by substitution
solve linear inequalities and represent the solution on a number line
find the values which satisfy both of two inequalities

Each topic is introduced with a theory section including examples and then some practice questions. At the end of the page there is an exercise where you can test your understanding of all the topics covered in this page.

Remember that you are NOT allowed to use calculators in this topic.

Simultaneous Equations

Basic Principles

Simultaneous equation questions have two equations, both of which have two unknown variables. The solution of a pair of simultaneous equation is the values of both variables which make both equations work at the same time.

For example, consider the equations: x + y = 8 and x – y = 2.
x = 1, y = 7 is a solution of the first equation but not the second.
x = 4, y = 2 is a solution of the second equation but not the first.
x = 5, y = 3 is a solution of both equations at the same time, so it is the solution of these simultaneous equations.

The simultaneous equations in this section are linear because the variables are not raised to any higher powers e.g x².
In this section the examples will mostly use x and y but the principles are the same for any pair of letters.

Notes

There will normally be one solution (a pair of values) to a pair of simultaneous linear equations.
Simultaneous equations can be solved using algebraic methods or graphically.
You can always check your answer by substituting the answer into both your original equations.

Solving Simultaneous Linear Equations by Elimination

Elimination Method

This example will illustrate the elimination method. The steps are shown below and the working on the right.	2x = 16 – 5y 7x – 3y = 15
1.	Number the equations and make sure matching letters are above each other in the pair of equations (usually we get the letters on the left-hand side and the numbers on the right-hand side.	(1) 2x + 5y = 16 (2) 7x – 3y = 15
2.	Get the coefficients of one of the variables to be the same in both equations by multiplying one or both of the equations by numbers (it doesn't matter if the signs of the coefficients are different).	Multiply equation (1) by 7 and equation (2) by 2: (1) 14x + 35y = 112 (2) 14x – 6y = 30
3.	Eliminate the variable with the matching coefficients by adding or subtracting the two equations (add if the signs on the matching coefficients are different, or subtract if the signs are the same).	Signs on the matching terms are the same, so subtract equation (2) from equation (1): 41y = 82 (always be careful with negative signs)
4.	Solve the resulting equation which now only has one variable to get the first part of the solution.	Divide by 41 on both sides: y = 2
5.	Substitute the first half of the solution into one of the original equations and solve to get the other part of the solution.	Substitute y = 2 into equation (1): (1) 2x + 10 = 16 2x = 6 x = 3
6.	Substitute both parts of the solution into the other original equation to check your answer.	Substitute x=3, y = 2 into equation (2): (2) 21 – 6 = 15
7.	Write the both parts of the solution clearly.	It works, so the solution is: x = 3 y = 2

Further Examples

Make sure you understand these examples before moving on:

Example 1

Example 2

10x = 3y + 4
5x – 6y + 1 = 0

3x + y = 1
5x – 4y = 47

(1) 10x – 3y = 4
(2) 5x – 6y = –1

(1) 3x + y = 1
(2) 5x – 4y = 47

Multiply (2) by 2 so the x terms match:
(1)   10x   – 3y = 4
(2)   10x – 12y = –2

Multiply (1) by 4 so the y terms match:
(1) 12x + 4y = 4
(2) 5x – 4y = 47

Do equation (1) – equation (2):
9y = 6

Signs different so do equation (1) + equation (2):
17x = 51

Divide by 9 on both sides:

Divide by 17 on both sides:
x = 3

Substitute y =	2	into equation (1):
3

(1) 10x – 2 = 4
10x = 6

Substitute x = 3 into equation (1):

(1) 9 + y = 1
y = –8

Check x =	3	, y =	2	in equation (2):
5	3

(2) 3 – 4 = –1

Check x=3, y = –8 in equation (2):

(2) 15 – –32 = 47

It works, so the solution is:

It works, so the solution is:
x = 3
y = –8

Note that there are other ways to solve these problems, but they will all give the same answers if done correctly.

Practice Questions

Work out the answer to each question then click on the button marked

to see if you are correct.

Solve these simultaneous equations using the elimination method:

(a) 4x = 10 – y
8x – 3y = 0

(1) 4x + y = 10
(2) 8x – 3y = 0

Multiply (1) by 3:
(1) 12x + 3y = 30
(2) 8x – 3y = 0

Add equations (1) and (2):
20x = 30

Substitute x =	3	into equation (1):
2

(1) 6 + y = 10
y = 4

Check x =	3	, y = 4 in equation (2):
2

(2) 12 – 12 = 0

It works, so the solution is:

(b) 3x – 5y = 4
3y – 4x = 2

(1) 3x – 5y = 4
(2) – 4x + 3y = 2

Multiply (1) by 4 and (2) by 3:
(1) 12x – 20y = 16
(2) – 12x + 9y = 6

Add equations (1) and (2):
–11y = 22
y = –2

Substitute y = –2 into equation (1):
(1)     3x + 10 = 4
                  3x = –6
                    x = –2

Check x = –2, y = –2 in equation (2):
(2) 8 + – 6 = 2

It works, so the solution is:

Solving Simultaneous Linear Equations by Substitution

Substitution Method

This example will illustrate the substitution method. The steps are shown below and the working on the right.	3x = 9 – 2y 2x – y = 20
1.	Number the equations (there is no need to rearrange them yet).	(1) 3x = 9 – 2y (2) 2x – y = 20
2.	Rearrange one of the equations so that one of the letters is the subject (on its own). Pick the one which is easiest to rearrange!	Rearrange equation (2) to make y the subject: (2) 2x – 20 = y
3.	Substitute the newly rearranged equation into the other original equation.	Substitute y = 2x – 20 into equation (1): (1) 3x = 9 – 2( 2x – 20 )
4.	Solve the resulting equation which now only has one variable to get the first part of the solution.	Multiply out the brackets: 3x = 9 – 4x + 40 Add 4x to both sides: 7x = 49 Divide by 7 both sides: x = 7
5.	Substitute the first half of the solution into the rearranged version of the equation from earlier, giving the second part of the solution.	Substitute x = 7 into y = 2x – 20: y = 14 – 20 y = –6
6.	Substitute both parts of the solution into the original equation which you didn't rearrange, to check your answer.	Substitute x = 7, y = –6 into equation (1): (1) 21 = 9 – –12
7.	Write the both parts of the solution clearly.	It works, so the solution is: x = 7 y = –6

Further Examples

Make sure you understand these examples before moving on:

Example 1	Example 2
2x = 3y + 2 x – y = 2	y = 4x – 5 y = 2x + 2
(1) 2x = 3y + 2 (2) x – y = 2	(1) y = 4x – 5 (2) y = 2x + 2
Make y the subject of (2): (2) x = 2 + y	y is already the subject of (1)
Substitute y = 2 + y into (1): (1) 2( 2 + y ) = 3y + 2	Substitute y = 4x – 5 into (2): (2) 4x – 5 = 2x + 2
4 + 2y = 3y + 2 4 = y + 2 2 = y	2x – 5 = 2 2x = 7 x = 3.5
Substitute y = 2 into x = 2 + y: x = 2 + 2 x = 4	Substitute x = 3.5 into y = 4x – 5: y = 14 – 5 y = 9
Substitute x = 4, y = 2 into equation (1): (1) 8 = 6 + 2	Substitute x = 3.5, y = 9 into equation (2): (2) 9 = 7 + 2
It works, so the solution is: x = 4 y = 2	It works, so the solution is: x = 3.5 y = 9

Note that both these problems could have been solved in other ways, including by elimination.

Choosing Elimination or Substitution

It doesn't matter which method you use for linear simultaneous equations, although one will usually be easier than the other for a particular question. If the corresponding terms are already lined up and maybe already matching, then elimination is usually easier. If there is already a letter on its own, maybe already the subject of one of the equations, then substitution should be easier.

In the A-Level Core 1 module you will need to do simultaneous equations where one is linear and the other is quadratic (it includes squared terms). In this case the only option is to use substitution, so you definitely need to know this skill for that situation.

Fraction or Decimal Answers

It doesn't matter whether you give your answers as fractions or decimals, but if you give your answers as fractions you should cancel them down and use top-heavy fractions rather than mixed numbers. Your answers should always be exact so avoid rounding decimals or using recurring decimals – for this reason it is best to mainly use fractions.

Practice Questions

Work out the answer to each question then click on the button marked

to see if you are correct.

Solve these simultaneous equations using the substitution method:

(a) 4x = 10 – y
8x – 3y = 0

(1) 4x = 10 – y
(2) 8x – 3y = 0

Make y the subject in (1):
(1) y = 10 – 4x

Substitute y = 10 – 4x into (2):
(2) 8x – 3( 10 – 4x ) = 0

Solve for x:
  8x – 30 + 12x = 0
          20x – 30 = 0
                  20x = 30

Substitute x =	3	in y = 10 – 4x:
2

y = 10 – 6
y = 4

Check x =	3	, y = 4 in equation (2):
2

(2) 12 – 12 = 0

It works, so the solution is:

(b) x = 5 + y
y = 6 – x

(1) x = 5 + y
(2) y = 6 – x

Substitute y = 6 – x into (1):
(1)     x = 5 + ( 6 – x )
         x = 11 – x
       2x = 11
         x = 5.5

Substitute x = 5.5 in y = 6 – x:
y = 6 – 5.5
y = 0.5

Check x = 5.5, y = 0.5 in equation (1):
(2) 5.5 = 5 + 0.5

It works, so the solution is:
x = 5.5
y = 0.5

Inequalities

Inequality Symbols

<	less than	further left on the number line
>	greater than	further right on the number line
≤	less than or equal to	further left or in the same position on the number line
≥	greater than or equal to	further right or in the same position on the number line
≠	not equal to	in a different position on the number line

The following statements are all true – check through them to make sure you agree with them:
3 < 5	–6 < –2	–4 ≤ 3	6 ≤ 6
4 ≠ –4	4 ≥ –4	3.4 > 2.7	0 ≥ –5
It's worth noting that when we write the ≤ and ≥ symbols by hand, we usually make the additional line parallel with the lower half of the < or > symbol. In the exercises below you will see examples of them written this way.

Representing Single Inequalities on a Number Line

The single inequality x > 5 means that x can take any value from 5 upwards, but not 5 itself. This includes all the decimals as well as whole numbers. We represent this on a number line as follows:

If we wanted to also include the number 5 we would use the inequality x ≥ 5, and would represent it as follows:

The filled circle above the 5 indicates that 5 is included in the inequality. Similarly, we can represent x ≤ 0 as follows:

The number line below represents the inequality x < 2:

Representing Double Inequalities on a Number Line

Sometimes we want to represent two inequalities at the same time. For example, if we wanted to represent all the values between –3 and +3 including the end values, we would use the inequality –3 ≤ x ≤ 3. This is a shortform for the following statement composed of two inequalities: " x ≥ –3 and x ≤ 3 " and would be represented as follows:

Imagine instead we wanted to represent all the values which are less than 0 or greater than 2. It's not possible in this situation to use just one statement, so we need to write both inequalities: " x < 0 or x > 2 ". We can still represent this on a number line as seen here:

It is not necessary for both ends to be both filled or both not filled. The number line below represents the double inequality 1 ≤ x < 3, where x can take any value between 1 and 3, including 1 but not including 3:

As before, 1 ≤ x < 3 is a shortform for the following statement composed of two inequalities: " x ≥ 1 and x < 3 "

Practice Questions

Work out the answer to each question then click on the button marked

to see if you are correct.

Represent these inequalities on a number line:

(a) x < 4

(b) 2 ≤ x < 7

State the inequalities represented by these number lines:

(a)

x ≥ 7

(b)

x < 2 or x > 6

Solving Algebraic Inequalities

Basic Principles

Solving inequalities is very similar to solving simple equations with two important differences:

You should avoid multiplying or dividing by negative numbers, or if if you do multiply or divide by a negative number you should then reverse the direction of the inequality.
As with equations we do operations to both sides of the inequality, but if we have a double inequality with three parts then we need to do the same operation to all three parts.

The main aim is still to get the unknown letter (e.g. x) on its own, leaving a simple inequality which can be represented easily on a number line.

Examples Combining Inequalities

Sometimes we are asked to find all the values which satisfy both of two inequalities. In this case it is helpful to represent both inequalities above each other on a number line and then look for those parts of the number line on which both inequalities are true. It is worth noting that sometimes there are no values which satisfy both inequalities. For example, there are no values which satisfy both " x ≤ 2 " and " x ≥ 5 ".

Examples

e.g. Find all the values which satisfy both
" x < 5 " and " x > 2 "

Start by representing both above the same number line:

Looking at these, we can see that both inequalities are true at the following points on the number line:

The numbers 2 and 5 are not included as they do not satisfy both inequalities.

The correct answer is " 2 < x < 5 ".

e.g. Find all the values which satisfy both
" x < 4 or x > 6 " and " x ≥ 2 "

Start by representing both above the same number line:

Looking at these, we can see that both inequalities are true at the following points on the number line:

Number 2 is included this time as it satisfies both inequalities. Numbers 4 and 6 do not satisfy both.

The correct answer is " 2 ≤ x < 4 or x ≥ 6 ".

Practice Questions

Work out the answer to each question then click on the button marked

to see if you are correct.

Solve these inequalities and represent the solution on a number line:
(a) 4( x – 3 ) ≤ 2( x + 2 ) 4( x – 3 ) ≤ 2( x + 2 ) 4x – 12 ≤ 2x + 4 2x ≤ 16 x ≤ 8	(b) 2 < 3x – 7 ≤ 8 9 < 3x ≤ 15 3 < x ≤ 5
State the values for which both the inequalities in the question are true:
(a) x < 7 x > 3 Individually: Combined: The correct answer is " 3 < x < 7 ".

Work out the answers to the questions below and fill in the boxes. Click on the

button to find out whether you have answered correctly. If you have then the answer will be ticked

and you should move on to the next question. If a cross

appears then your answer is wrong. Click on

to clear the incorrect answer and have another go, or you can click on

to get some advice on how to work out the answer and then have another go. If you still can't work out the answer after this then you can click on

to see the solution.

How do you solve equations and inequalities?

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.