Equations and Inequalities (Math 7 Curriculum - Unit 3) | All Things Algebra®

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This Equations and Inequalities Unit Bundle includes guided notes, homework assignments, three quizzes, a study guide, and a unit test that cover the following topics:

• One-Step Equations with Addition and Subtraction

• One-Step Equations with Multiplication and Division

• One-Step Equations with Rational Numbers (Decimals and Fractions)

• Translating One-Step Equations

• One-Step Equation Word Problems

• Two-Step Equations

• Translating Two-Step Equations

• Two-Step Equations Word Problems

• Multi-Step Equations (Variables on One Side)

• Multi-Step Equations (Variables on Both Sides)

• Writing and Graphing Inequalities

• Solving One-Step Inequalities

• Solving Two-Step Inequalities

• Translating One- and Two-Step Inequalities

• One- and Two-Step Inequality Word Problems

ADDITIONAL COMPONENTS INCLUDED:

(1) Links to Instructional Videos: Links to videos of each lesson in the unit are included. Videos were created by fellow teachers for their students using the guided notes and shared in March 2020 when schools closed with no notice.  Please watch through first before sharing with your students. Many teachers still use these in emergency substitute situations. (2) Editable Assessments: Editable versions of each quiz and the unit test are included. PowerPoint is required to edit these files. Individual problems can be changed to create multiple versions of the assessment. The layout of the assessment itself is not editable. If your Equation Editor is incompatible with mine (I use MathType), simply delete my equation and insert your own.

(3) Google Slides Version of the PDF: The second page of the Video links document contains a link to a Google Slides version of the PDF. Each page is set to the background in Google Slides. There are no text boxes;  this is the PDF in Google Slides.  I am unable to do text boxes at this time but hope this saves you a step if you wish to use it in Slides instead! 

How does this unit compare to your Pre-Algebra Equations and Inequalities Unit? This unit is similar in that it contains similar topics. However, all material has been rewritten to ensure there is no duplication. The biggest differences between this Math 7 unit and Pre-Algebra Curriculum include no square roots equations, special solutions, multi-step equation word problems, or multi-step inequalities. The material is also presented at a slighter slower pace.

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

Math 7 Curriculum

More Math 7 Units:

Unit 1 – Number Sense

Unit 2 – Expressions

Unit 4 – Ratios, Proportions, and Percents

Unit 5 – Functions and Graphing

Unit 6 – Geometry

Unit 7 – Measurement: Area and Volume

Unit 8 – 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 [email protected].

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.

© All Things Algebra (Gina Wilson), 2012-present

Questions & Answers

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ⓐ yes ⓑ yes ⓒ no ⓓ yes ⓔ no

ⓐ no ⓑ yes ⓒ yes ⓓ no ⓔ no

2 · 2 · 2 · 2 · 5 2 · 2 · 2 · 2 · 5

2 · 2 · 3 · 5 2 · 2 · 3 · 5

ⓐ 9 ⓑ 64 ⓒ 40

ⓐ 216 ⓑ 64 ⓒ 185

10 x 2 + 16 x + 17 10 x 2 + 16 x + 17

12 y 2 + 9 y + 7 12 y 2 + 9 y + 7

ⓐ 14 x 2 − 13 14 x 2 − 13 ⓑ 12 x ÷ 2 12 x ÷ 2 ⓒ z + 13 z + 13 ⓓ 8 x − 18 8 x − 18

ⓐ 17 y 2 + 19 17 y 2 + 19 ⓑ 7 y 7 y ⓒ x + 11 x + 11 ⓓ 11 a − 14 11 a − 14

ⓐ 4 ( p + q ) 4 ( p + q ) ⓑ 4 p + q 4 p + q

ⓐ 2 x − 8 2 x − 8 ⓑ 2 ( x − 8 ) 2 ( x − 8 )

w − 7 w − 7

l − 6 l − 6

4 q − 8 4 q − 8

7 n + 3 7 n + 3

ⓐ = = ⓑ > > ⓒ < < ⓓ = =

ⓐ > > ⓑ = = ⓒ > > ⓓ < <

ⓐ −6 −6 ⓑ 2 ⓒ −2 −2

ⓐ −7 −7 ⓑ 3 ⓒ −3 −3

ⓐ 2 ⓑ −2 −2 ⓒ −10 −10 ⓓ 10

ⓐ 3 ⓑ −3 −3 ⓒ −11 −11 ⓓ 11

ⓐ 8 , 8 8 , 8 ⓑ −18 , −18 , −18 −18 ⓒ 19 , 19 19 , 19 ⓓ −4 , −4 , −4 −4

ⓐ 8 , 8 8 , 8 ⓑ −22 , −22 , −22 −22 ⓒ 23 , 23 23 , 23 ⓓ 3 , 3 3 , 3

ⓐ 23 ⓑ 60 ⓒ −63 −63 ⓓ −9 −9

ⓐ 39 ⓑ 39 ⓒ −28 ⓓ −7

ⓐ 81 ⓑ −81 −81

ⓐ 49 ⓑ −49 −49

( 9 + ( −16 ) ) + 4 ; − 3 ( 9 + ( −16 ) ) + 4 ; − 3

( −8 + ( −12 ) ) + 7 ; − 13 ( −8 + ( −12 ) ) + 7 ; − 13

The difference in temperatures was 45 degrees.

The difference in temperatures was 9 degrees.

− 23 40 − 23 40

− 5 8 − 5 8

−33 a −33 a

−26 b −26 b

3 4 b 3 4 b

79 60 79 60

103 60 103 60

ⓐ 27 a − 32 36 27 a − 32 36 ⓑ 2 a 3 2 a 3

ⓐ 24 k − 5 30 24 k − 5 30 ⓑ 2 k 15 2 k 15

− 1 2 − 1 2

ⓐ 6.58 6.58 ⓑ 6.6 6.6 ⓒ 7

ⓐ 15.218 15.218 ⓑ 15.22 15.22 ⓒ 15.2 15.2

ⓐ −16.49 −16.49 ⓑ −0.42 −0.42

ⓐ −23.593 −23.593 ⓑ −12.58 −12.58

−27.4815 −27.4815

−87.6148 −87.6148

ⓐ 25.8 ⓑ 258 ⓒ 2,580

ⓐ 142 ⓑ 1,420 ⓒ 14,200

587.3 587.3

34.25 34.25

ⓐ 117 500 117 500 ⓑ −0.875 −0.875

ⓐ 3 125 3 125 ⓑ −0.375 −0.375

ⓐ 0.09, 0.87, 0.039 ⓑ 17%, 175%, 8.25%

ⓐ 0.03, 0.91, 0.083 ⓑ 41%, 225%, 9.25%

ⓐ 6 ⓑ 13 ⓒ −15 −15

ⓐ 4 ⓑ 14 ⓒ −10 −10

ⓐ 4 , 49 4 , 49 ⓑ −3 , 4 , 49 −3 , 4 , 49 ⓒ −3 , 0. 3 – , 9 5 , 4 , 49 −3 , 0. 3 – , 9 5 , 4 , 49 ⓓ − 2 − 2 ⓔ −3 , − 2 , 0. 3 – , 9 5 , 4 , 49 −3 , − 2 , 0. 3 – , 9 5 , 4 , 49

ⓐ 6 , 121 6 , 121 ⓑ − 25 , −1 , 6 , 121 − 25 , −1 , 6 , 121 ⓒ − 25 , − 3 8 , −1 , 6 , 121 − 25 , − 3 8 , −1 , 6 , 121 ⓓ 2.041975.. . 2.041975.. . ⓔ − 25 , − 3 8 , −1 , 6 , 121 , 2.041975.. . − 25 , − 3 8 , −1 , 6 , 121 , 2.041975.. .

32 r + 29 s 32 r + 29 s

41 m + 6 n 41 m + 6 n

1 7 15 1 7 15

1 2 9 1 2 9

−48 a −48 a

−92 x −92 x

11 25 11 25

ⓐ 0 ⓑ undefined

4 x + 8 4 x + 8

6 x + 42 6 x + 42

5 y + 3 5 y + 3

4 n + 9 4 n + 9

70 + 15 p 70 + 15 p

4 + 35 d 4 + 35 d

−10 + 15 a −10 + 15 a

−56 + 105 y −56 + 105 y

− z + 11 − z + 11

− x + 4 − x + 4

3 − 3 x 3 − 3 x

2 x − 20 2 x − 20

5 x − 66 5 x − 66

7 x − 13 7 x − 13

Section 1.1 Exercises

Divisible by 2, 3, 6

Divisible by 2

Divisible by 3, 5

2 · 43 2 · 43

5 · 7 · 13 5 · 7 · 13

2 · 2 · 2 · 2 · 3 · 3 · 3 2 · 2 · 2 · 2 · 3 · 3 · 3

ⓐ 64 ⓑ 16 ⓒ 7

10 x + 6 10 x + 6

22 a + 1 22 a + 1

17 x 2 + 20 x + 16 17 x 2 + 20 x + 16

ⓐ 5 x 2 − 6 x y 5 x 2 − 6 x y ⓑ 6 y 2 5 x 6 y 2 5 x ⓒ y 2 + 21 y 2 + 21 ⓓ 81 x 2 − 6 x 81 x 2 − 6 x

ⓐ 4 a b 2 + 3 a 2 b 4 a b 2 + 3 a 2 b ⓑ 20 x y 2 20 x y 2 ⓒ m + 15 m + 15 ⓓ 121 x 2 − 9 x 121 x 2 − 9 x

ⓐ 8 ( y − 9 ) 8 ( y − 9 ) ⓑ 8 y − 9 8 y − 9

ⓐ 5 ( 3 x + y ) 5 ( 3 x + y ) ⓑ 15 x + y 15 x + y

2 c + 14 2 c + 14

3 n − 7 3 n − 7

Answers will vary.

Section 1.2 Exercises

ⓐ > > ⓑ > > ⓒ > > ⓓ > >

ⓐ = = ⓑ = = ⓒ > > ⓓ = =

ⓐ −11 −11 ⓑ −3 −3 ⓒ 3 3

ⓐ 6 ⓑ −6 −6 ⓒ −20 −20 ⓓ 20 20

ⓐ −32 −32 ⓑ −65 −65 ⓒ −4 −4 ⓓ 13 13

ⓐ −4 −4 ⓑ −12 −12 ⓒ −39 −39 ⓓ 14 14

ⓐ 64 64 ⓑ −64 −64

ⓐ −47 −47 ⓑ 16 16

( 3 + ( −15 ) ) + 7 ; − 5 ( 3 + ( −15 ) ) + 7 ; − 5

ⓐ 10 − ( −18 ) ; 28 10 − ( −18 ) ; 28 ⓑ −25 − 11 ; − 36 −25 − 11 ; − 36

−6 a + b −6 a + b

− $ 28 − $ 28

Section 1.3 Exercises

− 12 7 − 12 7

10 21 10 21

2 x 2 3 y 2 x 2 3 y

− 21 a 2 11 b 2 − 21 a 2 11 b 2

− 21 50 − 21 50

11 30 11 30

33 4 x 33 4 x

− 4 9 − 4 9

10 u 9 v 10 u 9 v

− 1 16 − 1 16

− 10 9 − 10 9

− 2 5 − 2 5

2 m 3 n 2 m 3 n

29 24 29 24

17 105 17 105

− 53 40 − 53 40

4 x + 3 12 4 x + 3 12

ⓐ 5 6 5 6 ⓑ 4 4

ⓐ 25 n 16 25 n 16 ⓑ 25 n − 16 30 25 n − 16 30

ⓐ −8 x − 15 18 −8 x − 15 18 ⓑ − 10 k 27 − 10 k 27

ⓐ −5 ( a + 1 ) 3 −5 ( a + 1 ) 3 ⓑ a a

49 25 49 25

−28 − 15 y 60 −28 − 15 y 60

33 64 33 64

23 24 23 24

ⓐ 1 5 1 5 ⓑ 6 5 6 5

− 1 9 − 1 9

− 5 11 − 5 11

Section 1.4 Exercises

ⓐ 5.78 ⓑ 5.8 ⓒ 6

ⓐ 0.30 ⓑ 0.3 ⓒ 0

ⓐ 63.48 ⓑ 63.5 ⓒ 63

−40.91 −40.91

−7.22 −7.22

−27.5 −27.5

102.212 102.212

51.31 51.31

−4.89 −4.89

−1200.47982 −1200.47982

337.8914 337.8914

1.305 1.305

$ 2.44 $ 2.44

19 200 19 200

−12.4 −12.4

0.393 0.393

156 % 156 %

6.25 % 6.25 %

ⓐ 0 , 36 , 9 0 , 36 , 9 ⓑ −8 , 0 , 36 , 9 −8 , 0 , 36 , 9 ⓒ −8 , 0 , 12 5 , 36 , 9 −8 , 0 , 12 5 , 36 , 9 ⓓ 1.95286... , 1.95286... , ⓔ −8 , 0 , 1.95286... , 12 5 , 36 , 9 −8 , 0 , 1.95286... , 12 5 , 36 , 9

ⓐ none ⓑ − 100 , −7 , −1 − 100 , −7 , −1 ⓒ − 100 , −7 , − 8 3 , −1 , 0.77 , 3 1 4 − 100 , −7 , − 8 3 , −1 , 0.77 , 3 1 4 ⓓ none ⓔ − 100 , −7 , − 8 3 , −1 , 0.77 , 3 1 4 − 100 , −7 , − 8 3 , −1 , 0.77 , 3 1 4

Section 1.5 Exercises

27 m + ( −21 n ) 27 m + ( −21 n )

5 4 g + 1 2 h 5 4 g + 1 2 h

2.43 p + 8.26 q 2.43 p + 8.26 q

1 5 6 1 5 6

14.88 14.88

49 11 49 11

32 y + 72 32 y + 72

6 c − 78 6 c − 78

3 4 q + 3 3 4 q + 3

5 y − 3 5 y − 3

3 + 8 r 3 + 8 r

36 d + 90 36 d + 90

r s − 18 r r s − 18 r

y p + 4 p y p + 4 p

−28 p − 7 −28 p − 7

−3 x + 18 −3 x + 18

−3 x + 7 −3 x + 7

−3 y − 8 −3 y − 8

−33 c + 26 −33 c + 26

− a + 19 − a + 19

4 m − 10 4 m − 10

72 x − 25 72 x − 25

22 n + 9 22 n + 9

6 c + 34 6 c + 34

12 y + 63 12 y + 63

Review Exercises

Divisible by 2 , 3 , 5 , 6 , 10 2 , 3 , 5 , 6 , 10

6 x 2 − x + 5 6 x 2 − x + 5

ⓐ 11 ( y − 2 ) 11 ( y − 2 ) ⓑ 11 y − 2 11 y − 2

ⓐ 8 ⓑ −8 −8 ⓒ −22 −22 ⓓ 22

ⓐ −3 −3 ⓑ −15 −15 ⓒ −56 −56 ⓓ 17

( −4 + ( −9 ) ) + 23 ; 10 ( −4 + ( −9 ) ) + 23 ; 10

− 15 x 3 11 y 2 − 15 x 3 11 y 2

8 x 15 y 8 x 15 y

31 36 31 36

ⓐ 11 8 11 8 ⓑ 5 6 5 6

− 1 6 − 1 6

− 1 5 − 1 5

96.978 96.978

− 48 5 − 48 5

1. 27 ¯ 1. 27 ¯

4.75 % 4.75 %

no real number

3 4 x + y 3 4 x + y

1 11 15 1 11 15

8 b + 10 8 b + 10

x p − 5 p x p − 5 p

−6 x − 6 −6 x − 6

6 y + 16 6 y + 16

Practice Test

7 n + 7 7 n + 7

−8 − 11 ; − 19 −8 − 11 ; − 19

( −8 − ( −3 ) ) + 5 ; 0 ( −8 − ( −3 ) ) + 5 ; 0

ⓐ 28.15 28.15 ⓑ 28.146 28.146

15 17 15 17

− 5 3 − 5 3

− 7 6 − 7 6

−65.4 −65.4

1 8 13 1 8 13

13 y − 3 13 y − 3

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  • Book title: Intermediate Algebra 2e
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  • Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
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Mathematics LibreTexts

1.1: Solving Linear Equations and Inequalities

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  • Save as PDF
  • Page ID 48347

Learning Objectives

  • Verify linear solutions.
  • Use the properties of equality to solve basic linear equations.
  • Clear fractions from equations.
  • Identify linear inequalities and check solutions.
  • Solve linear inequalities and express the solutions graphically on a number line and in interval notation.

Prerequisite Skills

Before you get started, take this prerequisite quiz.

1. Simplify \(2-6(4-7)^2\) without using a calculator.

If you missed this problem, review here . (Note that this will open a different textbook in a new window.)

2. Evaluate \(6x−4\) when \(x=−2\).

3. Evaluate \(-5x^2−x+9\) when \(x=-3\).

4. Simplify \(7x−1−4x+5\).

\(3x+4\) 

Solving Basic Linear Equations

An equation 129 is a statement indicating that two algebraic expressions are equal. A linear equation with one variable 130 , \(x\), is an equation that can be written in the standard form \(ax + b = 0\) where \(a\) and \(b\) are real numbers and \(a ≠ 0\). For example

\(3 x - 12 = 0\)

A solution 131 to a linear equation is any value that can replace the variable to produce a true statement. The variable in the linear equation \(3x − 12 = 0\) is \(x\) and the solution is \(x = 4\). To verify this, substitute the value \(4\) in for \(x\) and check that you obtain a true statement.

\(\begin{aligned} 3 x - 12 & = 0 \\ 3 ( \color{Cerulean}{4}\color{Black}{ )} - 12 & = 0 \\ 12 - 12 & = 0 \\ 0 & = 0 \:\: \color{Cerulean}{✓} \end{aligned}\)

Alternatively, when an equation is equal to a constant, we may verify a solution by substituting the value in for the variable and showing that the result is equal to that constant. In this sense, we say that solutions “satisfy the equation.”

Example \(\PageIndex{1}\):

Is \(a=2\) a solution to \(−10a+5=−25\)?

Recall that when evaluating expressions, it is a good practice to first replace all variables with parentheses, and then substitute the appropriate values. By making use of parentheses, we avoid some common errors when working the order of operations.

\(\begin{align*} - 10 a + 5 &= -25 \\ - 10 ( \color{Cerulean}{2} \color{Black}{ ) +} 5 & = -25 \\ -20 + 5 & = -25 \\ -15 &\neq 25\:\: \color{red}{✗}\end{align*}\)  

No, \(a=2\) does not satisfy the equation and is therefore not a solution.

Developing techniques for solving various algebraic equations is one of our main goals in algebra. This section reviews the basic techniques used for solving linear equations with one variable. We begin by defining equivalent equations 132 as equations with the same solution set.

\(\left. \begin{aligned} 3 x - 5 & = 16 \\ 3 x & = 21 \\ x & = 7 \end{aligned} \right\} \quad \color{Cerulean}{Equivalent \:equations}\)

Here we can see that the three linear equations are equivalent because they share the same solution set, namely, \(\{7\}\). To obtain equivalent equations, use the following properties of equality 133 . Given algebraic expressions \(A\) and \(B\), where \(c\) is a nonzero number:

Table 1.1.1

Multiplying or dividing both sides of an equation by \(0\) is carefully avoided. Dividing by \(0\) is undefined and multiplying both sides by \(0\) results in the equation \(0 = 0\).

We solve algebraic equations by isolating the variable with a coefficient of 1. If given a linear equation of the form \(ax + b = c\), then we can solve it in two steps. First, use the appropriate equality property of addition or subtraction to isolate the variable term. Next, isolate the variable using the equality property of multiplication or division. Checking the solution in the following examples is left to the reader.

Example \(\PageIndex{2}\):

Solve: \(7x − 2 = 19\).

\(\begin{aligned} 7 x - 2 & = 19 \\ 7 x - 2 \color{Cerulean}{+ 2} & = 19 \color{Cerulean}{+ 2} & & \color{Cerulean}{Add\: 2\: to\: both\: sides.} \\ 7 x & = 21 \\ \frac { 7 x } { \color{Cerulean}{7} } & = \frac { 21 } { \color{Cerulean}{7} } & & \color{Cerulean}{Divide\: both\: sides\: by\: 7.} \\ x & = 3 \end{aligned}\)

The solution is \(3\).

Example \(\PageIndex{3}\):

Solve: \(56 = 8 + 12y\).

When no sign precedes the term, it is understood to be positive. In other words, think of this as \(56 = +8 + 12y\). Therefore, we begin by subtracting \(8\) on both sides of the equal sign.

\(\begin{aligned} 56 \color{Cerulean}{- 8} & = 8 + 12 y \color{Cerulean}{- 8} \\ 48 & = 12 y \\ \frac { 48 } { \color{Cerulean}{12} } & = \frac { 12 y } { \color{Cerulean}{12} } \\ 4 & = y \end{aligned}\)

It does not matter on which side we choose to isolate the variable because the symmetric property 134 states that \(4 = y\) is equivalent to \(y = 4\).

The solution is \(4\).

Example \(\PageIndex{4}\):

Solve: \(\frac { 5 } { 3 } x + 2 = - 8\).

Isolate the variable term using the addition property of equality, and then multiply both sides of the equation by the reciprocal of the coefficient \(\frac{5}{3}\) .

\begin{aligned} \frac { 5 } { 3 } x + 2 & = - 8 \\ \frac { 5 } { 3 } x + 2 \color{Cerulean}{- 2} & = - 8 \color{Cerulean}{- 2}\quad \color{Cerulean}{Subtract\: 2\: on\: both\: sides.} \\ \frac { 5 } { 3 } x & = - 10 \\ \color{Cerulean}{\frac { 3 } { 5 }} \color{Black}{ \cdot} \frac { 5 } { 3 } x & = \color{Cerulean}{\frac { 3 } { \cancel{5} }} \color{Black}{\cdot} ( \overset{-2}{\cancel{-10}} )\quad \color{Cerulean}{Multiply \:both \:sides\: by\: \frac{3}{5}.} \\ 1x & = 3 \cdot ( - 2 ) \\ x & = - 6 \end{aligned}

The solution is \(−6\).

In summary, to retain equivalent equations, we must perform the same operation on both sides of the equation.

Exercise \(\PageIndex{1}\)

Solve: \(\frac { 2 } { 3 } x + \frac { 1 } { 2 } = - \frac { 5 } { 6 }\).

Video Solution: www.youtube.com/v/cQwqXs9AD6M

General Guidelines for Solving Linear Equations

Typically linear equations are not given in standard form, and so solving them requires additional steps. When solving linear equations, the goal is to determine what value, if any, will produce a true statement when substituted in the original equation. Do this by isolating the variable using the following steps:

  • Step 1: Simplify both sides of the equation using the order of operations and combine all like terms on the same side of the equal sign.
  • Step 2a: Add or subtract as needed to isolate the variable.
  • Step 2b: Divide or multiply as needed to isolate the variable.
  • Step 3: Check to see if the answer solves the original equation.

We will often encounter linear equations where the expressions on each side of the equal sign can be simplified. If this is the case, then it is best to simplify each side first before solving. Normally this involves combining same-side like terms.

At this point in our study of algebra the use of the properties of equality should seem routine. Therefore, displaying these steps in this text, usually in blue, becomes optional.

Example \(\PageIndex{5}\):

Solve: \(- 4 a + 2 - a = 1\).

First combine the like terms on the left side of the equal sign.

\(\begin{aligned} - 4 a + 2 - a = 1 & \quad \color{Cerulean}{ Combine\: same-side\: like\: terms.} \\ - 5 a + 2 = 1 & \quad\color{Cerulean} { Subtract\: 2\: on\: both\: sides.} \\ - 5 a = - 1 & \quad\color{Cerulean} { Divide\: both\: sides\: by\: - 5.} \\ a = \frac { - 1 } { - 5 } = \frac { 1 } { 5 } \end{aligned}\)

Always use the original equation to check to see if the solution is correct.

\(\begin{aligned} - 4 a + 2 - a & = - 4 \left( \color{OliveGreen}{\frac { 1 } { 5 }} \right) + 2 - \color{OliveGreen}{\frac { 1 } { 5 }} \\ & = - \frac { 4 } { 5 } + \frac { 2 } { 1 } \cdot \color{Cerulean}{\frac { 5 } { 5 }}\color{Black}{ -} \frac { 1 } { 5 } \\ & = \frac { - 4 + 10 + 1 } { 5 } \\ & = \frac { 5 } { 5 } = 1 \:\:\color{Cerulean}{✓} \end{aligned}\)

The solution is \(\frac{1}{5}\) .

Given a linear equation in the form \(ax + b = cx + d\), we begin the solving process by combining like terms on opposite sides of the equal sign. To do this, use the addition or subtraction property of equality to place like terms on the same side so that they can be combined. In the examples that remain, the check is left to the reader.

Example \(\PageIndex{6}\):

Solve: \(−2y − 3 = 5y + 11\).

Subtract \(5y\) on both sides so that we can combine the terms involving y on the left side.

\(\begin{array} { c } { - 2 y - 3 \color{Cerulean}{- 5 y}\color{Black}{ =} 5 y + 11 \color{Cerulean}{- 5 y} } \\ { - 7 y - 3 = 11 } \end{array}\)

From here, solve using the techniques developed previously.

\(\begin{aligned} - 7 y - 3 & = 11 \quad\color{Cerulean}{Add\: 3\: to\: both\: sides.} \\ - 7 y & = 14 \\ y & = \frac { 14 } { - 7 } \quad\color{Cerulean}{Divide\: both\: sides\: by\: -7.} \\ y & = - 2 \end{aligned}\)

The solution is \(−2\).

Solving will often require the application of the distributive property.

Example \(\PageIndex{7}\):

Solve: \(- \frac { 1 } { 2 } ( 10 x - 2 ) + 3 = 7 ( 1 - 2 x )\).

Simplify the linear expressions on either side of the equal sign first.

\(\begin{aligned} - \frac { 1 } { 2 } ( 10 x - 2 ) + 3 = 7 ( 1 - 2 x ) & \quad\color{Cerulean} { Distribute } \\ - 5 x + 1 + 3 = 7 - 14 x & \quad\color{Cerulean} { Combine\: same-side\: like\: terms. } \\ - 5 x + 4 = 7 - 14 x & \quad\color{Cerulean} { Combine\: opposite-side\: like\: terms. } \\ 9 x = 3 & \quad\color{Cerulean} { Solve. } \\ x = \frac { 3 } { 9 } = \frac { 1 } { 3 } \end{aligned}\)

The solution is \(\frac{1}{3}\) .

Example \(\PageIndex{8}\):

Solve: \(5(3−a)−2(5−2a)=3\).

Begin by applying the distributive property.

\(\begin{aligned} 5 ( 3 - a ) - 2 ( 5 - 2 a ) & = 3 \\ 15 - 5 a - 10 + 4 a & = 3 \\ 5 - a & = 3 \\ - a & = - 2 \end{aligned}\)

Here we point out that \(−a\) is equivalent to \(−1a\); therefore, we choose to divide both sides of the equation by \(−1\).

\(\begin{array} { c } { - a = - 2 } \\ { \frac { - 1 a } { \color{Cerulean}{- 1} }\color{Black}{ =} \frac { - 2 } { \color{Cerulean}{- 1} } } \\ { a = 2 } \end{array}\)

Alternatively, we can multiply both sides of \(−a=−2\) by negative one and achieve the same result.

\(\begin{aligned} - a & = - 2 \\ \color{Cerulean}{( - 1 )}\color{Black}{ (} - a ) & = \color{Cerulean}{( - 1 )}\color{Black}{ (} - 2 ) \\ a & = 2 \end{aligned}\)

The solution is \(2\).

Exercise \(\PageIndex{2}\)

Solve: \(6 - 3 ( 4 x - 1 ) = 4 x - 7\).

Video Solution: www.youtube.com/v/NAIAZrFjU-o

The coefficients of linear equations may be any real number, even decimals and fractions. When this is the case it is possible to use the multiplication property of equality to clear the fractional coefficients and obtain integer coefficients in a single step. If given fractional coefficients, then multiply both sides of the equation by the least common multiple of the denominators (LCD).

Example \(\PageIndex{9}\):

Solve: \(\frac { 1 } { 3 } x + \frac { 1 } { 5 } = \frac { 1 } { 5 } x - 1\).

Clear the fractions by multiplying both sides by the least common multiple of the given denominators. In this case, it is the \(LCD (3, 5) = 15\).

\(\begin{aligned} \color{Cerulean}{15}\color{Black}{ \cdot} \left( \frac { 1 } { 3 } x + \frac { 1 } { 5 } \right) & = \color{Cerulean}{15}\color{Black}{ \cdot} \left( \frac { 1 } { 5 } x - 1 \right) \quad \color{Cerulean}{Multiply\: both\: sides\: by\: 15.} \\ \color{Cerulean}{15}\color{Black}{ \cdot} \frac { 1 } { 3 } x + \color{Cerulean}{15}\color{Black}{ \cdot} \frac { 1 } { 5 } & = \color{Cerulean}{15}\color{Black}{ \cdot} \frac { 1 } { 5 } x - \color{Cerulean}{15}\color{Black}{ \cdot} 1\quad\color{Cerulean}{Simplify.} \\ 5 x + 3 & = 3 x - 15\quad\quad\quad\color{Cerulean}{Solve.} \\ 2 x & = - 18 \\ x & = \frac { - 18 } { 2 } = - 9 \end{aligned}\)

The solution is \(−9\).

It is important to know that this technique only works for equations. Do not try to clear fractions when simplifying expressions. As a reminder:

Table 1.1.2

We simplify expressions and solve equations. If you multiply an expression by \(6\), you will change the problem. However, if you multiply both sides of an equation by \(6\), you obtain an equivalent equation.

Table 1.1.3

Applications Involving Linear Equations

Algebra simplifies the process of solving real-world problems. This is done by using letters to represent unknowns, restating problems in the form of equations, and by offering systematic techniques for solving those equations. To solve problems using algebra, first translate the wording of the problem into mathematical statements that describe the relationships between the given information and the unknowns. Usually, this translation to mathematical statements is the difficult step in the process. The key to the translation is to carefully read the problem and identify certain key words and phrases.

Table 1.1.4

When translating sentences into mathematical statements, be sure to read the sentence several times and parse out the key words and phrases. It is important to first identify the variable, “ let x represent… ” and state in words what the unknown quantity is. This step not only makes our work more readable, but also forces us to think about what we are looking for.

Example \(\PageIndex{10}\):

When \(6\) is subtracted from twice the sum of a number and \(8\) the result is \(5\). Represent this as an algebraic equation and find the number.

Let n represent the unknown number.

fb5b25de9d7267c4fdc3cf2953eae974.png

To understand why we included the parentheses in the set up, you must study the structure of the following two sentences and their translations:

Table 1.1.5

The key was to focus on the phrase “ twice the sum ,” this prompted us to group the sum within parentheses and then multiply by \(2\). After translating the sentence into a mathematical statement we then solve.

\(\begin{aligned} 2 ( n + 8 ) - 6 & = 5 \\ 2 n + 16 - 6 & = 5 \\ 2 n + 10 & = 5 \\ 2 n & = - 5 \\ n & = \frac { - 5 } { 2 } \end{aligned}\)

\(\begin{aligned} 2 ( n + 8 ) - 6 & = 2 \left( \color{Cerulean}{- \frac { 5 } { 2 }}\color{Black}{ +} 8 \right) - 6 \\ & = 2 \left( \frac { 11 } { 2 } \right) - 6 \\ & = 11 - 6 \\ & = 5 \quad\color{Cerulean}{✓}\end{aligned}\)       

The number is \(−\frac{5}{2}\).

General guidelines for setting up and solving word problems follow.

  • Step 1: Read the problem several times, identify the key words and phrases, and organize the given information.
  • Step 2: Identify the variables by assigning a letter or expression to the unknown quantities.
  • Step 3: Translate and set up an algebraic equation that models the problem.
  • Step 4: Solve the resulting algebraic equation.
  • Step 5: Finally, answer the question in sentence form and make sure it makes sense (check it).

For now, set up all of your equations using only one variable. Avoid two variables by looking for a relationship between the unknowns.

Linear Inequalities

A linear inequality 138 is a mathematical statement that relates a linear expression as either less than or greater than another. The following are some examples of linear inequalities, all of which are solved in this section:

A solution to a linear inequality 139 is a real number that will produce a true statement when substituted for the variable.

Example \(\PageIndex{11}\):

Are \(x=−4\) and \(x=6\) solutions to \(5x+7<22\)?

Substitute the values in for \(x\), simplify, and check to see if we obtain a true statement.

Table 1.1.6

\(x=−4\) is a solution and \(x=6\) is not

Linear inequalities have either infinitely many solutions or no solution. If there are infinitely many solutions, we graph the solution set on a number line and/or express the solution using interval notation.

Expressing Solutions to Linear Inequalities

What number would make the inequality \(x>3\) true? Are you thinking, "\(x\) could be four"? That’s correct, but \(x\) could be 6, too, or 137, or even 3.0001. Any number greater than three is a solution to the inequality \(x>3\). We show all the solutions to the inequality \(x>3\) on the number line by shading in all the numbers to the right of three, to show that all numbers greater than three are solutions. Because the number three itself is not a solution, we put an open parenthesis at three.

We can also represent inequalities using interval notation . There is no upper end to the solution to this inequality. In interval notation, we express \(x>3\) as \((3,\infty)\). The symbol \(\infty\) is read as “ infinity .” It is not an actual number. Figure \(\PageIndex{2}\) shows both the number line and the interval notation.

The figure shows the inquality, x is greater than 3, graphed on a number line from negative 5 to 5. There is shading that starts at 3 and extends to numbers to its right. The solution for the inequality is written in interval notation. It is the interval from 3 to infinity, not including 3.

We use the left parenthesis symbol, (, to show that the endpoint of the inequality is not included. The left bracket symbol, [, would show that the endpoint is included.

The inequality \(x\leq 1\) means all numbers less than or equal to one. Here we need to show that one is a solution, too. We do that by putting a bracket at \(x=1\). We then shade in all the numbers to the left of one, to show that all numbers less than one are solutions (Figure \(\PageIndex{3}\)). There is no lower end to those numbers. We write \(x\leq 1x\leq 1 \)in interval notation as \((−\infty,1]\). The symbol \(−\infty\) is read as “negative infinity.”

The figure shows the inquality, x is less than or equal to l, graphed on a number line from negative 5 to 5. There is shading that starts at 1 and extends to numbers to its left. The solution for the inequality is written in interval notation. It is the interval from negative infinity to one, including 1.

Figure \(\PageIndex{4}\) shows both the number line and interval notation.

INEQUALITIES, NUMBER LINES, AND INTERVAL NOTATION

The figure shows that the solution of the inequality x is greater than a is indicated on a number line with a left parenthesis at a and shading to the right, and that the solution in interval notation is the interval from a to infinity enclosed in parentheses. It shows the solution of the inequality x is greater than or equal to a is indicated on a number line with an left bracket at a and shading to the right, and that the solution in interval notation is the interval a to infinity within a left bracket and right parenthesis. It shows that the solution of the inequality x is less than a is indicated on a number line with a right parenthesis at a and shading to the left, and that the solution in interval notation is the the interval negative infinity to a within parentheses. It shows that the solution of the inequality x is less than or equal to a is indicated on anumber line with a right bracket at a and shading to the left, and that the solution in interval notation is negative infinity to a within a left parenthesis and right bracket.

The notation for inequalities on a number line and in interval notation use the same symbols to express the endpoints of intervals. Notice that \(\infty\) and \(-\infty\) always use parentheses in interval notation, never brackets.

Example \(\PageIndex{12}\)

Graph each inequality on the number line and write in interval notation.

  • \(x\geq −3\)
  • \(x<2.5\)
  • \(x\leq −\frac{3}{5}\)

Exercise \(\PageIndex{3}\)

Graph each inequality on the number line and write in interval notation:

  • \(x\leq −4\)
  • \(x\geq 0.5\)
  • \(x<−\frac{2}{3}\).

The graph of the inequality x is less than or equal to negative 4 is indicated on a number line with a right bracket at negative 4 and shading to the left. The solution in interval notation is the interval from negative infinity to negative 4 enclosed within an left parenthesis and right bracket.

Solving Linear Inequalities

All but one of the techniques learned for solving linear equations apply to solving linear inequalities. You may add or subtract any real number to both sides of an inequality, and you may multiply or divide both sides by any positive real number to create equivalent inequalities. For example:

\[\begin{align*} 10 &> - 5 \\[4pt] 10{\color{Cerulean}{-7}}\,&{\color{Black}{>}} -5{\color{Cerulean}{-7}} & & {\color{Cerulean}{Subtract\: 7\: on\: both\: sides.}}\\[4pt]  3 &> - 12 & & \color{Cerulean}{✓}\quad\color{Cerulean}{True.} \\[20pt] 10 &>-5\\[4pt] \frac{10}{\color{Cerulean}{5}}\,&\color{Black}{>}\frac{-5}{\color{Cerulean}{5}} & & \color{Cerulean}{Divide\: both\: sides\: by\: 5.}\\[4pt] 2 &>-1 & & \color{Cerulean}{✓\:\:True} \end{align*} \]

Subtracting \(7\) from each side and dividing each side by positive \(5\) results in an inequality that is true.

Example \(\PageIndex{13}\):

Solve and graph the solution set: \(5x+7<22\).

\(\begin{array} { c } { 5 x + 7 < 22 } \\ { 5 x + 7 \color{Cerulean}{- 7}\color{Black}{ < 22}\color{Cerulean}{ - 7} } \\ { 5 x < 15 } \\ { \frac { 5 x } {\color{Cerulean}{ 5} } < \frac { 15 } { \color{Cerulean}{5} } } \\ { x < 3 } \end{array}\)

a9da756c92955b8c6a5644a9b4418b89.png

It is helpful to take a minute and choose a few values in and out of the solution set, substitute them into the original inequality, and then verify the results. As indicated, you should expect \(x=0\) to solve the original inequality and that \(x=5\) should not.

Table 1.1.7

Checking in this manner gives us a good indication that we have solved the inequality correctly.

We can express this solution in two ways: using set notation and interval notation.

\(\begin{array} { r } { \{ x | x < 3 \} } &\color{Cerulean}{Set\: notation} \\ { ( - \infty , 3 ) } &\color{Cerulean}{Interval\: notation} \end{array}\)

In this text we will choose to present answers using interval notation.

\((−∞,  3) \)

When working with linear inequalities, a different rule applies when multiplying or dividing by a negative number. To illustrate the problem, consider the true statement \(10 > −5\) and divide both sides by \(−5\).

\(\begin{array} { l } { 10 > - 5 } \\ { \frac { 10 } { \color{Cerulean}{- 5} } \color{Black}{>} \frac { - 5 } { \color{Cerulean}{- 5} } } \quad \color{Cerulean}{Divide\: both\: sides\: by\: -5.} \\ { - 2 \color{red}{>}\color{Black}{ 1} \quad \color{red}{✗} \color{Cerulean}{ False } } \end{array}\)

Dividing by \(−5\) results in a false statement. To retain a true statement, the inequality must be reversed.

\(\begin{array} { l } { 10 \color{OliveGreen}{>}\color{Black}{ - 5} } \\ { \frac { 10 } { \color{Cerulean}{- 5} } \color{Black}{<} \frac { - 5 } { \color{Cerulean}{- 5} } } \quad \color{Cerulean}{Reverse\: the\: inequality.} \\ { - 2 \color{OliveGreen}{<}\color{Black}{ 1} \quad \color{Cerulean}{✓} \color{Cerulean}{ True } } \end{array}\)

The same problem occurs when multiplying by a negative number. This leads to the following new rule: when multiplying or dividing by a negative number, reverse the inequality . It is easy to forget to do this so take special care to watch for negative coefficients. In general, given algebraic expressions \(A\) and \(B\), where \(c\) is a positive nonzero real number, we have the following properties of inequalities 140 :

Table 1.1.8

We use these properties to obtain an equivalent inequality 141 , one with the same solution set, where the variable is isolated. The process is similar to solving linear equations.

Example \(\PageIndex{14}\):

Solve and graph the solution set: \(−2(x+8)+6≥20\).

\(\begin{aligned} - 2 ( x + 8 ) + 6 & \geq 20 \quad\color{Cerulean}{Distribute.} \\ - 2 x - 16 + 6 & \geq 20 \quad\color{Cerulean}{Combine\: like\: terms.} \\ - 2 x - 10 & \geq 20 \quad\color{Cerulean}{Solve\: for\: x.} \\ - 2 x & \geq 30 \quad\color{Cerulean}{Divide\: both\: sides\: by\: -2.} \\ \frac { - 2 x } { \color{Cerulean}{- 2} } & \color{OliveGreen}{\leq} \frac { \color{Black}{30} } { \color{Cerulean}{- 2} } \quad\color{Cerulean}{Reverse\: the\: inequality.} \\ x & \leq - 15 \end{aligned}\)

8c99e3a6a02d925e770328430c5de15c.png

Interval notation \((−∞,  −15] \)

Example \(\PageIndex{15}\):

Solve and graph the solution set: \(−2(4x−5)<9−2(x−2)\).

\(\begin{array} { c } { - 2 ( 4 x - 5 ) < 9 - 2 ( x - 2 ) } \\ { - 8 x + 10 < 9 - 2 x + 4 } \\ { - 8 x + 10 < 13 - 2 x } \\ { - 6 x < 3 } \\ { \frac { - 6 x } { \color{Cerulean}{- 6} } \color{OliveGreen}{>} \frac { \color{Black}{3} } { \color{Cerulean}{- 6} } }\color{Cerulean}{Reverse\:the\:inequality.} \\ { x > - \frac { 1 } { 2 } } \end{array}\)

c5fab2d63fef6f7f1c9fb0803070c7c6.png

Interval notation \((−\frac{1}{2}, ∞)\)

Example \(\PageIndex{16}\):

Solve and graph the solution set: \(\frac{1}{2}x−2≥\frac{1}{2}(\frac{7}{4}x−9)+1\).

\(\begin{array} { c } { \frac { 1 } { 2 } x - 2 \geq \frac { 1 } { 2 } \left( \frac { 7 } { 4 } x - 9 \right) + 1 } \\ { \frac { 1 } { 2 } x - 2 \geq \frac { 7 } { 8 } x - \frac { 9 } { 2 } + 1 } \\ { \frac { 1 } { 2 } x - \frac { 7 } { 8 } x \geq - \frac { 7 } { 2 } + 2 } \\ { - \frac { 3 } { 8 } x \geq - \frac { 3 } { 2 } } \\ { \left( \color{Cerulean}{- \frac { 8 } { 3 }} \right) \left(\color{Black}{ - \frac { 3 } { 8 } x} \right) \leq \left( \color{Cerulean}{- \frac { 8 } { 3 }} \right) \left( \color{Black}{-} \frac { 3 } { 2 } \right) \quad \color{Cerulean} { Reverse\: the\: inequality. } } \\ { x \leq 4 } \end{array}\)

147955e243a86aa7b1105004683059eb.png

Interval notation: \((−∞,  4]\)

Exercise \(\PageIndex{4}\)

Solve and graph the solution set: \(10 - 5 ( 2 x + 3 ) \leq 25\)

\([ - 3 , \infty )\);

3bea9d6d532f6059af024ddfb6b02549.png

Video Solution: www.youtube.com/v/COLLNtwYFm8

Translation of Linear Inequalities

Some of the key words and phrases that indicate inequalities are summarized below:

Table 1.1.9

Key Takeaways

  • Solving general linear equations involves isolating the variable, with coefficient \(1\), on one side of the equal sign. To do this, first use the appropriate equality property of addition or subtraction to isolate the variable term on one side of the equal sign. Next, isolate the variable using the equality property of multiplication or division. Finally, check to verify that your solution solves the original equation.
  • If solving a linear equation leads to a true statement like \(0 = 0\), then the equation is an identity and the solution set consists of all real numbers, \(ℝ\).
  • If solving a linear equation leads to a false statement like \(0 = 5\), then the equation is a contradiction and there is no solution, \(Ø\).
  • Clear fractions by multiplying both sides of an equation by the least common multiple of all the denominators. Distribute and multiply all terms by the LCD to obtain an equivalent equation with integer coefficients.
  • Simplify the process of solving real-world problems by creating mathematical models that describe the relationship between unknowns. Use algebra to solve the resulting equations.
  • Inequalities typically have infinitely many solutions. The solutions are presented graphically on a number line or using interval notation or both.
  • All but one of the rules for solving linear inequalities are the same as solving linear equations. If you divide or multiply an inequality by a negative number, reverse the inequality to obtain an equivalent inequality.

129 Statement indicating that two algebraic expressions are equal.

130 An equation that can be written in the standard form \(ax + b = 0\), where \(a\) and \(b\) are real numbers and \(a ≠ 0\).

131 Any value that can replace the variable in an equation to produce a true statement.

132 Equations with the same solution set.

133 Properties that allow us to obtain equivalent equations by adding, subtracting, multiplying, and dividing both sides of an equation by nonzero real numbers.

134 Allows you to solve for the variable on either side of the equal sign, because \(x = 5\) is equivalent to \(5 = x\).

135 Equations that are true for particular values.

136 An equation that is true for all possible values.

137 An equation that is never true and has no solution.

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    Unit test. Level up on all the skills in this unit and collect up to 1100 Mastery points! There are lots of strategies we can use to solve equations. Let's explore some different ways to solve equations and inequalities. We'll also see what it takes for an equation to have no solution, or infinite solutions.

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    6th grade 11 units · 148 skills. Unit 1 Ratios. Unit 2 Arithmetic with rational numbers. Unit 3 Rates and percentages. Unit 4 Exponents and order of operations. Unit 5 Negative numbers. Unit 6 Variables & expressions. Unit 7 Equations & inequalities. Unit 8 Plane figures.

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    Introduction; 4.1 Solve Systems of Linear Equations with Two Variables; 4.2 Solve Applications with Systems of Equations; 4.3 Solve Mixture Applications with Systems of Equations; 4.4 Solve Systems of Equations with Three Variables; 4.5 Solve Systems of Equations Using Matrices; 4.6 Solve Systems of Equations Using Determinants; 4.7 Graphing Systems of Linear Inequalities

  10. 3.1 Absolute Value Inequality

    A2.5.4 Solve equations and inequalities involving absolute values of linear expressions; Need a tutor? Click this link and get your first session free! Packet. a2_3.1_packet.pdf: File Size: 599 kb: File Type: pdf: Download File. Practice Solutions. a2_3.1_practice_solutions.pdf: File Size: 464 kb: File Type: pdf: Download File. Corrective ...

  11. Unit 5

    Unit 5 - Systems of Linear Equations and Inequalities. This unit begins by ensuring that students understand that solutions to equations are points that make the equation true, while solutions to systems make all equations (or inequalities) true. Graphical and substitution methods for solving systems are reviewed before the development of the ...

  12. 1.1: Solving Linear Equations and Inequalities

    Solving Basic Linear Equations. An equation 129 is a statement indicating that two algebraic expressions are equal. A linear equation with one variable 130, \(x\), is an equation that can be written in the standard form \(ax + b = 0\) where \(a\) and \(b\) are real numbers and \(a ≠ 0\).For example \(3 x - 12 = 0\) A solution 131 to a linear equation is any value that can replace the ...

  13. Algebra 1 Curriculum

    This curriculum is divided into the following units: Unit 1 - Algebra Basics. Unit 2 - Multi-Step Equations & Inequalities. Unit 3 - Relations & Functions. Unit 4 - Linear Equations. Mini-Unit - Direct & Inverse Variation. Unit 5 - Systems of Equations & Inequalities. Unit 6 - Exponents & Exponential Functions.

  14. PDF ALG2 Guided Notes

    Algebra 2 -55 - Systems of Equations SECTION 3.3: SYSTEMS OF INEQUALITIES MACC.912.A-REI.D.12: Graph the solutions to a linear inequality in two variables as a half plane (excluding the boundary in the case of a strict inequality), and graph the solution set to a system of linear inequalities in two variables as the intersection of the

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

    Unit 4 Percentages. Unit 5 Exponents intro and order of operations. Unit 6 Variables & expressions. Unit 7 Equations & inequalities introduction. Unit 8 Percent & rational number word problems. Unit 9 Proportional relationships. Unit 10 One-step and two-step equations & inequalities. Unit 11 Roots, exponents, & scientific notation.

  16. Algebra 1, Unit 5

    In this lesson, we reviewed strategies for solving systems of linear equations, including using tables, graphs, and algebra. We examined two different algebraic methods, both of which involve substitution of expressions for variables. These methods work with systems of linear equations that are linear functions, as well as systems of linear ...

  17. Unit 3: Equations and Inequalities Flashcards

    difference, minus, less than, subtracted from, decreased by, take away. product, times, multiplied by, of, per, twice, triple, double. The form px+q=r. the relationship between two numbers that are not equal. Uses symbols like less than ( < ) and greater than ( > ) Key words from unit 3 Learn with flashcards, games, and more — for free.

  18. Unit 3: Systems of Linear Equations and Inequalities Flashcards

    Two or more linear inequalities using the same variables. The solutions that make both inequalities true lie in the region where the shading overlaps. So in the visual example, the purple region. Study with Quizlet and memorize flashcards containing terms like Solve by graphing, Solve by elimination, Solve by substitution and more.

  19. Unit 3 Equations And Inequalities Worksheets

    Worksheets are Unit 3 linear systems, Algebra i, Equations inequalities, Algebra 1 unit 2 equations and inequalities answer key, 8th math unit 3, Unit 1 equations and inequalities answers, Solving inequalities 4 directions x examples, Georgia standards of excellence curriculum frameworks. *Click on Open button to open and print to worksheet.

  20. Unit 3--Equations and Inequalities Flashcards

    A mathematical process that combines two or more numbers such that its product or sum equals the identity. States that when both sides of an equation are multiplied by the same number, the remaining expressions are still equal. The set of all values which, when substituted for unknowns, make an equation true.

  21. Solved Name: Date: Unit 2: Equations & Inequalitles Homework

    Algebra questions and answers. Name: Date: Unit 2: Equations & Inequalitles Homework 10: Multh-Step Inequalities Bell:_ Directions: Solve and express the fol lowing inequaities in interval L. 3 interval Notation: Interval Notation: 3.5+8)-7s 23 sx + 33 느 23 3373 Interval Notation: Interval Notation: 6. 2x +5 s 3x- 10 Interval Notation ...