unit 4 linear equations homework 8 writing linear equations review

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Standard Form to Slope Intercept

7th -  9th  , slope & slope formula, 7th -  10th  , slopey slope, 10.7k plays, slope intercept form, graphing linear equations.

Unit 4 Linear Equations Review

9th - 12th grade, mathematics.

19 questions

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Graph The Linear relationship with a y intercept is (0,5) and a slope is 2

Graph The Linear relationship with a y intercept is (0,-2) and a slope is –3

Find the slope an y intercept of the line graphed and write the equation of the line

Equation: y=2x-1

Equation: y=(1/2)x+1

Equation: y=(1/2)x-1

Equation: y=-1x+(1/2)

Find the slope an y intercept of the line graphed and write the equation of the line in slope-intercept form

y Intercept= -1

Equation: x=-1

Equation: y=-1

Slope= Undefined

y Intercept= 0

Equation: y=-1x+0

y Intercept= N/A

Find the slope of a line that contains the points (0, 3) and (−4, −1) then write the equation of the slope-intercept form.

Equation: y=1x+3

Equation: y=2x+3

Equation: y=1x+4

Equation: y=(1/2)x+4

Find the slope of a line that contains the points (−4, 5) and (8, −10) and then write the equation of the slope-intercept form.(Choose Multiple Answers)

Equation: y=(-5/4)x

Equation: y= x-(5/4)

Find the slope and x and y intercepts of the graph of the equation: 11x − 8y = −48

Slope= (11/8)

y intercept= 6

x Intercept= (-48/11)

Slope= (8/11)

y intercept=(-48/11)

x Intercept= 6

Find the slope and x and y intercepts of the graph of the equation: y = 4x − 1

y intercept= -1

x intercept= (1/4)

y intercept=(1/4)

x intercept= -1

y intercept= 1

x intercept= (-1/4)

x intercept=(1/4)

Solve the system of equations:

−3x − 2y = −12

−3x + 3y = 3

−5x + y = 13

No solution

Graph the inequality 4x + 2y > 16 (be sure to plot at least two points and graph the appropriate line and shading)

Graph the inequality y ≤ 6x – 3 (be sure to plot at least two points and graph the appropriate line and shading)

Simplify The Expression

Simplify The Expression:

−9(6m − 3) + 6(1 + 4m)

Determine which of the following relations is a function.

a. {(10,7), (−2,4), (5,3), (4,10)}

b. {(-2,-1) , (1,-4) , (7,-10) , (1,-11)}

Use the graph to identify domain and range. Classify if the graph is continuous or discrete (Choose all that apply)

Domain: (-2,2)

Range:[-4,0]

Range:(-4,0)

For a rectangle with length of 2x + 5 and perimeter of 12x+30, what is the width of the rectangle? What is the value of x when the perimeter is 50cm?

Width: 4x+10

Width: 2x+5

You can buy bananas at Trader Joe’s for 31 cents each. The cost if a function of the number of bananas bought. Use the rule C(b) = .31b.

a)Label the dependent and Independent

b) Is the data discrete or continuous? Explain.

c) How many bananas were bought if the total bill was $2.79( 279 cents)?

Select all that Apply

Dependent: Cost, Independent: Bananas

Dependent: Bananas, Independent: Cost

Has the following radical been simplified correctly

Justify answer (2-3 Sentences)

Simplified Correctly

There is an error in the work provided

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Unit 4 - Linear Equations and Linear Systems

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Unit 4 – Linear Functions

Introduction to Linear Functions

LESSON/HOMEWORK

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SMART NOTEBOOK

More Work Graphing Linear Functions (Lines)

Writing the Equation of a Line

Working with Linear Functions in Table Form

Modeling with Linear Functions

More Modeling with Linear Functions

Equations of Horizontal and Vertical Lines

Piecewise Linear Functions

Step Functions

Absolute Value Functions

The Truth About Graphs

Linear Inequalities with Two Variables

Introduction to Sequences

Arithmetic Sequences

Unit Review

Unit 4 Review

UNIT REVIEW

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Unit 4 Assessment – Form A

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Unit 4 Assessment – Form B

Unit 4 Exit Tickets

Unit 4 Mid-Unit Quiz – Form A

U04.AO.01 – Modeling the Boundaries of a Shed

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Linear Equations (Algebra 1 Curriculum - Unit 4) | All Things Algebra®

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Description

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

• Slope from a Graph

• Slope from Ordered Pairs (The Slope Formula)

• Linear Equations: Slope Intercept Form vs. Standard Form

• Graphing by Slope Intercept Form

• Writing Linear Equations Given a Graph

• Graphing by Intercepts

• Vertical vs. Horizontal Lines

• Writing Linear Equations given Point and Slope

• Writing Linear Equations given Two Points

• Linear Equation Word Problems

• Parallel vs. Perpendicular Lines

• Scatter Plots & Line of Best Fit

• Linear Regression

Please download the preview to see a sample outline along with a collage of some of the pages.

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! 

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

Algebra 1 First Semester Notes Bundle

Algebra 1 Curriculum Algebra 1 Curriculum (with Activities)

More Algebra 1 Units:

Unit 1 – Algebra Basics

Unit 2 – Multi-Step Equations & Inequalities

Unit 3 – Relations & Functions

Direct & Inverse Variation (Mini-Unit)

Unit 5 – Systems of Equations & Inequalities

Unit 6 – Exponents and Exponential Functions

Unit 7 – Polynomials & Factoring

Unit 8 – Quadratic Equations

Unit 9 – Linear, Quadratic, and Exponential Functions

Unit 10 – Radical Expressions & Equations

Unit 11 – Rational Expressions & Equations

Unit 12 – 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

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Big Ideas Math Answers Grade 8 Chapter 4 Graphing and Writing Linear Equations

Hello Students!!! Are you looking for the Big Ideas Math Book 8th Grade Solution Key Chapter 4 Graphing and Writing Linear Equations on various websites? Stop your search now, because you are on the right page. Here the students can get the best study material to practice math in a correct way. You can learn the simple tricks to solve the problems with the help of Big Ideas Math Answers Grade 8 Chapter 4 Graphing and Writing Linear Equations.

Big Ideas Math Book 8th Grade Answer Key Chapter 4 Graphing and Writing Linear Equations

Practice makes your preparation perfect. You can score the maximum marks with reference to Big Ideas Math Answers Grade 8 Chapter 4 Graphing and Writing Linear Equations. We make you understand the concept of drawing the graphs with linear equations. Download Big Ideas Math Book 8th Grade Answer Key Chapter 4 Graphing and Writing Linear Equations for free.

Performance Task

Graphing and Writing Linear Equations STEAM Video/Performance Task

Graphing and writing linear equations getting ready for chapter 4.

Lesson: 1 Graphing Linear Equations

Lesson 4.1 Graphing Linear Equations

Graphing linear equations homework & practice 4.1.

Lesson: 2 Slope of a Line

Lesson 4.2 Slope of a Line

Slope of a line homework & practice 4.2.

Lesson: 3 Graphing Proportional Relationships

Lesson 4.3 Graphing Proportional Relationships

Graphing proportional relationships homework & practice 4.3.

Lesson: 4 Graphing Linear Equations in Slope-Intercept Form

Lesson 4.4 Graphing Linear Equations in Slope-Intercept Form

Graphing linear equations in slope-intercept form homework & practice 4.4.

Lesson: 5 Graphing Linear Equations in Standard Form

Lesson 4.5 Graphing Linear Equations in Standard Form

Graphing linear equations in standard form homework & practice 4.5.

Lesson: 6 Writing Equations in Slope-Intercept Form

Lesson 4.6 Writing Equations in Slope-Intercept Form

Writing equations in slope-intercept form homework & practice 4.6.

Lesson: 7 Writing Equations in Point-Slope Form

Lesson 4.7 Writing Equations in Point-Slope Form

Writing equations in point-slope form homework & practice 4.7.

Chapter: 4 – Graphing and Writing Linear Equations

Graphing and Writing Linear Equations Connecting Concepts

Graphing and writing linear equations chapter review, graphing and writing linear equations practice test, graphing and writing linear equations cumulative practice.

STEAM Video

Vocabulary The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts. linear equation slope y-intercept solution of a linear equation x-intercept

EXPLORATION 1

Draw the line through the points

Self-Assessment for Concepts & Skills Solve each exercise. Then rate your understanding of the success criteria in your journal.

Self-Assessment for Problem Solving Solve each exercise. The rate your understanding of the success criteria in your journal.

Question 13. A game show contestant earns y dollars for completing a puzzle in x minutes. This situation is represented by the equation y = – 250x + 5000. How long did a contestant who earned $500 take to complete the puzzle? Justify your answer. Answer: Given, A game show contestant earns y dollars for completing a puzzle in x minutes. This situation is represented by the equation y = – 250x + 5000. y = -250x + 5000 500 = -250x + 5000 500 – 5000 = -250x + 5000 – 5000 -4500 = -250x x = 18

b. You have $75 to spend. How many competitions can you attend? Answer: 75 ≤ 10x + 50 75 – 50 ≤ 10x 25 ≤ 10x 2.5 ≥ x By this I can say that I can attend 2 competitions if I have $75 to spend.

Question 15. The seating capacity for a banquet hall is represented by y = 8x + 56, where x is the number of extra tables you need. How many extra tables do you need to double the original seating capacity? Answer: Given, The seating capacity for a banquet hall is represented by y = 8x + 56, where x is the number of extra tables you need. y = 8x + 56 2 × 56 = 8x + 56 112 = 8x + 56 8x = 112 – 56 8x = 56 x = 7 tables

Review & Refresh

Describe the translation of the point to its image. Question 3. (1, – 4) → (3, 0) Answer: A(1, -4) = A'(1 + 2, -4) = (3, -4) A'(3, 4) = B(3, -4 + 4) = (3, 0) Translate 2 units right and 4 units up.

Question 4. (6, 4) → (- 4, – 6) Answer: We are given the points (6, 4) → (- 4, – 6) A(6, 4) = A'(6 – 10, 4) = (-4, 4) A'(-4, -4) = B(-4, 4 – 10) = (-4, -6)

Question 5. (4, – 2) → (- 9, 3) Answer: We are given the points A(4, -2) B(-9, 3) A(4, -2) = A'(4 – 13, -2) = (-9, -2) A'(-9, -2) = B(-9, -2 + 4) = (-9, 3)

Concepts, Skills, & Problem Solving

Question 30. MODELING REAL LIFE The amount y (in dollars) of money in your savings account after x months is represented by the equation y = 12.5x + 100. a. Graph the linear equation.

Question 32. GEOMETRY The sum S of the interior angle measures of a polygon with n sides is S = (n – 2) • 180°. a. Plot four points (n, S) that satisfy the equation. Is the equation a linear equation? Explain your reasoning.

Question 33. DIG DEEPER! One second of video on your cell phone uses the same amount of memory as two pictures. Your cell phone can store 2500 pictures. a. Create a graph that represents the number y of pictures your cell phone can store when you take x seconds of video.

Measuring the Steepness of a Line Work with a partner. Draw any nonvertical line in a coordinate plane. a. Develop a way to measure the steepness of the line. Compare your method with other pairs. b. Draw a line that is parallel to your line. What can you determine about the steepness of each line? Explain your reasoning. Answer:

EXPLORATION 2

Find the slope of the line through the given points. Question 3. (1, -2), (7, -2) Answer: (x1, y1) = (1, -2) (x2, y2) = (7, -2) m = (y2 – y1)/(x2 – x1) m = (-2 – (-2))/(7 – 1) m = 0/6 Thus slope = 0

Question 4. (-3, -3), (-3, -5) Answer: (x1, y1) = (-3, -3) (x2, y2) = (-3, -5) m = (y2 – y1)/(x2 – x1) m = (-5 + 3)/(-3 + 3) m = -2/0 Thus slope = undefined

Question 5. WHAT IF The blue line passes through (-4, -3) and (-3, 2). Are any of the lines parallel? Explain. Answer: (x1, y1) = (-4, -3) (x2, y2) = (-3, 2) m = (y2 – y1)/(x2 – x1) m = (2 + 3)/(-3 + 4) m = 5/1 m = 5 The slpe of the blue line is 5 and the slope of the red line is also 5. The blue lines and red lines have same slopes so they are parallel.

Question 6. VOCABULARY What does it mean for a line to have a slope of 4? Answer: If a line have a slope of 4 it means that the line rises 4 units for every 1 units it runs.

FINDING THE SLOPE OF A LINE Find the slope of the line through the given points. Question 7. (1, -1), (6, 2) Answer: (x1, y1) = (1, -1) (x2, y2) = (6, 2) m = (y2 – y1)/(x2 – x1) m = (2 – (-1))/(6 – 1) m = 3/5

Question 8. (2, -3), (5, -3) Answer: (x1, y1) = (2, -3) (x2, y2) = (5, -3) m = (y2 – y1)/(x2 – x1) m = (5 – 2)/(-3 + 3) m = 3/0 m = undefined

Self-Assessment for Problem Solving Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 11. A customer pays an initial fee and a daily fee to rent a snowmobile. The total payment for 3 days is 92 dollars. The total payment for 5 days is 120 dollars. What is the daily fee? Justify your answer. Answer: Given, A customer pays an initial fee and a daily fee to rent a snowmobile. The total payment for 3 days is 92 dollars. The total payment for 5 days is 120 dollars. m = (120 – 92)/5 – 3 m = 28/2 m = 14

Concepts, Skills, &Problem Solving

FINDING THE SLOPE OF A LINE Find the slope of the line through the given points. Question 15. (4, -1), (-2, -1) Answer: (x1, y1) = (4, -1) (x2, y2) = (-2, -1) m = (y2 – y1)/(x2 – x1) m = (-1 – (-1))/(-2 – 4) m = 0/-6 m = 0

Question 16. (5, -3), (5, 8) Answer: (x1, y1) = (5, -3) (x2, y2) = (5, 8) m = (y2 – y1)/(x2 – x1) m = (8 – 3)/(5 – 5) m = 5/0 m = undefined

Question 17. (-7, 0), (-7, -6) Answer: (x1, y1) = (-7, 0) (x2, y2) = (-7, -6) m = (y2 – y1)/(x2 – x1) m = (-6 – 0)/(-7 – (-7)) m = -6/0 m = undefined

Question 18. (-3, 1), (-1, 5) Answer: (x1, y1) = (-3, 1) (x2, y2) = (-1, 5) m = (y2 – y1)/(x2 – x1) m = (5 – 1)/(-1 + 3) m = 4/2 m = 2

Question 19. (10, 4), (4, 15) Answer: (x1, y1) = (10, 4) (x2, y2) = (4, 15) m = (y2 – y1)/(x2 – x1) m = (15 – 4)/(4 – 10) m = 11/-6 m = -11/6

Question 20. (-3, 6), (2, 6) Answer: (x1, y1) = (-3, 6) (x2, y2) = (2, 6) m = (y2 – y1)/(x2 – x1) m = (6 – 6)/(2 – (-3)) m = 0/5 m = 0

Answer: rise/run < 1/12 m = 0.06 1/12 = 0.0833 0.06 < 0.0833 As m < 1/12 the wheelchair ramp follows the guides.

b. Design a wheelchair ramp that provides access to a building with a front door that is 2.5 feet above the sidewalk. Illustrate your design. Answer: AC/AB = 1/12 2.5/AB = 1/12 AB = 2.5 × 12 AB = 30 So the end of the ramp should be placed at least 30 feet from the front door.

USING AN EQUATION Use an equation to find the value of k so that the line that passes through the given points has the given slope. Question 32. (1, 3), (5, k); m = 2 Answer: A(1, 3) B(5, k) m = 2 2 = (k – 3)/(5 – 1) 2 × 4 = k – 3 8 = k – 3 k = 8 + 3 k = 11

Question 33. (-2, k), (2, 0); m = -1 Answer: Given, A(-2, k) B(2, 0) m = -1 -1 = (0 – k)/2 – (-2) -1 = -k/4 -4 = -k k = 4

Question 34. (-4, k), (6, -7); m = –\(\frac{1}{5}\) Answer: Given, A(-4, k) B(6, -7) m = –\(\frac{1}{5}\) –\(\frac{1}{5}\) = (-7 – k)/6 – (-4) -2 = -7 – k -2 + 7 = -k 5 = -k k = -5

Question 35. (4, -4), (k, -1); m = \(\frac{3}{4}\) Answer: \(\frac{3}{4}\) = (-1 – (-4))/(k – 4) 4 = k – 4 k = 4 + 4 k = 8

Question 38. REASONING Do the points A(-2, -1), B(1, 5), and C(4, 11) lie on the same line? Without using a graph, how do you know? Answer: Given, A(-2, -1), B(1, 5), and C(4, 11) mAB = (5 – (-1))/(1 – (-2)) = 6/3 = 2 mBC = (11 – 5)/(4 – 1) = 6/3 = 2 By seeing the slopes we can say that the points A, B, C lie on the same line.

Question 39. PROBLEM SOLVING A small business earns a profit of $6500 in January and $17,500 in May. What is the rate of change in profit for this time period? Justify your answer. Answer: Pjan = 6500 Pmay = 17,500 Pmay – Pjan/5 – 1 = (17,500 – 6500)/4 = 11,000/4 = 2750

Question 40. STRUCTURE Choose two points in the coordinate plane. Use the slope formula to find the slope of the line that passes through the two points. Then find the slope using the formula \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\). Compare your results. Answer: P1(2, 5) P2(3, 10) m1 = (10 – 5)/(3 – 2) = 5/1 = 5 m2 = (5 – 10)/(1 – 3) = -5/-1 = 5 m1 = m2

Question 2. How much would a spacecraft that weighs 3500 kilograms on Earth weigh on Titan? Answer: y = 1/7 x y = 1/7 × 3500 y = 500 kg So a spacecraft would weigh 500 kg on Titan.

Question 6. WRITING AND USING AN EQUATION The number of objects a x machine produces is proportional to the time (in minutes) that the machine runs. The machine produces five objects in four minutes. a. Write an equation that represents the situation.

Answer: As 5 objects are produced in 4 minutes, the slope of the line is m = 5/4. The equation that represents the situation is y = 5/4 x y = 1.25 x

b. Graph the equation in part (a) and interpret the slope.

Answer: Use the slope. The equation shows that the slope m is 1.25. So the graph passes through the points (0, 0) and (1, 1.25)

Question 8. The speed of sound in air is 343 meters per second. You see lightning and hear thunder 12 seconds later. a. Is there a proportional relationship between the amount of time that passes and your distance from a lightning strike? Explain.

Answer: y = kx where k is the speed of sound, x the time and y the distance. Yes, there is a proportional relationship between the amount of time that passes and your distance from the lightning strike as the further you are, the more time will pass until the sound reaches you.

b. Estimate your distance from the lightning strike. Answer: y = 343 × 12 = 4116 meters

Solve the equation. Check your solution. Question 4. 2x + 3x = 10 Answer: Given the equation 2x + 3x = 10 5x = 10 x = 10/5 x = 2

Question 5. x + \(\frac{1}{6}\) = 4 – 2x Answer: Given the equation x + \(\frac{1}{6}\) = 4 – 2x x + 2x = 4 – \(\frac{1}{6}\) 3x = 4 – \(\frac{1}{6}\) 3x = \(\frac{23}{6}\) x = \(\frac{23}{18}\)

Question 6. 2(1 – x) = 11 Answer: 2(1 – x) = 11 2 – 2x = 11 2 – 11 = 2x 2x = -9 x = -9/2

Answer: y = 18x (0, 0), (2, 50) m = (50 – 0)/(2 – 0) m = 50/2 m = 25 25 > 18 Therefore the car has better mileage.

b. How much farther can the vehicle you chose in part(a) travel on 8 gallons of gasoline? Answer: y = 25 × 8 – 18 × 8 = 200 – 144 = 56 miles

Answer: y = 0.25x m = (1.4 – 0.7)/(2 – 1) m = 0.7 y = 0.7x Because 0.7 > 0.25, the fingernails grow faster.

Question 17. REASONING The quantities and are in a proportional relationship. What do you know about the ratio of y to x for any point (x, y) on the graph of x and y? Answer: y = kx where k is constant y/x = k This means the ratio of y to x is constant.

Answer: The relationship is proportional because the graph is linear and passes through the origin.

b. Write an equation of the line. Interpret the slope.

Answer: (0,0), (10, -35) m = (-35 – 0)/(10 – 0) = -35/10 = -3.5 y = -3.5x

c. You are at the bottom of a mountain where the temperature is 74°F. The top of the mountain is 5500 feet above you. What is the temperature at the top of the mountain? Justify your answer. Answer: x = 5.5 – 0 = 5.5 thousand feet y = -3.5x = -3.5(5.5) = -19.25 74 – 19.25 = 54.75°F

Find the slope and the y-intercept of the graph of the linear equation. Question 1. y = 3x – 7 Answer: Given the equation y = 3x – 7 Write the equation in slope – intercept form: y = mx + b The slope of the line is m and the y – intercept of the line is b. y = 3x – 7 Slope = 3 and y – intercept = -7

Question 2. y – 1 = –\(\frac{2}{3}\)x Answer: Write the equation in slope – intercept form: y = mx + b The slope of the line is m and the y – intercept of the line is b. y – 1 = –\(\frac{2}{3}\)x y = –\(\frac{2}{3}\)x + 1 Slope = –\(\frac{2}{3}\) and y – intercept = 1

Question 5. IN YOUR OWN WORDS Consider the graph of the equation y = mx + b. a. How does changing the value of m affect the graph of the equation?

Answer: The value of m is the slope of the graph. If the value of m changes it means the slope of the graph is changing, whether it will rise or fall from left or right is dependent on the value of m.

b. How does changing the value of b affect the graph of the equation? Answer: The value of b is the y-intercept of the graph. If the value of b changes it means it affects where the graph crosses the y – axis.

IDENTIFYING SLOPE AND y-INTERCEPT Find the slope and the y-intercept of the graph of the linear equation. Question 6. y = -x + 0.25 Answer: y = mx + c slope = -1 and y – intercept = 0.25

Question 7. y – 2 = –\(\frac{3}{4}\)x Answer: Given the equation y – 2 = –\(\frac{3}{4}\)x y = –\(\frac{3}{4}\)x + 2 slope = –\(\frac{3}{4}\) and y – intercept = 2

Solve the equation for y. Question 3. x = 4y – 2 Answer: Given the equation x = 4y – 2 x – 2 = 4y y = x/4 + 1/2

Question 4. 3y = -6x + 1 Answer: Given the equation 3y = -6x + 1 y = -2x + 1/3

Question 5. 1 + y = –\(\frac{4}{5}\)x – 2 Answer: Given the equation 1 + y = –\(\frac{4}{5}\)x – 2 y = –\(\frac{4}{5}\)x – 3

Question 6. 2.5y = 5x – 5 Answer: Given the equation 2.5y = 5x – 5 y = 2x – 2

Question 7. 1.3y + 5.2 = -3.9x Answer: Given the equation 1.3y + 5.2 = -3.9x 1.3y = -3.9x – 5.2 y = -3x – 4

Question 8. y – \(\frac{2}{3}\)x = -6 Answer: Given the equation y – \(\frac{2}{3}\)x = -6 y = \(\frac{2}{3}\)x -6

Question 13. y = –\(\frac{2}{3}\)x + 1 Answer:

Answer: Slope = -2/3 and y – intercept = 1 The graph which passes through the point (0, 1) and has a negative slope is the matching graph of the given equation.

IDENTIFYING SLOPES AND y-INTERCEPTS Find the slope and the y-intercept of the graph of the linear equation. Question 14. y = 4x – 5 Answer: y = mx + b slope = 4 and y — intercept = -5

Question 15. y = -7x + 12 Answer: y = -7x + 12 y = mx + b slpoe = -7 and y – intercept = 12

Question 16. y = –\(\frac{4}{5}\)x – 2 Answer: y = mx + b slope = -4/5 y – intercept = -2

Question 17. y = 2.25x + 3 Answer: y = mx + b slope = 2.25 and y – intercept = 3

Question 18. y + 1 = \(\frac{4}{3}\)x Answer: y = mx + b y + 1 = \(\frac{4}{3}\)x y = \(\frac{4}{3}\)x – 1 slope = \(\frac{4}{3}\), y – intercept = -1

Question 19. y – 6 = \(\frac{3}{5}\)x Answer: y = mx + b y – 6 = \(\frac{3}{5}\)x y = \(\frac{3}{5}\)x + 6 slope = 3/8 and y – intercept = 6

Question 20. y – 3.5 = -2x Answer: y = mx + b y – 3.5 = -2x y = -2x + 3.5 slope = -2 and y – intercept = 3.5

Question 21. y = -5 – \(\frac{1}{2}\)x Answer: y = mx + b y = -5 – \(\frac{1}{2}\)x y =- \(\frac{1}{2}\)x – 5 slope = – \(\frac{1}{2}\) and y – intercept = -5

Question 22. y = 11 + 1.5x Answer: y = mx + b y = 1.5x + 11 slope = 1.5 and y – intercept = 11

So, the intercept is 3.

Question 2. –\(\frac{1}{2}\)x + 2y = 6 Answer: –\(\frac{1}{2}\)x + 2y = 6 2y = 6 + \(\frac{1}{2}\)x y = 0.25x + 3 Comparing the value of b and m from y = mx + b m = 0.25 and b = 3 Plot y – intercept = (0, b) = (0, 3) Slope = 0.25 run/rise = 0.25/1 Plot the point 0.25 unit up and 1 unit to the right = (1, 3.25) Now plot the points and draw the graph

STRUCTURE Determine whether the equation is in standard form. If not, rewrite the equation in standard form. Question 7. y = x – 6 Answer: y = x – 6 The standard form of equation is: Ax + By = C The given equation is not in the standard form. y = x – 6 x – y = 6

Question 8. y – \(\frac{1}{6}\)x + 5 = 0 Answer: The standard form of equation is: Ax + By = C The given equation is not in the standard form. y – \(\frac{1}{6}\)x + 5 = 0 \(\frac{1}{6}\)x – y = 5

Question 9. 4x + y = 5 Answer: The standard form of equation is: Ax + By = C The given equation is in the form of the standard form.

Question 10. WRITING Describe two ways to graph the equation 4x + 2y = 6. Answer: The two ways to graph the equation: 1. Graph the equation using standard form 2. Graph the equation using intercept.

b. You spend twice as much time assembling the birdhouse as you do writing the paper. How much time do you spend writing the paper? Justify your answer. Answer: We are given, y = 2x 2x = -x + 60 2x + x = 60 3x = 60 x = 20 y = 2 (20) y = 40

Find the slope and the y-intercept of the graph of the linear equation. Question 1. y = x – 1 Answer: y = mx + b Slope = -1 and y – intercept = -1

Question 2. y = -2x + 1 Answer: y = -2x + 1 y = mx + b Slope = -2 and y – intercept = 1

Question 3. y = \(\frac{8}{9}\)x – 8 Answer: y = \(\frac{8}{9}\)x – 8 y = mx + b Slope = \(\frac{8}{9}\) and y – intercept = -8

REWRITING AN EQUATION Write the linear equation in slope-intercept form. Question 9. 2x + y = 17 Answer: Given the equation 2x + y = 17 y = 17 – 2x y = -2x + 17

Question 10. 5x – y = \(\frac{1}{4}\) Answer: Given the equation 5x – y = \(\frac{1}{4}\) -y = \(\frac{1}{4}\) – 5x y = 5x – \(\frac{1}{4}\)

Question 11. –\(\frac{1}{2}\)x + y = 10 Answer: Given the equation –\(\frac{1}{2}\)x + y = 10 y = \(\frac{1}{2}\)x + 10

MATCHING Match the equation with its graph. Question 15. 15x – 12y = 60 Answer: y = 0 15x – 12(0) = 60 15x = 60 x = 60/15 x = 4 x = 0 15(0) – 12y = 60 -12y = 60 y = -5 The graph having the x – intercept 4 and y – intercept -5

Question 16. 5x + 4y = 20 Answer: Given the linear equation 5x + 4y = 20 y = 0 5x + 4(0) = 20 5x = 20 x = 4 x = 0 5(0) + 4y = 20 4y = 20 y = 5

Question 17. 10x + 8y = -40 Answer:

Question 26. LOGIC Does the graph of every linear equation have an x-intercept? Justify your reasoning. Answer: y = mx + b y = 0 0 = mx + b mx = -b x = -b/m for m ≠ 0 If m = 0 the equation has no solution. Therefore the equation y = b has no x – intercept.

Writing Equations of Lines Work with a partner.For each part, answer the following questions.

• What are the slopes and the y-intercepts of the lines?
• What are equations that represent the lines?
• What do the lines have in common?

Question 7. WRITING AN EQUATION Write an equation of the line that passes through (0, -5) and (2, -5). Answer: m = (y2 – y1)/(x2 – x1) = (-5 – (-5))/(2 – 0) = 0/2 = 0 Because y = -5 when x = 0, the y – intercept is -5 y = mx + b y = (0)x + -5 y = -5

Question 8. You load boxes onto an empty truck at a constant rate. After 3 hours, there are 100 boxes on the truck. How much longer do you work if you load a total of 120 boxes? Justify your answer. Answer: Let x be the number of hours you work if you load a total of 120 boxes. 100/3 = 120/x 100x = 3 × 120 x = 360/100 x = 3.6 hours 3.6 – 3 = 0.6 hours

Question 10. DIG DEEPER! A lifetime subscription to a website costs $250. A monthly subscription to the website costs $10 to join and $15 per month. Write equations to represent the costs of each plan. If you want to be a member for one year, which plan is less expensive? Explain. Answer: Given, A lifetime subscription to a website costs $250. A monthly subscription to the website costs $10 to join and $15 per month. Total cost for plan 1 = the lifetime subscription y = 250 Total cost for Plan 2 = Fixed tax + Number of months . monthly cost y = 10 + 15x Plan 1: y = 250 Plan 2: y = 10 + 15(12) = 190 As 190 < 250, plan 1 is less expensive.

Write the linear equation in slope-intercept form. Question 1. 4x + y = 1 Answer: Given the equation 4x + y = 1 y = -4x + 1

Question 2. x – y = \(\frac{1}{5}\) Answer: Given the equation x – y = \(\frac{1}{5}\) x – \(\frac{1}{5}\) = y

Question 3. –\(\frac{2}{3}\)x + 2y = -7 Answer: Given the equation –\(\frac{2}{3}\)x + 2y = -7 2y = -7 + \(\frac{2}{3}\)x y = \(\frac{1}{3}\)x – \(\frac{7}{2}\)

Question 9. Write an equation that represents the graph. Answer: m = 10 b = 15 y = mx + b y = 10x + 15

Question 10. How can you determine the total cost of opening an account and buying 6 games? Answer: y = 10x + 15 y = 10(6) + 15 y = 60 + 15 y = 75

WRITING EQUATIONS Write an equation of the line that passes through the given points. Question 17. (-1, 4), (0, 2) Answer: m = (y2 – y1)/(x2 – x1) = (2 – 4)/(0 – (-1)) = -2/1 = -2 Because y = 2 when x = 0, the y – intercept is 2 y = mx + b y = -2x + 2

Question 18. (-1, 0), (0, 0) Answer: m = (y2 – y1)/(x2 – x1) = (0 – 0)/(0 – (-1)) = 0/1 = 0 Because y = 0 when x = 0, the y – intercept is 0 y = mx + b y = 0

Question 19. (0, 4), (0, -3) Answer: Both points belong to the y-axis. Therefore the equation of the line passing through them is x = 0

Question 24. MODELING REAL LIFE One of your friends gives you $10 for a charity walkathon. Another friend gives you an amount per mile. After 5 miles, you have raised $13.50 total. Write an equation that represents the amount y of money you have raised after x miles. Answer: Given, One of your friends gives you $10 for a charity walkathon. Another friend gives you an amount per mile. After 5 miles, you have raised $13.50 total. y = mx + b b = 10 13.50 = 5m + 10 13.50 – 10 = 5m 3.50 = 5m m = 3.50/5 m = 0.7 y = 0.7x + 10

Question 25. PROBLEM SOLVING You have 500 sheets of notebook paper. After 1 week, you have 72% of the sheets left. You use the same number of sheets each week. Write an equation that represents the number y of sheets remaining after x weeks. Answer: y = mx + b 500 – 0.72 × 500 = 500 – 360 = 140 sheets m = -140 b = 500 y = -140x + 500

Try It Write an equation in point -slope form of the line that passes through the given point and has the given slope. Question 1. (1, 2); m = -4 Answer: y – y1 = m(x – x1) y – 2 = -4(x – (1)) y – 2 = -4(x – 1)

Question 2. (7, 0); m = 1 Answer: y – y1 = m(x – x1) y – 0 = 1(x – (7)) y – 0 = 1(x – 7)

Question 3. (-8, -5); m = –\(\frac{3}{4}\) Answer: y – y1 = m(x – x1) y – (-5) = –\(\frac{3}{4}\)(x – (-8)) y + 5 = –\(\frac{3}{4}\)(x + 8)

Write an equation in slope-intercept form of the line that passes through the given points. Question 4. (-2, 1), (3, -4) Answer: Slope(m) = (-4 – 1)/(3 – (-2)) = -5/5 m = -1 y – y1 = m(x – x1) y – 1 = -1(x – (-2)) y – 1 = -1(x + 2) y – 1 = -x – 2 y = -x – 1

WRITING AN EQUATION Write an equation in point-slope form of the line that passes through the given point and has the given slope. Question 6. (2, 0); m = 1 Answer: y – y1 = m(x – x1) y – 0 = 1(x – (2)) y – 0 = 1(x – 2)

Question 7. (-3, -1); m = –\(\frac{1}{3}\) Answer: y – y1 = m(x – x1) y – (-1) = –\(\frac{1}{3}\)(x – (-3)) y + 1 = –\(\frac{1}{3}\)(x + 3)

Question 8. (5, 4); m = 3 Answer: y – y1 = m(x – x1) y – 4 = 3(x – (5)) y – 4 = 3(x – 5)

Question 12. DIG DEEPER! You and your friend begin to run along a path at different constant speeds.After 1 minute,your friend is 45 meters ahead of you. After 3 minutes, your friend is 105 meters ahead of you. a. Write and graph an equation for the distance y (in meters) your friend is ahead of you after x minutes. Justify your answer.

b. Did you and your friend start running from the same spot? Explain your reasoning. Answer: The distance between you and your friend in the initial moment is b = 15 meters. So you are ahead your friend by 15 meters at the starting point.

Solve the equation. Check your solution, if possible. Question 3. 2x + 3 = 2x Answer: Given the equation 2x + 3 = 2x 3 = 2x – 2x 3 ≠ 0

Question 4. 6x – 7 = 1 – 3x Answer: Given the equation 6x – 7 = 1 – 3x 6x + 3x = 1 + 7 9x = 8 x = 8/3

Question 5. 0.1x – 1 = 1.2x – 5.4 Answer: Given the equation 0.1x – 1 = 1.2x – 5.4 0.1x – 1.2x = 1 – 5.4 -1.1x = -4.4 x = 4

Question 7. Write an equation that represents the value V of the car after t years. Answer: y = -4000t + b where b is the original price 18,000 = -4000(3) + b 18,000 + 12,000 = b b = 30,000 y = -4000t + 30,000

WRITING AN EQUATION Write an equation in point-slope form of the line that passes through the given point and has the given slope. Question 8. (3, 0); m = –\(\frac{2}{3}\) Answer: y – y1 = m(x – x1) y – (0) = -2/3(x – 3) y – 0 = -2/3(x – 3)

Question 9. (4, 8); m = \(\frac{3}{4}\) Answer: y – y1 = m(x – x1) y – (8) = 3/4(x – 4) y – 8 = 3/4(x – 4)

Question 10. (1, -3); m = 4 Answer: y – y1 = m(x – x1) y – (-3) = 4(x – 1) y + 3 = 4(x – 1)

Question 11. (7, -5); m = –\(\frac{1}{7}\) Answer: y – y1 = m(x – x1) y – (-5) = –\(\frac{1}{7}\)(x – 7) y + 5 = –\(\frac{1}{7}\)(x – 7)

Question 12. (3, 3); m = \(\frac{5}{3}\) Answer: y – y1 = m(x – x1) y – (3) = \(\frac{5}{3}\)(x – 3) y – 3 = \(\frac{5}{3}\)(x – 3)

Question 13. (-1, -4); m = -2 Answer: y – y1 = m(x – x1) y – (-4) = -2(x – (-1)) y + 4 = -2(x + 1)

WRITING AN EQUATION Write an equation in slope-intercept form of the line that passes through the given points. Question 14. (-1, -1), (1, 5) Answer: Slope(m) = (5 – (-1))/(2 – (-1)) = (5 + 1)/(1 + 1) m = 6/2 m = 3 y – y1 = m(x – x1) y – (5) = 3(x – (1)) y – 5 = 3x – 3 y = 3x + 2

Question 15. (2, 4), (3, 6) Answer: Slope(m) = (6 – 4)/(3 – 2) m = 2/1 m = 2 y – y1 = m(x – x1) y – (4) = 2(x – (2)) y – 4 = 2x – 4 y = 2x

Question 16. (-2, 3), (2, 7) Answer: Slope(m) = (7 – (3))/(2 – (-2)) = (7 – 3)/(2 + 2) m = 4/4 m = 1 y – y1 = m(x – x1) y – (3) = 1(x – (-2)) y – 3 = x + 2 y = x + 5

Question 17. (4, 1), (8, 2) Answer: Slope(m) = (2 – (1))/(8 – (4)) = (2 – 1)/(8 – 4) m = 1/4 y – y1 = m(x – x1) y – (1) = 1/4(x – (4)) y – 1 = 1/4 x – 1 y = 1/4 x

Question 18. (-9, 5), (-3, 3) Answer: Slope(m) = (3 – (5))/(-3 – (-9)) = (3 – 5)/(-3 + 9) m = -2/6 m = -1/3 y – y1 = m(x – x1) y – (3) = -1/3(x + 3) y – 3 = -1/3 x – 1 y = -1/3 x + 2

Question 19. (1, 2), (-2, -1) Answer: Slope(m) = (2 – (1))/(8 – (4)) = (-1 – 2)/(-2 – 1) m = -3/-3 m = 1 y – y1 = m(x – x1) y – (2) = 1(x – (1)) y – 2 = x – 1 y = x + 1

Question 20. MODELING REAL LIFE At 0° C, the volume of a gas is 22 liters. For each degree the temperature T (in degrees Celsius) increases, the volume V (in liters) of the gas increases by \(\frac{2}{25}\). Write an equation that represents the volume of the gas in terms of the temperature. Answer: The equation modeling the situation has the form: V = mT + b m = 2/25 22 = 2/25(0) + b b = 22 V = 2/25 T + 22

Question 27. REASONING Write an equation of the line that passes through the point (8, 2) and is parallel to the graph of the equation y = 4x – 3. Answer: y = 4x – 3 Comparing the given equation with y = mx + b, we get m = 4 y – y1 = m(x – x1) y – 2 = 4(x – 8) y – 2 = 4x – 32 y = 4x – 32 + 2 y = 4x – 30

Question 29. DIG DEEPER! According to Dolbear’s law, you can predict the temperature T (in degrees Fahrenheit) by counting the number x of chirps made by a snowy tree cricket in 1 minute.When the temperature is 50°F, a cricket chirps 40 times in 1 minute. For each rise in temperature of 0.25°F, the cricket makes an additional chirp each minute. a. You count 100 chirps in 1 minute. What is the temperature? b. The temperature is 96°F.How many chirps do you expect the cricket to make? Justify your answer. Answer:

Question 2. Two supplementary angles have angle measures of x° and y°. Write and graph an equation that represents the relationship between the measures of the angles. Answer:

Chapter Self-Assessment

4.1 Graphing Linear Equations (pp. 141–146) Learning Target: Graph linear equations.Graph the linear equation.

Question 9. Is y = x 2 a linear equation? Explain your reasoning. Answer: y = x 2 The graph of the given equation passes through the origin, but is not linear, therefore it is not a linear equation. So, the answer is no.

4.2 Slope of a Line (pp. 147–154) Learning Target: Find and interpret the slope of a line.

Find the slope of the line through the given points. Question 13. (-5, 4), (8, 4) Answer: (x1, y1) = (-5, 4) (x2, y2) = (8, 4) m = (y2 – y1)/(x2 – x1) m = (4 – 4)/(8 – (-5)) m = 0/13 m = 0

Question 14. (-3, 5), (-3, 1) Answer: (x1, y1) = (-3, 5) (x2, y2) = (-3, 1) m = (y2 – y1)/(x2 – x1) m = (1 – 5)/(-3 + 3) m = -4/0 m = undefined

Question 17. How do you know when two lines are parallel? Use an example to justify your answer. Answer: Two lines are parallel when their slopes are the same. In order for the two lines not to coincide, we must add the condition that their y – intercepts. Example 1: d1: y = 3x – 6 d2: 3x – y = 6 The lines d1 and d2 have the same slope and the same y – intercept, therefore they coincide.

4.3 Graphing Proportional Relationships (pp. 155–160) Learning Target: Graph proportional relationships.

Question 21. The cost y (in dollars) to provide food for guests at a dinner party is proportional to the number x of guests attending the party. It costs $30 to provide food for 4 guests. a. Write an equation that represents the situation. b. Interpret the slope of the graph of the equation. c. How much does it cost to provide food for 10 guests? Justify your answer. Answer: y = kx 30 = 4k k = 30/4 k = 7.5 y = 7.5x b. The slope 7.5 represents the unit cost for a guest. y = 7.5 × 10 y = 75 c. Determine y for x = 10 So it costs $75 to provide food for 10 guests.

Answer: The rate of growth on average is 15/12 = 1.25 cm/month The slope/rate of growth for your friend is (6 – 3)/(4 – 2) = 3/2 = 1.5 cm/month As 1.5 > 1.25, your friends hair grows faster than average.

4.4 Graphing Linear Equations in Slope-Intercept Form (pp. 161–166) Learning Target: Graph linear equations in slope-intercept form.

Find the slope and the -intercept of the graph of the linear equation. Question 24. y = -4x + 1 Answer: y = mx + b slope = -4 and y – intercept = 1

Question 25. y = \(\frac{2}{3}\)x – 12 Answer: y = mx + b slope = \(\frac{2}{3}\) and y – intercept = -12

Question 26. y – 7 = 0.5x Answer: Given the equation y – 7 = 0.5x y = 0.5x + 7 slope = 0.5 and y – intercept = 7

4.5 Graphing Linear Equations in Standard Form (pp. 167–172) Learning Target: Graph linear equations in standard form.

Write the linear equation in slope-intercept form. Question 32. 4x + 2y = -12 Answer: 4x + 2y = -12 2y = -12 – 4x y = -6 – 2x y = -2x – 6

Question 33. x – y = \(\frac{1}{4}\) Answer: Given the equation x – y = \(\frac{1}{4}\) y = x – \(\frac{1}{4}\)

4.6 Writing Equations in Slope-Intercept Form (pp. 173–178) Learning Target: Write equations of lines in slope-intercept form.

Question 43. Write an equation of the line that passes through (0, 8) and (6, 8). Answer: m = (y2 – y1)/(x2 – x1) m = (8 – 8)/(6 – 0) m = 0/6 m = 0 We have to find the y – intercept because y = 8 when x = 0, the y – intercept is 8. y = mx + b y = (0) x + 8 y = 8

Question 44. Write an equation of the line that passes through (0, -5) and (-5, -5). Answer: m = (y2 – y1)/(x2 – x1) m = (-5 – (-5))/(-5 – 0) m = 0/-5 m = 0 We have to find the y – intercept because y = -5 when x = 0, the y – intercept is -5 y = mx + b y = (0) x + (-5) y = -5

Question 45. A construction crew is extending a highway sound barrier that is 13 miles long. The crew builds \(\frac{1}{2}\) of a mile per week. Write an equation in slope -intercept form that represents the length y (in miles) of the barrier after x weeks. Answer: Given, A construction crew is extending a highway sound barrier that is 13 miles long. The crew builds \(\frac{1}{2}\) of a mile per week. y = mx + b m = \(\frac{1}{2}\) b = 13 y = \(\frac{1}{2}\)x + 13

4.7 Writing Equations in Point-Slope Form (pp. 179–184) Learning Target: Write equations of lines in point-slope form.

Write an equation in point-slope form of the line that passes through the given point and has the given slope. Question 46. (4, 4); m = 3 Answer: y – y1 = m(x – x1) Substitute m value, x and y value in the equation y – (4) = 3(x – 4) y – 4 = 3(x – 4)

Question 47. (2, -8); m = –\(\frac{2}{3}\) Answer: y – y1 = m(x – x1) Substitute m value, x and y value in the equation y – (-8) = –\(\frac{2}{3}\)(x – 2) y + 8 = –\(\frac{2}{3}\)(x – 2)

Write an equation in slope-intercept form of the line that passes through the given points. Question 48. (-4, 2), (6, -3) Answer: m = (y2 – y1)/(x2 – x1) m = (-3 – 2)/(6 – (-4)) m = -5/10 m = -1/2 y – y1 = m(x – x1) Substitute m value, x and y value in the equation y – 2 = –\(\frac{1}{2}\)(x – (-4)) y – 2 = –\(\frac{1}{2}\)(x + 4) y = –\(\frac{1}{2}\)x

Answer: m = (600 – 800)/(2 – 1) m = -200 800 = -200(1) + b 800 = -200 + b 800 + 200 = b b = 1000 feet

b. What is your starting elevation?

Answer: The starting elevation is the y – intercept b = 1000 feet

c. After how many minutes do you reach the bottom of the ski slope? Justify your answer. Answer: 0 = -200x + 1000 0 – 1000 = -200x -1000 = -200x 200x = 1000 x = 5 minutes

Question 51. A company offers cable television at$29.95 per month plus a one-time installation fee. The total cost for the first six months of service is $214.70. a. Write an equation in point-slope form that represents the total cost you pay for cable television after x months. b. How much is the installation fee? Justify your answer. Answer: y – y1 = m(x – x1) m = 29.95 y – 214.70 = 29.95(x – 6) y – 214.70 + 214.70 = 29.95x – 179.97 + 2147.70 y = 29.95x + 35 b = 35

Question 52. When might it be better to represent an equation in point-slope form rather than slope-intercept form? Use an example to justify your answer. Answer: When we are given the slope and a point that is the y – intercept, then the easiest way is to use the slope – intercept form y = mx + b Example: m = 2 (0, 5) y = 2x + 5 m = 2 (1, 3) y – 3 = 2(x – 1) Easier when given the slope and a point that is not the y – intercept.

Find the slope and the -intercept of the graph of the linear equation. Question 1. y = 6x – 5 Answer: y = 6x – 5 Slope = 6 and y – intercept = -5

Question 2. y – 1 = 3x + 8.4 Answer: Given the equation y – 1 = 3x + 8.4 y = 3x + 8.4 + 1 y = 3x + 9.4 Slope = 3 and y – intercept = 9.4

Question 3. –\(\frac{1}{2}\)x + 2y = 7 Answer: Given the equation –\(\frac{1}{2}\)x + 2y = 7 y = \(\frac{1}{4}\)x + \(\frac{7}{2}\) Slope = \(\frac{1}{4}\) and y – intercept = \(\frac{7}{2}\)

Question 11. Write an equation in point-slope form of the line that passes through (-4, 1) and (4, 3). Answer: m = (y2 – y1)/(x2 – x1) m = (3 – 1)/(4 – (-4)) m = 2/8 m = 1/4 y – y1 = m(x – x1) y – 1 = 1/4(x – (-4)) y – 1 = 1/4(x + 4)

Question 2. Which point lies on the graph of 6x – 5y = 14? F. (-4, -1) G. (-2, 4) H. (-1, -4) I. (4, -2) Answer: 6x – 5y = 14 F. 6(-4) – 5(-1) = 14 -24 + 5 = 14 -19 ≠ 14 G. 6(-2) – 5(4) = 14 -12 – 20 = 14 -32 ≠ 14 H. 6(-1) – 5(-4) = 14 -6 + 20 = 14 14 = 14 Thus the correct answer is option H.

Question 6. An emergency plumber charges $49.00 plus $70.00 per hour of the repair. A bill to repair your sink is $241.50. This can be modeled by 70.00 h + 49.00 = 241.50, where h represents the number of hours for the repair. How many hours did it take to repair your sink? A. 2.75 hours B. 3.45 hours C. 4.15 hours D. 13,475 hours Answer: 70.00 h + 49.00 = 241.50 70h = 241.50 – 49 70h = 192.5 h = 2.75 hours Thus the correct answer is option A.

Question 10. Solve the formula K = 3M – 7. A. M = K + 7 B. M = \(\frac{K+7}{3}\) C. M = \(\frac{K}{3}\) + 7 D. M = \(\frac{K-7}{3}\) Answer: K = 3M – 7 K + 7 = 3M M = \(\frac{K+7}{3}\) Thus the correct answer is option B.

Conclusion:

All the solutions in the above article are beneficial for all the students of middle school students. All the solutions are prepared by the math professionals. The solutions are given clearly with step by step explanations. If you have any doubts regarding the chapter we are always ready to clarify your doubts. All you have to do is to post the comments in the below comment box.

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Maneuvering the Middle

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Linear Equations Activity Ideas

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Linear equations require lots of practice as the skill progresses to include more steps and increases in complexity. Any of these activities can be used from the basics of simplifying expressions to solving equations with variables on both sides. 

Model by Using Algebra Tiles

Are you surprised I am starting with Algebra tiles? They are foundational for concrete understanding! You can learn more about using Algebra tiles in your classroom by grabbing our free Getting Started with Algebra Tiles guide, which can be found by checking out our post on Solving Equations . This Modeling Equations with Variables on Both Sides activity is a great way to practice solving equations with Algebra tiles. You just need some Algebra tiles (or your students can draw them!).

Student Teacher

Put students into pairs and show an equation on the board. Have one student instruct the other on how to solve as the student listening writes each step and solution. Then, show a new equation and have students switch roles. This gives students a chance to teach and reinforce what they remember about linear equations. I love this activity because it is simple and it makes every student explain their thinking. You, as the teacher, can circulate listening to each pair.

Round Table

Give students individual white boards and have them work in teams of 2-4.  With one equation written on the board, the first person will solve step one.  The second person will complete the second step in solving and the third will complete the next step. Keep rotating until the problem is solved and the last person checks the solution.  Have groups hold up their boards when they are finished.  If you want something like this, we have this Solving Equations with Variables on Both Sides Round Table available! 

Board Races

After students have had time to practice, implement “Board Races.” Two students will come up to the board and race to solve an equation shown on the board. The person who solves it correctly first stays up at the board for the next equation with a new competitor. I like to have the students who aren’t at the board working the equations on notebook paper to help check the solutions. An element of competition makes repetitive practice more fun! For race type activities, I like to have teams compete (boys v. girls or one side of the room v. the other side of the room).

Digital Activities 

I love our digital activities! This linear equations set of digital activities includes simplifying expressions, solving one and two-step equations, solving multi-step equations, and equations with variables on both sides, making it the perfect review before a linear equations test. 

What are some fun ways that students practice solving linear equations in your class? You can check out how to turn any worksheet into an activity which can easily be used for linear equations too!

Getting Started with Algebra Tiles

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