lesson 4 problem solving practice answer key

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Eureka Math Grade 8 Module 4 Lesson 4 Answer Key

Engage ny eureka math 8th grade module 4 lesson 4 answer key, eureka math grade 8 module 4 lesson 4 exercise answer key.

For each problem, show your work, and check that your solution is correct.

Exercise 1. Solve the linear equation x+x+2+x+4+x+6=-28. State the property that justifies your first step and why you chose it. Answer: The left side of the equation can be transformed from x+x+2+x+4+x+6 to 4x+12 using the commutative and distributive properties. Using these properties decreases the number of terms of the equation. Now we have the equation: 4x+12=-28 4x+12-12=-28-12 4x=-40 \(\frac{1}{4}\)∙4x=-40∙\(\frac{1}{4}\) x=-10. The left side of the equation is equal to (-10)+(-10)+2+(-10)+4+(-10)+6, which is -28. Since the left side is equal to the right side, then x=-10 is the solution to the equation. Note: Students could use the division property in the last step to get the answer.

Exercise 2. Solve the linear equation 2(3x+2)=2x-1+x. State the property that justifies your first step and why you chose it. Answer: Both sides of equation can be rewritten using the distributive property. I have to use it on the left side to expand the expression. I have to use it on the right side to collect like terms. The left side is 2(3x+2)=6x+4. The right side is 2x-1+x=2x+x-1 =3x-1. The equation is 6x+4=3x-1 6x+4-4=3x-1-4 6x=3x-5 6x-3x=3x-3x-5 (6-3)x=(3-3)x-5 3x=-5 \(\frac{1}{3}\)∙3x=\(\frac{1}{3}\)∙(-5) x=-\(\frac{5}{3}\). The left side of the equation is 2(3x+2). Replacing x with –\(\frac{5}{3}\) gives 2(3(-\(\frac{5}{3}\))+2)=2(-5+2)=2(-3)=-6. The right side of the equation is 2x-1+x. Replacing x with –\(\frac{5}{3}\) gives 2(-\(\frac{5}{3}\))-1+(-\(\frac{5}{3}\))=-\(\frac{10}{3}\)-1-\(\frac{5}{3}\)=-6. Since both sides are equal to -6, then x=-\(\frac{5}{3}\) is a solution to 2(3x+2)=2x-1+x. Note: Students could use the division property in the last step to get the answer.

Exercise 3. Solve the linear equation x-9=\(\frac{3}{5}\) x. State the property that justifies your first step and why you chose it. Answer; I chose to use the subtraction property of equality to get all terms with an x on one side of the equal sign. x-9=\(\frac{3}{5}\) x x-x-9=\(\frac{3}{5}\) x-x (1-1)x-9=(\(\frac{3}{5}\)-1)x -9=-\(\frac{2}{5}\) x –\(\frac{5}{2}\)∙(-9)=-\(\frac{5}{2}\)∙-\(\frac{2}{5}\) x 4\(\frac{5}{2}\)=x The left side of the equation is 4\(\frac{5}{2}\)– \(\frac{18}{2}\)=\(\frac{27}{2}\). The right side is \(\frac{3}{5}\)∙4\(\frac{5}{2}\)=\(\frac{3}{1}\)∙\(\frac{9}{2}\)=\(\frac{27}{2}\). Since both sides are equal to the same number, then x=4\(\frac{5}{2}\) is a solution to x-9=\(\frac{3}{5}\) x.

Exercise 4. Solve the linear equation 29-3x=5x+5. State the property that justifies your first step and why you chose it. Answer: I chose to use the addition property of equality to get all terms with an x on one side of the equal sign. 29-3x=5x+5 29-3x+3x=5x+3x+5 29=8x+5 29-5=8x+5-5 24=8x \(\frac{1}{8}\)∙24=\(\frac{1}{8}\)∙8x 3=x The left side of the equal sign is 29-3(3)=29-9=20. The right side is equal to 5(3)+5=15+5=20. Since both sides are equal, x=3 is a solution to 29-3x=5x+5. Note: Students could use the division property in the last step to get the answer.

Exercise 5. Solve the linear equation \(\frac{1}{3}\) x-5+171=x. State the property that justifies your first step and why you chose it. Answer: I chose to combine the constants -5 and 171. Then, I used the subtraction property of equality to get all terms with an x on one side of the equal sign. \(\frac{1}{3}\) x-5+171=x \(\frac{1}{3}\) x+166=x \(\frac{1}{3}\) x-\(\frac{1}{3}\) x+166=x-\(\frac{1}{3}\) x 166=\(\frac{2}{3}\) x 166∙\(\frac{3}{2}\)=\(\frac{3}{2}\)∙\(\frac{2}{3}\) x 83∙3=x 249=x The left side of the equation is \(\frac{1}{3}\)∙249-5+171=83-5+171=78+171=249, which is exactly equal to the right side. Therefore, x=249 is a solution to \(\frac{1}{3}\) x-5+171=x.

Eureka Math Grade 8 Module 4 Lesson 4 Exit Ticket Answer Key

Question 6. Guess a number for x that would make the equation true. Check your solution. 5x-2=8 Answer: When x=2, the left side of the equation is 8, which is the same as the right side. Therefore, x=2 is the solution to the equation.

Question 7. Use the properties of equality to solve the equation 7x-4+x=12. State which property justifies your first step and why you chose it. Check your solution. Answer: I used the commutative and distributive properties on the left side of the equal sign to simplify the expression to fewer terms. 7x-4+x=12 7x+x-4=12 (7+1)x-4=12 8x-4=12 8x-4+4=12+4 8x=16 \(\frac{8}{8}\)x=\(\frac{16}{8}\) x=2 The left side of the equation is 7(2)-4+2=14-4+2=12. The right side is also 12. Since the left side equals the right side, x=2 is the solution to the equation.

Question 8. Use the properties of equality to solve the equation 3x+2-x=11x+9. State which property justifies your first step and why you chose it. Check your solution. Answer: I used the commutative and distributive properties on the left side of the equal sign to simplify the expression to fewer terms. 3x+2-x=11x+9 3x-x+2=11x+9 (3-1)x+2=11x+9 2x+2=11x+9 2x-2x+2=11x-2x+9 (2-2)x+2=(11-2)x+9 2=9x+9 2-9=9x+9-9 -7=9x –\(\frac{7}{9}\)=\(\frac{9}{9}\) x –\(\frac{7}{9}\)=x The left side of the equation is 3(\(\frac{-7}{9}\))+2-\(\frac{-7}{9}\)=-\(\frac{21}{9}\)+\(\frac{18}{9}\)+\(\frac{7}{9}\)=\(\frac{4}{9}\). The right side is 11(-\(\frac{7}{9}\))+9=\(\frac{-77}{9}\)+\(\frac{81}{9}\)=\(\frac{4}{9}\). Since the left side equals the right side, x=-\(\frac{7}{9}\) is the solution to the equation.

Eureka Math Grade 8 Module 4 Lesson 4 Problem Set Answer Key

Students solve equations using properties of equality.

Question 1. Solve the linear equation x+4+3x=72. State the property that justifies your first step and why you chose it. Answer: I used the commutative and distributive properties on the left side of the equal sign to simplify the expression to fewer terms. x+4+3x=72 x+3x+4=72 (1+3)x+4=72 4x+4=72 4x+4-4=72-4 4x=68 \(\frac{4}{4}\) x=\(\frac{68}{4}\) x=17 The left side is equal to 17+4+3(17)=21+51=72, which is equal to the right side. Therefore, x=17 is a solution to the equation x+4+3x=72.

Question 2. Solve the linear equation x+3+x-8+x=55. State the property that justifies your first step and why you chose it. Answer: I used the commutative and distributive properties on the left side of the equal sign to simplify the expression to fewer terms. x+3+x-8+x=55 x+x+x+3-8=55 (1+1+1)x+3-8=55 3x-5=55 3x-5+5=55+5 3x=60 \(\frac{3}{3}\) x=\(\frac{60}{3}\) x=20 The left side is equal to 20+3+20-8+20=43-8+20=35+20=55, which is equal to the right side. Therefore, x=20 is a solution to x+3+x-8+x=55.

Question 3. Solve the linear equation \(\frac{1}{2}\) x+10=\(\frac{1}{4}\) x+54. State the property that justifies your first step and why you chose it. Answer; I chose to use the subtraction property of equality to get all of the constants on one side of the equal sign. \(\frac{1}{2}\) x+10=\(\frac{1}{4}\) x+54 \(\frac{1}{2}\) x+10-10=\(\frac{1}{4}\) x+54-10 \(\frac{1}{2}\) x=\(\frac{1}{4}\) x+44 \(\frac{1}{2}\) x-\(\frac{1}{4}\) x=\(\frac{1}{4}\) x-\(\frac{1}{4}\) x+44 \(\frac{1}{4}\) x=44 4∙\(\frac{1}{4}\) x=4∙44 x=176 The left side of the equation is \(\frac{1}{2}\) (176)+10=88+10=98. The right side of the equation is \(\frac{1}{4}\) (176)+54=44+54=98. Since both sides equal 98, x=176 is a solution to the equation \(\frac{1}{2}\) x+10=\(\frac{1}{4}\) x+54.

Question 4. Solve the linear equation \(\frac{1}{4}\) x+18=x. State the property that justifies your first step and why you chose it. I chose to use the subtraction property of equality to get all terms with an x on one side of the equal sign. \(\frac{1}{4}\) x+18=x \(\frac{1}{4}\) x-\(\frac{1}{4}\) x+18=x-\(\frac{1}{4}\) x 18=\(\frac{3}{4}\) x \(\frac{4}{3}\)∙18=\(\frac{4}{3}\)∙\(\frac{3}{4}\) x 24=x The left side of the equation is \(\frac{1}{4}\) (24)+18=6+18=24, which is what the right side is equal to. Therefore, x=24 is a solution to \(\frac{1}{4}\) x+18=x.

Question 5. Solve the linear equation 17-x=\(\frac{1}{3}\)∙15+6. State the property that justifies your first step and why you chose it. Answer: The right side of the equation can be simplified to 11. Then, the equation is 17-x=11, and x=6. Both sides of the equation equal 11; therefore, x=6 is a solution to the equation 17-x=\(\frac{1}{3}\)∙15+6. I was able to solve the equation mentally without using the properties of equality.

Question 6. Solve the linear equation \(\frac{x+x+2}{4}\)=189.5. State the property that justifies your first step and why you chose it. Answer: I chose to use the multiplication property of equality to get all terms with an x on one side of the equal sign. \(\frac{x+x+2}{4}\)=189.5 x+x+2=4(189.5) 2x+2=758 2x+2-2=758-2 2x=756 \(\frac{2}{2}\) x=\(\frac{756}{2}\) x=378 The left side of the equation is \(\frac{378+378+2}{4}\)=\(\frac{758}{4}\)=189.5, which is equal to the right side of the equation. Therefore, x=378 is a solution to \(\frac{x+x+2}{4}\)=189.5.

Question 7. Alysha solved the linear equation 2x-3-8x=14+2x-1. Her work is shown below. When she checked her answer, the left side of the equation did not equal the right side. Find and explain Alysha’s error, and then solve the equation correctly. 2x-3-8x=14+2x-1 -6x-3=13+2x -6x-3+3=13+3+2x -6x=16+2x -6x+2x=16 -4x=16 \(\frac{-4}{-4}\) x=\(\frac{16}{-4}\) x=-4 Answer: Alysha made a mistake on the fifth line. She added 2x to the left side of the equal sign and subtracted 2x on the right side of the equal sign. To use the property correctly, she should have subtracted 2x on both sides of the equal sign, making the equation at that point: -6x-2x=16+2x-2x -8x=16 \(\frac{-8}{-8}\) x=\(\frac{16}{-8}\) x=-2.

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A: ( 5 , 1 ) ( 5 , 1 )  B: ( −2 , 4 ) ( −2 , 4 )  C: ( −5 , −1 ) ( −5 , −1 )  D: ( 3 , −2 ) ( 3 , −2 )  E: ( 0 , −5 ) ( 0 , −5 )  F: ( 4 , 0 ) ( 4 , 0 )

A: ( 4 , 2 ) ( 4 , 2 )  B: ( −2 , 3 ) ( −2 , 3 )  C: ( −4 , −4 ) ( −4 , −4 )  D: ( 3 , −5 ) ( 3 , −5 )  E: ( −3 , 0 ) ( −3 , 0 )  F: ( 0 , 2 ) ( 0 , 2 )

Answers will vary.

ⓐ yes, yes  ⓑ yes, yes

ⓐ no, no  ⓑ yes, yes

x - intercept: ( 2 , 0 ) ( 2 , 0 ) ; y - intercept: ( 0 , −2 ) ( 0 , −2 )

x - intercept: ( 3 , 0 ) ( 3 , 0 ) , y - intercept: ( 0 , 2 ) ( 0 , 2 )

x - intercept: ( 4 , 0 ) ( 4 , 0 ) , y - intercept: ( 0 , 12 ) ( 0 , 12 )

x - intercept: ( 8 , 0 ) ( 8 , 0 ) , y - intercept: ( 0 , 2 ) ( 0 , 2 )

x - intercept: ( 4 , 0 ) ( 4 , 0 ) , y - intercept: ( 0 , −3 ) ( 0 , −3 )

x - intercept: ( 4 , 0 ) ( 4 , 0 ) , y - intercept: ( 0 , −2 ) ( 0 , −2 )

− 2 3 − 2 3

− 4 3 − 4 3

− 3 5 − 3 5

− 1 36 − 1 36

− 1 48 − 1 48

slope m = 2 3 m = 2 3 and y -intercept ( 0 , −1 ) ( 0 , −1 )

slope m = 1 2 m = 1 2 and y -intercept ( 0 , 3 ) ( 0 , 3 )

2 5 ; ( 0 , −1 ) 2 5 ; ( 0 , −1 )

− 4 3 ; ( 0 , 1 ) − 4 3 ; ( 0 , 1 )

− 1 4 ; ( 0 , 2 ) − 1 4 ; ( 0 , 2 )

− 3 2 ; ( 0 , 6 ) − 3 2 ; ( 0 , 6 )

ⓐ intercepts  ⓑ horizontal line  ⓒ slope–intercept  ⓓ vertical line

ⓐ vertical line  ⓑ slope–intercept  ⓒ horizontal line  ⓓ intercepts

• ⓐ 50 inches
• ⓑ 66 inches
• ⓒ The slope, 2, means that the height, h , increases by 2 inches when the shoe size, s , increases by 1. The h -intercept means that when the shoe size is 0, the height is 50 inches.
• ⓐ 40 degrees
• ⓑ 65 degrees
• ⓒ The slope, 1 4 1 4 , means that the temperature Fahrenheit ( F ) increases 1 degree when the number of chirps, n , increases by 4. The T -intercept means that when the number of chirps is 0, the temperature is 40 ° 40 ° .
• ⓒ The slope, 0.5, means that the weekly cost, C , increases by $0.50 when the number of miles driven, n, increases by 1. The C -intercept means that when the number of miles driven is 0, the weekly cost is $60
• ⓒ The slope, 1.8, means that the weekly cost, C, increases by $1.80 when the number of invitations, n , increases by 1.80. The C -intercept means that when the number of invitations is 0, the weekly cost is $35.;

not parallel; same line

perpendicular

not perpendicular

y = 2 5 x + 4 y = 2 5 x + 4

y = − x − 3 y = − x − 3

y = 3 5 x + 1 y = 3 5 x + 1

y = 4 3 x − 5 y = 4 3 x − 5

y = 5 6 x − 2 y = 5 6 x − 2

y = 2 3 x − 4 y = 2 3 x − 4

y = − 2 5 x − 1 y = − 2 5 x − 1

y = − 3 4 x − 4 y = − 3 4 x − 4

y = 8 y = 8

y = 4 y = 4

y = 5 2 x − 13 2 y = 5 2 x − 13 2

y = − 2 5 x + 22 5 y = − 2 5 x + 22 5

y = 1 3 x − 10 3 y = 1 3 x − 10 3

y = − 2 5 x − 23 5 y = − 2 5 x − 23 5

x = 5 x = 5

x = −4 x = −4

y = 3 x − 10 y = 3 x − 10

y = 1 2 x + 1 y = 1 2 x + 1

y = − 1 3 x + 10 3 y = − 1 3 x + 10 3

y = −2 x + 16 y = −2 x + 16

y = −5 y = −5

y = −1 y = −1

x = −5 x = −5

ⓐ yes  ⓑ yes  ⓒ yes  ⓓ yes  ⓔ no

ⓐ yes  ⓑ yes  ⓒ no  ⓓ no  ⓔ yes

y ≥ −2 x + 3 y ≥ −2 x + 3

y < 1 2 x − 4 y < 1 2 x − 4

x − 4 y ≤ 8 x − 4 y ≤ 8

3 x − y ≤ 6 3 x − y ≤ 6

Section 4.1 Exercises

A: ( −4 , 1 ) ( −4 , 1 )  B: ( −3 , −4 ) ( −3 , −4 )  C: ( 1 , −3 ) ( 1 , −3 )  D: ( 4 , 3 ) ( 4 , 3 )

A: ( 0 , −2 ) ( 0 , −2 )  B: ( −2 , 0 ) ( −2 , 0 )  C: ( 0 , 5 ) ( 0 , 5 )  D: ( 5 , 0 ) ( 5 , 0 )

ⓑ Age and weight are only positive.

Section 4.2 Exercises

ⓐ yes; no  ⓑ no; no  ⓒ yes; yes  ⓓ yes; yes

ⓐ yes; yes  ⓑ yes; yes  ⓒ yes; yes  ⓓ no; no

$722, $850, $978

Section 4.3 Exercises

( 3 , 0 ) , ( 0 , 3 ) ( 3 , 0 ) , ( 0 , 3 )

( 5 , 0 ) , ( 0 , −5 ) ( 5 , 0 ) , ( 0 , −5 )

( −2 , 0 ) , ( 0 , −2 ) ( −2 , 0 ) , ( 0 , −2 )

( −1 , 0 ) , ( 0 , 1 ) ( −1 , 0 ) , ( 0 , 1 )

( 6 , 0 ) , ( 0 , 3 ) ( 6 , 0 ) , ( 0 , 3 )

( 0 , 0 ) ( 0 , 0 )

( 4 , 0 ) , ( 0 , 4 ) ( 4 , 0 ) , ( 0 , 4 )

( −3 , 0 ) , ( 0 , 3 ) ( −3 , 0 ) , ( 0 , 3 )

( 8 , 0 ) , ( 0 , 4 ) ( 8 , 0 ) , ( 0 , 4 )

( 2 , 0 ) , ( 0 , 6 ) ( 2 , 0 ) , ( 0 , 6 )

( 12 , 0 ) , ( 0 , −4 ) ( 12 , 0 ) , ( 0 , −4 )

( 2 , 0 ) , ( 0 , −8 ) ( 2 , 0 ) , ( 0 , −8 )

( 5 , 0 ) , ( 0 , 2 ) ( 5 , 0 ) , ( 0 , 2 )

( 4 , 0 ) , ( 0 , −6 ) ( 4 , 0 ) , ( 0 , −6 )

( 3 , 0 ) , ( 0 , 1 ) ( 3 , 0 ) , ( 0 , 1 )

( −10 , 0 ) , ( 0 , 2 ) ( −10 , 0 ) , ( 0 , 2 )

ⓐ ( 0 , 1000 ) , ( 15 , 0 ) ( 0 , 1000 ) , ( 15 , 0 ) ⓑ At ( 0 , 1000 ) ( 0 , 1000 ) , he has been gone 0 hours and has 1000 miles left. At ( 15 , 0 ) ( 15 , 0 ) , he has been gone 15 hours and has 0 miles left to go.

Section 4.4 Exercises

−3 2 = − 3 2 −3 2 = − 3 2

− 1 3 − 1 3

− 3 4 − 3 4

− 5 2 − 5 2

− 8 7 − 8 7

ⓐ 1 3 1 3   ⓑ 4 12 pitch or 4-in-12 pitch

3 50 3 50 ; rise = 3, run = 50

ⓐ 288 inches (24 feet)  ⓑ Models will vary.

When the slope is a positive number the line goes up from left to right. When the slope is a negative number the line goes down from left to right.

A vertical line has 0 run and since division by 0 is undefined the slope is undefined.

Section 4.5 Exercises

slope m = 4 m = 4 and y -intercept ( 0 , −2 ) ( 0 , −2 )

slope m = −3 m = −3 and y -intercept ( 0 , 1 ) ( 0 , 1 )

slope m = − 2 5 m = − 2 5 and y -intercept ( 0 , 3 ) ( 0 , 3 )

−9 ; ( 0 , 7 ) −9 ; ( 0 , 7 )

4 ; ( 0 , −10 ) 4 ; ( 0 , −10 )

−4 ; ( 0 , 8 ) −4 ; ( 0 , 8 )

− 8 3 ; ( 0 , 4 ) − 8 3 ; ( 0 , 4 )

7 3 ; ( 0 , −3 ) 7 3 ; ( 0 , −3 )

horizontal line

vertical line

slope–intercept

• ⓒ The slope, 2.54, means that Randy’s payment, P , increases by $2.54 when the number of units of water he used, w, increases by 1. The P –intercept means that if the number units of water Randy used was 0, the payment would be $28.
• ⓒ The slope, 0.32, means that the cost, C , increases by $0.32 when the number of miles driven, m, increases by 1. The C -intercept means that if Janelle drives 0 miles one day, the cost would be $15.
• ⓒ The slope, 0.09, means that Patel’s salary, S , increases by $0.09 for every $1 increase in his sales. The S -intercept means that when his sales are $0, his salary is $750.
• ⓒ The slope, 42, means that the cost, C , increases by $42 for when the number of guests increases by 1. The C -intercept means that when the number of guests is 0, the cost would be $750.

not parallel

• ⓐ For every increase of one degree Fahrenheit, the number of chirps increases by four.
• ⓑ There would be −160 −160 chirps when the Fahrenheit temperature is 0 ° 0 ° . (Notice that this does not make sense; this model cannot be used for all possible temperatures.)

Section 4.6 Exercises

y = 4 x + 1 y = 4 x + 1

y = 8 x − 6 y = 8 x − 6

y = − x + 7 y = − x + 7

y = −3 x − 1 y = −3 x − 1

y = 1 5 x − 5 y = 1 5 x − 5

y = − 2 3 x − 3 y = − 2 3 x − 3

y = 2 y = 2

y = −4 x y = −4 x

y = −2 x + 4 y = −2 x + 4

y = 3 4 x + 2 y = 3 4 x + 2

y = − 3 2 x − 1 y = − 3 2 x − 1

y = 6 y = 6

y = 3 8 x − 1 y = 3 8 x − 1

y = 5 6 x + 2 y = 5 6 x + 2

y = − 3 5 x + 1 y = − 3 5 x + 1

y = − 1 3 x − 11 y = − 1 3 x − 11

y = −7 y = −7

y = − 5 2 x − 22 y = − 5 2 x − 22

y = −4 x − 11 y = −4 x − 11

y = −8 y = −8

y = −4 x + 13 y = −4 x + 13

y = x + 5 y = x + 5

y = − 1 3 x − 14 3 y = − 1 3 x − 14 3

y = 7 x + 22 y = 7 x + 22

y = − 6 7 x + 4 7 y = − 6 7 x + 4 7

y = 1 5 x − 2 y = 1 5 x − 2

x = 4 x = 4

x = −2 x = −2

y = −3 y = −3

y = 4 x y = 4 x

y = 1 2 x + 3 2 y = 1 2 x + 3 2

y = 5 y = 5

y = 3 x − 1 y = 3 x − 1

y = −3 x + 3 y = −3 x + 3

y = 2 x − 6 y = 2 x − 6

y = − 2 3 x + 5 y = − 2 3 x + 5

x = −3 x = −3

y = −4 y = −4

y = x y = x

y = − 3 4 x − 1 4 y = − 3 4 x − 1 4

y = 5 4 x y = 5 4 x

y = 1 y = 1

y = x + 2 y = x + 2

y = 3 4 x y = 3 4 x

y = 1.2 x + 5.2 y = 1.2 x + 5.2

Section 4.7 Exercises

ⓐ yes  ⓑ no  ⓒ no  ⓓ yes  ⓔ no

ⓐ yes  ⓑ no  ⓒ no  ⓓ yes  ⓔ yes

ⓐ no  ⓑ no  ⓒ no  ⓓ yes  ⓔ yes

y < 2 x − 4 y < 2 x − 4

y ≤ − 1 3 x − 2 y ≤ − 1 3 x − 2

x + y ≥ 3 x + y ≥ 3

x + 2 y ≥ −2 x + 2 y ≥ −2

2 x − y < 4 2 x − y < 4

4 x − 3 y > 12 4 x − 3 y > 12

• ⓑ Answers will vary.

Review Exercises

ⓐ ( 2 , 0 ) ( 2 , 0 )   ⓑ ( 0 , −5 ) ( 0 , −5 )   ⓒ ( −4.0 ) ( −4.0 )   ⓓ ( 0 , 3 ) ( 0 , 3 )

ⓐ yes; yes  ⓑ yes; no

( 6 , 0 ) , ( 0 , 4 ) ( 6 , 0 ) , ( 0 , 4 )

− 1 2 − 1 2

slope m = − 2 3 m = − 2 3 and y -intercept ( 0 , 4 ) ( 0 , 4 )

5 3 ; ( 0 , −6 ) 5 3 ; ( 0 , −6 )

4 5 ; ( 0 , − 8 5 ) 4 5 ; ( 0 , − 8 5 )

plotting points

ⓐ −$250  ⓑ $450  ⓒ The slope, 35, means that Marjorie’s weekly profit, P , increases by $35 for each additional student lesson she teaches. The P –intercept means that when the number of lessons is 0, Marjorie loses $250.  ⓓ

y = −5 x − 3 y = −5 x − 3

y = −2 x y = −2 x

y = −3 x + 5 y = −3 x + 5

y = 3 5 x y = 3 5 x

y = −2 x − 5 y = −2 x − 5

y = 1 2 x − 5 2 y = 1 2 x − 5 2

y = − 2 5 x + 8 y = − 2 5 x + 8

y = 3 y = 3

y = − 3 2 x − 6 y = − 3 2 x − 6

ⓐ yes  ⓑ no  ⓒ yes  ⓓ yes  ⓔ no

y > 2 3 x − 3 y > 2 3 x − 3

x − 2 y ≥ 6 x − 2 y ≥ 6

Practice Test

ⓐ yes  ⓑ yes  ⓒ no

( 3 , 0 ) , ( 0 , −4 ) ( 3 , 0 ) , ( 0 , −4 )

y = − 3 4 x − 2 y = − 3 4 x − 2

y = 1 2 x − 4 y = 1 2 x − 4

y = − 4 5 x − 5 y = − 4 5 x − 5

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• Book title: Elementary Algebra
• Publication date: Feb 22, 2017
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• Section URL: https://openstax.org/books/elementary-algebra/pages/chapter-4

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• Kindergarten
• Greater Than Less Than
• Measurement
• Multiplication
• Place Value
• Subtraction
• Punctuation
• 1st Grade Reading
• 2nd Grade Reading
• 3rd Grade Reading
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Lesson 4 Problem Solving Practice Answer Key

Lesson 4 Problem Solving Practice Answer Key - Displaying top 8 worksheets found for this concept.

Some of the worksheets for this concept are Problem solution grade 4 lesson, Abeged mathematics activities student work, Solving quadratic inequalities algebraically work, Geometry lesson 10 7 practice a answers, Lesson 6 problem solving practice answers, Ready mathematics practice and problem solving teacher, Subtracting integers, Lesson applying gcf and lcm to fraction operations 4 1.

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1. Problem Solution Grade 4 Lesson -

2. abe/ged mathematics activities & student worksheets, 3. solving quadratic inequalities algebraically worksheet ..., 4. geometry lesson 10 7 practice a answers, 5. lesson 6 problem solving practice answers, 6. ready mathematics practice and problem solving teacher ..., 7. subtracting integers, 8. lesson applying gcf and lcm to fraction operations 4-1 ....

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