IM 6.6.3

Select all the equations that represent the hanger.

• x + x + x + x = 1 + 1 + 1 + 1 + 1 + 1

• x • x • x • x = 6

• 3x = 6

• x + 3 = 6

• x • x • x • x = 1 • 1 • 1 • 1 • 1 • 1

Tags

CCSS.HSA.CED.A.3

Write an equation to represent hanger A.

• 2x = 3

• x + x = 1 + 1 + 1

• x = 1.5

• x + 2 = 3

Tags

CCSS.HSA.CED.A.3

Write an equation to represent hanger B.

• w + 1.3 = 2.7

• w + w = 1.3

• 2w = 1.3

• w = 2.6

Tags

CCSS.HSA.CED.A.3

Write an equation to represent hanger C.

• y + y + y= 5.1

• 3y = 5.1

• y = 5.1

• y = 15.3

Tags

CCSS.HSA.CED.A.3

Write an equation to represent hanger D.

• z = 1 + 1/2 + 1/3

• z = 1.833

• z + 1/3 = 1/2

• z = 1/2 + 1/3

Tags

CCSS.HSA.CED.A.3

Write an equation to represent the hanger shown in the image.

• x + x = 14.62

• 2x = 14.62

• x - x = 14.62

• x / 2 = 14.62

Tags

CCSS.HSA.CED.A.3

Do you agree with Andre's statement that x is 7 because he can move the two 1s with the x to the other side?

• Yes, because moving the 1s to the other side does not change the value of x.

• No, because you cant move 1s to the other side but you can cancel them out so x is 3

• Yes, because the total weight on both sides of the hanger must be equal.

• No, because the value of x is not dependent on the position of the 1s.

Tags

CCSS.6.EE.B.6

Match the equation 12 - m = 4 to one of the diagrams.

• Diagram A

• Diagram B

• Diagram C

• Diagram D

Tags

CCSS.HSA.CED.A.3

Match the equation 12 = 4 · m to one of the diagrams.

• Diagram A

• Diagram B

• Diagram C

• Diagram D

Match the equation m - 4 = 12 to one of the diagrams.

• Diagram A

• Diagram B

• Diagram C

• Diagram D

Tags

CCSS.8.EE.C.7B