unit 2 post assessment

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Unit 2 Post Assessment Missed Problems

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Page 1: Unit 2 post assessment

Unit 2 Post Assessment

Missed Problems

Page 2: Unit 2 post assessment

Question 1

Solve for x: 3x = 30

3x = 30

3x/3 = 30/3

x = 10

Multiplication is the operation being performed on the variable (x) so we need to undo multiplication by dividing each side of the equation by 3.

Check your equation!

3(10)

30

Page 3: Unit 2 post assessment

Question 2

Solve for a: -2a = 16

-2a = 16

-2a/-2 = 16/-2

a = -8

Multiplication is the operation being performed on the variable (a) so we need to undo multiplication by dividing each side of the equation by -2.

Check your equation!

-2(-8)

16

Page 4: Unit 2 post assessment

Question 5

Solve for d: d – 12 = 10

d – 12 = 10

d – 12 + 12 = 10 + 12

d = 22

Subtraction is the operation being performed on the variable (d) so we need to undo subtraction by adding each side of the equation by 12

Check your equation!

22 – 12

10

Page 5: Unit 2 post assessment

Question 6

Solve for f: f – 1.5 = 3.5

f – 1.5 = 3.5

f – 1.5 + 1.5 = 3.5 + 1.5

f = 5

Subtraction is the operation being performed on the variable (f) so we need to undo subtraction by adding each side of the equation by 1.5

Check your equation!

5 – 1.5

3.5

Page 6: Unit 2 post assessment

Question 7

Solve for t: 3 + t = 17

3 + t = 17

3 + t – 3 = 17 – 3

t = 14

Addition is the operation being performed on the variable (t) so we need to undo addition by subtracting each side of the equation by 3.

Check your equation!

3 + 14

17

Page 7: Unit 2 post assessment

Question 9

Solve for l: s = l + w

s = l + w

s – w = l + w – w

s – w = l

Addition is the operation being performed between l and w so we need to subtract w to the other side to get l alone.

Page 8: Unit 2 post assessment

Question 11Tenisha is twice as tall as her little brother. If Tenisha is 60 inches

tall, how old is her little brother?

Brother’s height is ½ of Tenisha’s

60/2 is 30 so Tenisha’s little brother is 30 inches tall.

Page 9: Unit 2 post assessment

Question 12

You and Michael have a total of $19.75. If Michael has $8.25, how much money do you have.

You + Michael = $19.75

You + $8.25 = $19.75

Subtract $8.25 on each side.

You = $11.50

Page 10: Unit 2 post assessment

Question 13

Solve for x: 7x – 10 = 10

7x – 10 = 10

7x – 10 + 10 = 10 + 10

7x = 20

7x/7 = 20/7

x = 20/7

This is a two-step equation. We are multiplying 7 to the variable and subtraction 10 so we need to undo these two operations.

Step 1: Undo subtraction by adding

Add 10 to each side

Step 2: Undo multiplication by dividing

Divide by 7 on each side.

Page 11: Unit 2 post assessment

Question 14

At the grocery store you purchase 2 gallons of milk and loaf of bread. You spent a total of $8. If you know a loaf of bread costs $2, how much does a gallon of milk cost?

2x + 2 = 8

2x = 6

x = 3

This is a two-step equation. We are multiplying 2 to the variable and adding 2 so we need to undo these two operations.

Step 1: Undo addition by subtracting 2 on each side

Step 2: Undo multiplication by dividing by 2 on each side.

Page 12: Unit 2 post assessment

Question 16

Solve for a: 4a + a – 11 = 24

4a + a – 11 = 24

5a – 11 = 24

5a = 35

a = 7

Multiple Step Equation

1) Combine like terms

4a + a = 5a

2) Add 11 to each side.

3) Divide by 5 on each side.

Page 13: Unit 2 post assessment

Question 17

Solve for y: 4x – 8y = 16

4x – 8y = 16

-8y = -4x + 16

y = ½ x – 2

Multiple Step Equation

Subtract 4x on each side.

Divide by -8 on each side.

Page 14: Unit 2 post assessment

Question 18

Solve for p: 6p + 2(p – 1) = 22

6p + 2(p – 1) = 22

6p + 2p – 2 = 22

8p – 2 = 22

8p = 24

p = 3

Multiple Step Equation

1) Distribute 2

2) Combine like terms 6p + 2p

3) Add 2 to each side

4) Divide by 8 on each side.

Page 15: Unit 2 post assessment

Question 19

Solve for r: 5 + 7r + 3r = 8(r – 4)

5 + 7r + 3r = 8(r – 4)

5 + 7r + 3r = 8r – 32

5 + 10r = 8r – 32

5 + 2r = -32

2r = -37

r = -37/2

Multiple Step Equation

1) Distribute 8

2) Combine like terms 7r + 3r

3) Subtract 8r to each side

4) Subtract 5 on each side.

Page 16: Unit 2 post assessment

Question 20

Solve for n: 3(n -2) + 1 = n + 4 + 2n

3(n – 2) + 1 = n + 4 + 2n

3n – 6 + 1 = n + 4 + 2n

3n – 5 = 3n + 4

-5 = 4

No Solution

Multiple Step Equation

1) Distribute 3

2) Combine like terms

3) Subtract 3n on each side

-5 does not equal 4 so this equation does not have a solution!