## Integers Worksheet

1Write the following integers in increasing order, represent them graphically, and calculate the additive inverse and absolute values of the numbers:

8, −6, 5, 3, −2, 4, −4, 0, 7

2 Write the follwing integers graphically in increasing order and calculate the additive inverse and absolute values.

−4, 6, −2, 1, 5, 0, 9

3Remove the common factor in the follwing expressions:

1 3 · 2 + 3 · (−5) =

2(−2) · 12 + (−2) · (−6) =

38 · 5 + 8 = 8 · (5 + 1) =

4(−3) · (−2) + (−3) · (−5) =

4 Calculate:

1 (3 − 8) + [5 − (−2)] =

2 5 − [6 − 2 − (1 − 8) − 3 + 6] + 5 =

3 9 : [6 : (− 2)] =

4 [(−2)5 − (−3)3]2 =

5 (5 + 3 · 2 : 6 − 4 ) · (4 : 2 − 3 + 6) : (7 − 8 : 2 − 2)2 =

6 [(17 − 15)3 + (7 − 12)2] : [(6 − 7) · (12 − 23)] =

5 Calculate:

1 (7 − 2 + 4) − (2 − 5) =

2 1 − (5 − 3 + 2) − [5 − (6 − 3 + 1) − 2]=

3 −12 · 3 + 18 : (−12 : 6 + 8) =

6Solve, if it exists:

1

2

3

4

5

6

7 Calculate:

1 (−2)2 · (−2)3 · (−2)4 =

2 (−8) · (−2)2 · (−2)0 (−2) =

3 (−2)−2 · (−2)3 · (−2)4 =

4 2−2 · 2−3 · 24 =

5 22 : 23 =

6 2−2 : 23 =

7 22 : 2−3 =

8 2−2 : 2−3 =

9 [(−2)− 2] 3 · (−2)3 · (−2)4 =

10 [(−2)6 : (−2)3 ]3 · (−2) · (−2)−4 =

8 Calculate:

1(−3)1 · (−3)3 · (−3)4 =

2 (−27) · (−3) · (−3)2 · (−3)0=

3 (−3)2 · (−3)3 · (−3)−4 =

4 3−2 · 3−4 · 34 =

5 52 : 53 =

6 5−2 : 53 =

7 52 : 5 −3 =

8 5−2 : 5−3 =

9 (−3)1 · [(−3)3]2 · (−3)−4 =

10 [(−3)6 : (−3)3] 3 · (−3)0 · (−3)−4 =

## 1

Write the following integers in increasing order, represent them graphically, and calculate the additive inverse and absolute values of the numbers:

8, −6, −5, 3, − 2, 4, −4, 0, 7

− 6 < − 5 < − 4 < − 2 < 0 < 3 < 4 < 7 < 8

op(−6) = −(−6) = 6                             |−6| = 6

op(−5) = −(−5) = 5                             |−5| = 5

op(−4) = −(−4) = 4                             |−4| = 4

op(−2) = −(−2) = 2                             |−2| = 2

op(0) = 0                                      |0| = 0

op(3) = −3                                     |3| = 3

op(4) = −4                                     |4| = 4

op(7) = −7                                     |7| = 7

op(8) = −8                                     |8| = 8

## 2

Write the follwing integers graphically in increasing order and calculate the additive inverse and absolute values.

op(−4) = −(−4) = 4                          |−4| = 4

op(6) = −6                                      |6| = 6

op(−2) = −(−2) = 2                          |−2| = 2

op(1) = − 1                                      |1| = 1

op(−5) = −(−5) = 5                         |−5| = 5

op(0) = 0                                       |0| = 0

op(9) = −9                                     |9| = 9

## 3

Remove the common factor in the following expressions:

1. 3 · 2 + 3 · (−5) =

= 3 · [2 + (−5)] = 3 · (2 − 5) = 3 · (−3) = −9

2. (−2) · 12 + (−2) · (−6) =

= (−2) · [12 + (−6)] = (−2) · (12 − 6) = (−2) · 6 = −12

3.8 · 5 + 8 = 8 · (5 + 1) =

= 8 · 6 = 48

4.(−3) · (−2) + (−3) · (−5) =

= (−3) · [(−2) + (−5)] = (−3) · (−2 − 5) = (− 3) · (−7) = 21

## 4

Calculate:

1 (3 − 8) + [5 − (−2)] = −5 + (5 + 2) = −5 + 7= 2

2 5 − [6 − 2 − (1 − 8) − 3 + 6] + 5 =

= 5 − [6 − 2 − (−7) − 3 + 6] + 5 =

= 5 − [6 − 2 + 7 − 3 + 6] + 5 =

= 5 − 14 + 5 = −4

3 9 : [6 : (−2)] = 9 : (−3) = −3

4 [(−2)5 − (−3)3]2 =

= [− 32 − (−27)] = (−32 + 27)2 =

= (−5)2 = 25

5 (5 + 3 · 2 : 6 − 4 ) · (4 : 2 − 3 + 6) : (7 − 8 : 2 − 2)2 =

= (5 + 6 : 6 − 4 ) · (4 : 2 − 3 + 6) : (7 − 8 : 2 − 2)2 =

= (5 + 1 − 4 ) · (2 − 3 + 6) : (7 − 4 − 2)2 =

= 2 · 5 : 12 =

= 2 · 5 : 1 = 10 : 1 = 10

6 [(17 − 15)3 + (7 − 12)2] : [(6 − 7) · (12 − 23)] =

= [(2)3 + (−5)2] : [(−1) · (−11)] =

= (8 + 25) : [(−1) · (−11)] =

= (8 + 25) : 11 =

= 33: 11 = 3

## 5

Calculate:

1 (7 − 2 + 4) − (2 − 5) = 9 − (−3) = 9 + 3 =12

2 1 − (5 − 3 + 2) − [5 − (6 − 3 + 1) − 2] =

= 1 − (4) − [5 − (4) − 2] =

= 1 − (4) − (5 − 4 − 2)=

= 1 − (4) − (−1) =

= 1 − 4 + 1 = −2

3 −12 · 3 + 18 : (−12 : 6 + 8) =

= −12 · 3 + 18 : (−12 : 6 + 8) =

− 12 · 3 + 18 : (−2 + 8) =

= −12 · 3 + 18 : 6 =

= −36 + 3 = −33

4 2 · [( −12 + 36) : 6 + (8 − 5) : (−3)] − 6 =

= 2 · [24 : 6 +3 : (−3)] − 6 =

= 2 · [ 4 + (−1)] − 6 =

2 · 3 − 6 = 6 − 6 = 0

5 [(−2)5 · (−3)2] : (−2)2 =

(−32 · 9) : 4 = −288 : 4 = −72

66 + {4 − [(17 − (4 · 4)] + 3} − 5 =

= 6 + {4 − [(17 − (4 · 4)] + 3} − 5 =

6 + [4 − (17 − 16) + 3] − 5 =

= 6 + (4 − 1 + 3) − 5 =

6 + 6 − 5 = 7

## 6

Solve, if it exists:

1

2

3

4

5

6

## 7

7 Calculate:

(−2)2 · (−2)3 · (−2)4 = (−2)9 = −512

2 (−8) · (−2)2 · (−2)0 (−2) =

= (−2)3 · (−2)2 · (−2)0 · (−2) = (−2)6 = 64

3 (−2)−2 · (−2)3 · (−2)4 = (−2)5 = −32

4 2−2 · 2−3 · 24 = 2−1 = 1/2

5 22 : 23 = 2−1 = 1/2

6 2−2 : 23 = 2−5 = (1/2)5 = 1/32

7 22 : 2−3 = 25 = 32

8 2−2 : 2−3 = 2

9 [( −2 )− 2] 3 · (−2)3 · (−2)4 =

= (−2)−6 · (−2)3 · (−2)4 = −2

10 [(−2) 6 : (−2)3] 3 · (−2) · (−2)−4 =

[(−2)3] 3 · (−2) · (−2)−4 =

= (−2)9 · (−2) · (−2)−4 = (−2)6 = 64

## 8

Calculate:

1 (−3)1 · (−3)3 · (−3)4 = (−3)8 = 6561

2 (−27) · (−3) · (−3)2 · (−3)0=

(−3)3 · (−3) · (−3)2 · (−3)0 = (−3)6 = 729

3 (−3)2 · (−3)3 · (−3)−4 = −3

4 3−2 · 3−4 · 34 = 3−2 = (1/3)2 = 1/9

5 52 : 53 = 5−1 = 1/5

6 5−2 : 53 = 5−5 = (1/5)5 = 1/3125

7 52 : 5−3 = 55 = 3125

8 5−2 : 5−3 = 5

9 (−3)1 · [(−3)3]2 · (−3)−4 =

(−3)1 · (−3)6· (−3)−4 = (−3)3 = −27

10 [(−3)6 : (−3)3]3 · (−3)0 · (−3)−4 =

[(−3)3]3 · (−3)0· (−3)−4 =

(−3)9 · (−3)0 · (−3)−4 = (−3)5 = −243

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