INVERSE OPERATIONS WORKSHEET

Problem 1 :

Solve for k : 

7k  =  35

Solution :

Here 7 and k are multiplied. 

So, we have to use the inverse operation of multiplication to solve for k. 

Inverse operation of multiplication is division.

7k  =  35

Divide both sides by 7. 

k = 5

Problem 2 :

Solve for k ;

k + 7  =  9

Solution :

Here 7 is added to k. 

So, we have to use the inverse operation of addition to solve for k.

Inverse operation of addition is subtraction.

k + 7  =  9

Subtract 7 from each side. 

k  =  2

Problem 3 :

Solve for b : 

b/8  =  7

Solution :

Here b is divided by 8. 

So, we have to use the inverse operation of division to solve for b.

Inverse operation of division is multiplication.

b/8  =  7

Multiply each side by 8.

b  =  56

Problem 4 :

Solve for m : 

m - 15  =  9

Solution :

Here 15 is subtracted from m. 

So, we have to use the inverse operation of subtraction to solve for b.

Inverse operation of subtraction is addition.

m - 15  =  9

Add 15 to each side. 

 m  =  24

Problem 5 :

Solve for z : 

z + 17  =  23

Solution :

Here 17 is added with z. 

So, we have to use the inverse operation of addition to solve for k.

Inverse operation of addition is subtraction.

z + 17  =  23

Subtract 17 from each side.

z  =  6

Problem 6 :

Solve for x : 

2x + 5  =  23 

Solution :

Here 5 is added to 2x. 

So, we have to use the inverse operation of addition to solve for 2x.

Inverse operation of addition is subtraction.

2x + 5  =  23

Subtract 5 from each side. 

 2x  =  18 

Here 2 and x are multiplied.

Then, we have to use the inverse operation of multiplication to solve for x.

Inverse operation of multiplication is division. 

2x  =  18

Divide each side by 2.

x  =  9

Problem 7 :

Solve for p :

2p - 7  =  3 

Solution :

Here 7 is subtracted from p. 

So, we have to use the inverse operation of subtraction to solve for 2x.

Inverse operation of subtraction is addition.

2p - 7  =  3

Add 7 to each side.

2p  =  10 

Here 2 and p are multiplied.

Then, we have to use the inverse operation of multiplication to solve for x.

Inverse operation of multiplication is division. 

2p  =  10

Divide each side by 2.

p  =  5

Problem 8 :

Solve for r : 

(r - 6)/2  =  3

Solution :

Here (r - 6) is divided by 2.

So, we have to use the inverse operation of division to solve for (r - 6).

Inverse operation of division is multiplication.

(r - 6)/2  =  3

Multiply each side by 2.

r - 6  =  6 

Here 6 is subtracted from r.

Then, we have to use the inverse operation of subtraction to solve for r.

Inverse operation of subtraction is addition. 

r - 6  =  6

Add 6 to each side.

r  =  12

Problem 9 :

Solve for p : 

(p/3) - 4  =  0 

Solution :

Here 4 is subtracted from (p/3).

So, we have to use the inverse operation of subtraction to solve for (p/3).

Inverse operation of subtraction is addition.

(p/3) - 4  =  0

Add 4 to each side.

p/3  =  4

Here p is divided by 3. 

Then, we have to use the inverse operation of division to solve for p.

Inverse operation of division is multiplication. 

p/3  =  4

Multiply each side by 3.

p  =  12

Problem 10 :

Solve for m : 

10 - 3m  =  -5

Solution :

10 - 3m  =  -5

Add 3m to each side. 

10  =  -5 + 3m

Add 5 to each side.

15  =  3m

Divide each side by 3.

5  =  m

Problem 11 :

The second angle of a triangle is three times the first, and the third is 12 degrees less than twice the first. Find the measures of the angles. 

Solution :

Let x be the first angle

Second angle = 3x

Third angle = 2x - 12

Sum of three angles = 180 

x + 3x + 2x - 12 = 180

6x = 180 + 12

6x = 192

x = 192/6

x = 32

3x ==> 3(12) ==> 36

2x - 12 ==> 2(32) - 12 ==> 52

So, the required angles are 32, 36 and 52.

Problem 12 :

The second angle of a triangle is four times the first and the third is 5 degrees more than twice the first. Find the measures of the angles.

Solution :

Let x be the first angle. Second angle = 4x

Third angle = 2x + 5

x + 4x + 2x + 5 = 180

7x = 180 - 5

7x = 175

x = 175/7

x = 25

4x = 4(25) ==> 50

2x + 5 = 2(25) + 5 ==> 55

So, the required angle measures are 25, 50 and 55.

Problem 13 :

The perimeter of a rectangle is 150 cm. The length is 15 cm greater than the width. Find the dimensions.

Solution :

Perimeter of the rectangle = 150 cm

Let x be the width and length will be x + 15

2(length + width) = 150

2(x + x + 15) = 150

2x + 15 = 150/2

2x + 15 =75

2x = 60

x = 30

length = 30 + 15 ==> 45 cm

So, the length and width of the rectangle are 45 cm and 30 cm respectively.

Problem 14 :

The perimeter of a rectangle is 304 cm. The length is 40 cm longer than the width. Find the length and width.

Solution :

Perimeter of the rectangle = 304 cm

Let x be the width and length will be x + 40.

2(length + width) = 304

2(x + x + 40) = 304

2x + 40 = 304/2

2x + 40 = 152

2x = 112

x = 112/2

x = 56

length = 56 + 40 ==> 96 cm

So, the length and width of the rectangle are 96 cm and 56 cm respectively.

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