Solving Equations with Inverse Operations

Math 97 Supplement 2

LEARNING OBJECTIVES

1. Solve equations by using inverse operations, including squares, square roots, cubes,

and cube roots.

The Definition of Inverse Operations

A pair of inverse operations is defined as two operations that will be performed on a number or

variable, that always results in the original number or variable. Another way to think of this is

that the two inverse operations “undo” each other. For example, addition and subtraction are

inverse operations since we can say 2 2x x . If we start with x, then add 2 and subtract 2,

we are left with the original starting variable x.

There are several inverse operations you should be familiar with: addition and subtraction,

multiplication and division, squares and square roots (for positive numbers), as well as cubes and

cube roots. The following examples summarize how to undo these operations using their

inverses.

Using Inverse Operations with the 4 Basic Operations

Addition Subtraction Multiplication Division

Solve: 2 3x .

x has 2 added to it, so

we subtract 2 from

both sides.

2 3

2 2

x

Solution:

1x

Solve: 2 3x .

x has 2 subtracted

from it, so we add 2 to

both sides.

2 3

2 2

x

Solution:

5x

Solve: 2 8x .

x has 2 multiplied to

it, so we divide 2 from

both sides.

2 8

2 2

x

Solution:

4x

Solve: 82

x .

x is divided by 2, so

we multiply by 2 on

both sides.

2 8 22

x

Solution:

16x

2

Using Inverse Operations with Powers and Roots

Square Root Square Cube Root Cube

Solve: 4x .

x is being square

rooted, so we square

both sides.

2

24x

Solution:

16x

Solve: 2 4x .

x is being squared, so

we square root both

sides. (using root)

2 4x

Solution:

2x or 2x

Solve: 3 2x .

x is being cube-

rooted, so we cube

both sides.

3

33 2x

Solution:

8x

Solve: 3 8x .

x is being cubed, so

we cube root both

sides.

3 33 8x

Solution:

2x

Note that undoing the square with a square root required both a positive and a negative in front

of the root. That is because when we square a positive or a negative number we get a positive.

We can’t be sure if the x in 2 4x should be a +2 or a -2 since both of these make the original

equation true: 2

2 4 and 2

2 4 . So, we include both +2 and -2 as an answer.

Also note that we don’t need the with the cube root since only a positive cubed would give us

a positive. In other words, 32 8 , but

32 8 , so we just need the positive cube root.

Example 1

Solve the following:

a. 9x

b. 2 9x

c. 3 9x

Solution:

a. 29 81x

b.   9 3x

c. 3 9x (this won’t simplify, so we leave it as is)

3

Consider the following equation: 3 2 12x

There are two ways to solve this problem, and both of them require eliminating the parentheses.

One method is to use the distributive property, and the other is to use inverse operations. The

chart below shows a comparison of these techniques.

Using Distributive Property Using Inverse Operations Only

Solve 3 2 12x .

Distribute the 3 through the parentheses

3 6 12x

Now use inverse operations by adding 6 to get

the 3x isolated on the left side

3 6 12

6 6

3 18

x

x

Isolate the x by dividing both sides by 3

3 18

3 3

x

Solution: 6x

Solve 3 2 12x .

Divide both sides by 3 to isolate the parentheses

3 2 12

3 3

2 4

x

x

Now we can remove the parentheses since it is

alone on the left, then add 2 on each side

2 4

2 2

x

Solution: 6x

Now consider this similar problem: Solve 2

3 2 12x .

This one cannot be solved by distributing the 3, we have to use inverse operations on this one.

Example 2

Solve 2

3 2 12x .

2

2

3 2 12

3 3

2 4

2 4

2 2

xDivide both sides by 3

x Undo the square with a square root

x Add 2 on both sides, simplify the root

x Simplify the + and the

4 or 0 x x Final answers!

4

To see why you can’t distribute a square (or other powers), think about the following

computations:

2 2 2

2

3 4 3 4

7 9 16

49 25

2 2

2 2

2

2 3 4 2 3 2 4

2 7 6 8

2 49 14

98 196

The bottom line is obviously false, and so are all of the previous lines. The same is true for

roots:

9 16 9 16

25 3 4

5 7

4 9 16 4 9 4 16

4 25 36 64

4 5 100

20 10

It’s important to remember that you CANNOT distribute a number through a power or a root,

and you cannot distribute a power or a root to each term inside. This means we will only be

using inverse operations to solve equations with powers or roots for now.

Example 3

Solve 32 7 1x .

Solution:

3

3

3

2 7 1

7 7

2 6

3

x

Subtract 7 on both sides to isolate the root term

x Divide both sides by 2 to isolate the root

x Cube both sides to undo the cube - root

x

3

3

27

Simplify

x Final answer!

Try this! Solve: 2 5 3x

Answer: 7x

5

Example 4

Solve 3

3 2 1 192x .

Solution:

3

3

3

3 2 1 192

3 3

2 1 = 64

2 1 64

2 1 4

xDivide by 3 on both sides to isolate the square

x Take the cube root of both sides

x Simplify the root if possible

x

2 3

3

2

Add 1 on both sides to isolate the x - term

x Divide by 2 on both sides to isolate the x

x Final Answer!

Example 5 and 6 are a couple of tougher examples where the roots don’t simplify to nice whole

numbers.

Example 5

Solve 2

2 3 100x .

Solution:

2

2

2 3 100

2 2

3 =50

3 50

3 50

xDivide by 2 on both sides to isolate the square

x Take the square root of both sides (+ and -)

x Add 3 to both sides to isolate the x

x

3 25 2

3 5 2

Simplify the root if possible

x

x Final Answer!

6

Example 6

Solve 3 5 48x .

Solution:

2

3 5 48

3 3

5=16

5 16

5 256

xDivide by 3 on both sides to isolate the square - root

x Square both sides

x Simplify the square

x Subtract 5 on both sides to isolate

256 5

251

the x

x Simplify

x Final Answer!

The answers given above are exact answers since they are not rounded. You could also be asked

for approximate answers as well, rounded to a certain number of decimals. The answers to

Example 5 rounded to 2 decimal places is shown below:

Example 5: 3 5 2x gives 10.07x or 4.07x if you are asked for an answer rounded

to two decimal places.

KEY TAKEAWAYS

Although we can’t distribute like usual with a power or a root, we can solve some of

these types of equations by undoing operations until we have isolated our variable.

When solving a square by using a square-root, be sure to include the + and – in front of

the root.

TOPIC EXERCISES

Solve the following equations.

1. 5 2x

2. 2 4x

3. 2

3 16x

4. 2

1 36x

5. 3 4 2x

6. 3 5 3x

7

7. 3

1 8x

8. 3

6 1x

9. 2

2 4 50x

10. 2

3 1 48x

11. 4 2 8x

12. 2 3 8x

13. 3

2 1 64x

14. 3

5 4 3 5x

15. 319 3

2x

16. 38 4 8x

17. 22

5 63

x

18. 2

2 4 98x

19. 3 5 3 1x

20. 2 3 4 7x

21. 3 1

18

x

22. 3

3 5 2 81x

23. 3 8 5 4x

24. 3 5 1 10x

25. 2

4 2 3 22x

26. 2

2 3 1 77x

27. 2 3 2 8x

28. 3 2 2 8x

29. 32 2 130x

30. 33 3 122x

Find the exact answer, then use a calculator to approximate to the nearest hundredth.

31. 2

4 33x

32. 2

5 15x

33. 2

8 7x

34. 2

10 18x

35. 2

3 20x

36. 2

1 75x

37. 2

2 27x

38. 2

2 24x

39. 2

2 5 64x

40. 2

3 7 54x

8

ANSWERS

1. 1x 2. 3. 7x , 1x 4. 5. 12x 6. 7. 3x 8.

9. 1x , 9x 10. 11. 2x 12.

13. 5

2x

14. 15. 207x 16. 17. 8x , 2x 18.

19. 11

3x

20.

21. 3

2x

22. 23. 3x 24. 25. 0x , 6x 26. 27. 6x 28. 29. 4x 30.

31. 4 33x ,

Approx: 1.74x , 9.74x 32.

33. 8 7x ,

Approx: 10.65x , 5.35x 34.

35. 3 2 5x ,

Approx: 7.47x , 1.47x 36.

37. 2 3 3x

Approx: 7.20x , 3.20x 38.

39. 5 4 2x ,

Approx: 0.66x , 10.66x 40.