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 x  2  2  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
x
 8.
2
Solve: x  2  3 .
Solve: x  2  3 .
Solve: 2x  8 .
Solve:
x has 2 added to it, so
we subtract 2 from
both sides.
x has 2 subtracted
from it, so we add 2 to
both sides.
x has 2 multiplied to
it, so we divide 2 from
both sides.
x is divided by 2, so
we multiply by 2 on
both sides.
x23
x23
2x 8

2 2
2
Solution:
x4
Solution:
x  16
2 2
2 2
Solution:
x 1
Solution:
x 5
x
 82
2
Using Inverse Operations with Powers and Roots
Square Root
x  4.
Solve:
x is being square
rooted, so we square
both sides.
 x
2
Square
Solve: x 2  4 .
Solve:
x is being squared, so
we square root both
sides. (using  root)
x is being cuberooted, so we cube
both sides.
 42
Solution:
x  16
Cube Root
x  4
2
Solution:
x  2 or x  2
 x
3
3
3
x  2.
 23
Solution:
x 8
Cube
Solve: x3  8 .
x is being cubed, so
we cube root both
sides.
3
x  
3
3
8
Solution:
x2
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 x 2  4 should be a +2 or a -2 since both of these make the original
2
2
equation true:  2  4 and  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
3
a positive. In other words, 23  8 , but  2   8 , so we just need the positive cube root.
Example 1
Solve the following:
a. x  9
b. x 2  9
c. x3  9
Solution:
a. x  92  81
b. x   9   3
c. x  3 9 (this won’t simplify, so we leave it as is)
2
Consider the following equation: 3 x  2  12
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 x  2  12 .
Solve 3 x  2  12 .
Distribute the 3 through the parentheses
Divide both sides by 3 to isolate the parentheses
3x  6  12
Now use inverse operations by adding 6 to get
the 3x isolated on the left side
3  x  2  12

3
3
 x  2  4
3 x  6  12
6 6
Now we can remove the parentheses since it is
alone on the left, then add 2 on each side
3 x  18
Isolate the x by dividing both sides by 3
x2  4
2 2
Solution: x  6
3x 18

3
3
Solution: x  6
Now consider this similar problem: Solve 3  x  2   12 .
This one cannot be solved by distributing the 3, we have to use inverse operations on this one.
2
Example 2
Solve 3  x  2   12 .
2
3  x  2  12

3
3
2
 x  2
2
4
x2  4
x  22
x  4 or x  0
Divide both sides by 3
Undo the square with a square root
Add 2 on both sides, simplify the root
Simplify the + and the 
Final answers!
3
To see why you can’t distribute a square (or other powers), think about the following
computations:
3  4  3  4
2
 7   9  16
2
2
2 3  4   2  3  2  4
2
2
49  25
2  7    6  8
2
2  49  14 
2
2
2
98  196
The bottom line is obviously false, and so are all of the previous lines. The same is true for
roots:
4 9  16  4  9  4 16
9  16  9  16
4 25  36  64
25  3  4
4  5  100
57
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 2 3 x  7  1 .
Solution:
2 3 x  7 1
7 7
2 3 x  6
3
x  3
x   3
3
x  27
Try this! Solve:
Subtract 7 on both sides to isolate the root term
Divide both sides by 2 to isolate the root
Cube both sides to undo the cube - root
Simplify
Final answer!
2x  5  3
Answer: x  7
4
Example 4
Solve 3  2 x  1  192 .
3
Solution:
3  2 x  1
192

3
3
Divide by 3 on both sides to isolate the square
 2 x  1
Take the cube root of both sides
3
3
=  64
2 x  1  3 64
Simplify the root if possible
2 x  1  4
2 x  3
x
3
2
Add 1 on both sides to isolate the x - term
Divide by 2 on both sides to isolate the 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  x  3  100 .
2
Solution:
2  x  3 100

2
2
Divide by 2 on both sides to isolate the square
 x  3
Take the square root of both sides (+ and -)
2
2
=50
x  3   50
x  3  50
Add 3 to both sides to isolate the x
Simplify the root if possible
x  3  25  2
x  35 2
Final Answer!
5
Example 6
Solve 3 x  5  48 .
Solution:
3 x  5 48

3
3
Divide by 3 on both sides to isolate the square - root
x  5=16
Square both sides
x  5  162
Simplify the square
x  5  256
Subtract 5 on both sides to isolate the x
x  256  5
Simplify
x  251
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: x  3  5 2 gives x  10.07 or x  4.07 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.
 x 1
1.
x5  2
4.
2.
x2  4
5.
3
x4  2
6.
3
x5 3
3.
 x  3
2
 16
2
 36
6
7.
 x  1
8.
 x  6
3
3
 8
19.
3x  5  3  1
 1
20.
2x  3  4  7
 x  1
3

1
8
9. 2  x  4  50
21.
10. 3  x  1  48
22. 3  5 x  2   81
11. 4 x  2  8
23.
3
8  x  5  4
24.
3
5  x  1  10
2
2
12. 2 x  3  8
13.   2 x  1  64
3
3
14. 5  4 x  3  5
25. 4  2  x  3  22
2
3
15.
13
x9  3
2
26. 2  3  x  1  77
2
27. 2  3 x  2  8
16. 8 3 x  4  8
28. 3  2 x  2  8
2
2
17.  x  5   6
3
29. 2  2 x3  130
30. 3  3x3  122
18. 2  x  4   98
2
Find the exact answer, then use a calculator to approximate to the nearest hundredth.
31.  x  4  33
36.  x  1  75
32.  x  5  15
37.  x  2  27
33.  x  8  7
38.  x  2  24
34.  x  10   18
39. 2  x  5  64
2
2
2
2
35.  x  3  20
2
2
2
2
2
40. 3  x  7   54
2
7
ANSWERS
1. x  1
2.
3. x  7 , x  1
4.
5. x  12
6.
7. x  3
8.
9. x  1 , x  9
10.
11. x  2
12.
5
13. x  
2
14.
15. x  207
16.
17. x  8 , x  2
18.
11
19. x 
3
20.
3
21. x 
2
22.
23. x  3
24.
25. x  0 , x  6
26.
27. x  6
28.
29. x  4
30.
31. x  4  33 ,
Approx: x  1.74 , x  9.74
32.
33. x  8  7 ,
Approx: x  10.65 , x  5.35
34.
35. x  3  2 5 ,
Approx: x  7.47 , x  1.47
36.
37. x  2  3 3
Approx: x  7.20 , x  3.20
38.
39. x  5  4 2 ,
Approx: x  0.66 , x  10.66
40.
8