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 xx22   . 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 Solve: x 23. Solve: x 23. Solve: 28x  . Solve:  8 . 2 x has 2 added to it, so x has 2 subtracted x has 2 multiplied to x is divided by 2, so we subtract 2 from from it, so we add 2 to it, so we divide 2 from we multiply by 2 on both sides. both sides. both sides. both sides. x 23 x 23 28x x  2  8  2 22 22 22 2

Solution: Solution: Solution: Solution: x 16 x 1 x  5 x  4

Using Inverse Operations with Powers and Roots

Square Root Square Cube Root Cube

Solve: x  4 . Solve: x2  4 . Solve: 3 x  2 . Solve: x3  8 .

x is being square x is being squared, so x is being cube- x is being cubed, so rooted, so we square we square root both rooted, so we cube we cube root both both sides. sides. (using  root) both sides. sides.

2 3 3 3 2 2 3 3 3 x  8  x   4 x  4  x   2  

Solution: Solution: Solution: Solution: x 16 x  2 or x 2 x 8 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 should be a +2 or a -2 since both of these make the original equation true: 242  and 242 . 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, 283  , but 283 , so we just need the positive cube root.

Example 1

Solve the following: a. x  9

b. x2  9 c. x3  9

Solution: a. x 92 81

b. x  93  

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

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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 32 x   12  Now use inverse operations by adding 6 to get 33 the 3x isolated on the left side  x 24

3x  6 12 Now we can remove the parentheses since it is  6 6 alone on the left, then add 2 on each side 3x  18 x 24

Isolate the x by dividing both sides by 3  2 2

3x 18 Solution: x  6  33

Solution: x  6

Now consider this similar problem: Solve 3x  22 12. This one cannot be solved by distributing the 3, we have to use inverse operations on this one.

Example 2

Solve 3x  22 12.

2 32 x   12  Divide both sides by 3 33  x22 4 Undo the square with a square root x2   4 Add 2 on both sides, simplify the root x2 2 Simplify the + and the  x4 or x 0 Final answers!

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To see why you can’t distribute a square (or other powers), think about the following computations: 22 2 2 3 4  2  3  2  4 3 4  322  4     22 2 2 7 6 8 7  9 16     2 49 25 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: 4 9 16  4  9  4  16 9 16  9  16 4 25 36 64 25 3 4 57 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

3 Solve 2 x  7 1.

Solution:

3 2 x  7 1  7 7 Subtract 7 on both sides to isolate the root term

3 2 x 6 Divide both sides by 2 to isolate the root

3 x3 Cube both sides to undo the cube - root x  33 Simplify x27 Final answer!

Try this! Solve: 2x  5 3

Answer: x  7

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Example 4

3 Solve 3 2x  1   192 .

Solution:

3 3 2x  1 192  Divide by 3 on both sides to isolate the square 33 2x 13 = 64 Take the cube root of both sides

3 2x 1   64 Simplify the root if possible 2x  1   4 Add 1 on both sides to isolate the x - term 2x 3 Divide by 2 on both sides to isolate the x 3 x Final Answer! 2

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

Example 5

2 Solve 2x  3 100.

Solution:

2 23 x   100  Divide by 2 on both sides to isolate the square 22  x 32 =50 Take the square root of both sides (+ and -) x3   50 Add 3 to both sides to isolate the x x 3 50 Simplify the root if possible x 3  25  2 x3 5 2 Final Answer!

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Example 6

Solve 3x  5 48.

Solution:

3x  5 48  Divide by 3 on both sides to isolate the square - root 33 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.

2 1. x 52 4. x 1 36

2. x 24 5. 3 x 42 2 3. x 3 16 6. 3 x 53

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3 7. x 18   19. 3x  5  3  1

3 8. x 61   20. 2x  3  4  7

3 1 9. 2 21. x 1 2x  4 50 8

2 3 10. 3x  1 48 22. 3 5x  2 81

11. 4x  2 8 23. 3 8x  5 4 12. 2x  3   8 24. 3 5x  1 10 3 13. 2x  1  64 2 25. 4 2x  3  22 3 14. 5 4x  3 5 2 26. 2 3x  1  77 1 15. 3 x 93 2 27. 2 3x  2  8

3 16. 8 x  4   8 28. 3 2x  2  8

3 2 2 29. 2 2x 130 17.  x 56 3 3 30. 3 3x   122 2 18. 2x  4   98

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

2 2 31. x 4 33 36. x 1 75

2 2 32. x 5 15 37. x 2 27

2 2 33. x 87 38. x 2 24

2 2 34. x 10 18 39. 2x  5 64 2 2 40. 3x  7 54 35. x 3 20  

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ANSWERS

1. x 1 2. 3. x 7 , x 1 4. 5. x 12 6. 7. x 3 8. 9. , 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  87  , 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.

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