Solving Equations with Inverse Operations
Transcript of Solving Equations with Inverse Operations
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.