Grade 9 Algebra by Elaine Ernst Schneider
A Beginning Look at Basic Algebra – Lesson 1
Outline:
Algebra provides the basics for all higher math. You will work with numbers and letters (variables) to form sentences (expressions) that you can solve. The best way to learn math is by practicing it, so each lesson will include exercises using the skills learned.
A place to begin:
Letters in math are called variables. They can stand for different numbers at different times.
A mathematical sentence is called an expression. It can include numbers, variables, signs of operation, and symbols of inclusion.
Signs of operation tell you what to do to the sentence. The four operations are addition, subtraction, multiplication, and division.
Symbols of inclusion are parentheses ( ) and brackets [ ].
An important caution:
Be very neat in your calculations. Many an algebra problem is missed because the student misread what he or she had written or did not “line up” the column correctly for subtraction or division. Always double check operations. You don’t want to miss a problem because you added incorrectly.
Let’s Get Started:
To “evaluate” an expression means to find its value, or to solve it. The first rule to learn about algebra is “what to do when.” The order in which an expression’s operations are done can completely change the answer.
When evaluating an algebraic expression, first look for the symbols which show the innermost work. That can be expressed by use of parentheses or brackets. If BOTH parentheses and brackets are present, the parentheses are usually the innermost and should be worked first.
Here is an example:
24 + [46 – (2 X 11)]
24 + [46 – 22]
24 + 24
48
Now it’s time for you to try a few.
EXERCISE:
9 – (4 X 2)
(9 – 4) X 2
(9 – 4) X (2 X 1)
48 – [42 – (3 X 9)]
63 – [8/2 + (14 – 10)]
(Note: 8/2 is the same as 8 divided by 2, just like in fractions.)
[800/ (200 X 4)]
28 + [ 10 – (4 + 2) ]
(115) X (10 + 14)
125 / ( 5 X 5) (Remember from number 5? / = divided by.)
[28 – (4 X 5)] – 4
ANSWER KEY
Question#  Answer 
1… 2… 3… 4… 5… 6… 7… 8… 9… 10…  1 10 10 33 55 1 32 144 5 4 
Basic Algebra – Lesson 2
Outline:
In the first lesson, you learned that numbers and variables form sentences, or algebraic “expressions.” When you take information from a sentence and turn it into a mathematical expression, it is called “translating.”
A place to begin:

 When you write algebraic expressions, use +, , and = signs. For division, use / , the same way you know that when you see a fraction, it means to divide the top number by the bottom number.
 For multiplication, write the expression with no symbol or sign between them as the X (multiplication) symbol can be confused with the variable x. For example 3 times the variable y should be written 3y. You can also use parentheses to indicate multiplication. This is especially useful in longer problems such as (3y)(42x).
 When you want to multiply something AFTER another expression has been done first, use parenthesis. For example, if you want to add x and y and THEN multiply the result by 7, write it this way: 7(x + y).
 To translate from language to a math expression, read the sentence carefully. Then decide what operations it will take to reach a solution. Write this into an algebraic expression.
Let’s Get Started:
Here is the word problem:
I am stocking shelves at a Tshirt store. I have been given eight boxes of Tshirts. There are 25 Tshirts in each box. My instructions are to take one Tshirt from every box and set it aside to be given to the poor. How many Tshirts will be left for me to put on the store shelves after I sort out the Tshirts to be given to the poor?
Here is the algebraic expression:
8(251)
8(24)
192
If you don’t know the number of Tshirts in each box, then substitute the variable x for that. Then the expression would look like this:
8(x – 1)
Now it’s time for you to try a few.
Assignment(s) including Answer key:
EXERCISE:
 Katie is 21 years old. Write an algebraic expression that tells how old she will be in five years. Let x represent that age. Solve from your expression.
 One hamburger costs x cents. Write an algebraic expression that tells how much 4 hamburgers will cost.
 Sandy is 21 years old. Al is five years older. How old will Al be in five years? Let x represent Al’s age in five years. Solve for x.
 I have ten hats that cost $47.50 for all of them. Write an algebraic expression to show how much one of those hats cost, letting y represent the answer. Solve for y.
 I have ten hats that cost $47.50 for all of them. How much will I have to sell EACH hat for to make a profit of $1 on each? Set it up in an algebraic expression to solve.
ANSWER KEY
Question  Answer 
1..  21 + 5 = x 26 = x 
2..  4x 
.  
3..  (21 + 5 = Al’s age now) (21 + 5) + 5 = x (Al’s age in 5 years) 
.  
4..  $47.50 / 10 = y $4.75 = y 
.  
5..  (47.50 / 10 ) + $ 1 = x $4.75 + $ 1 = x $5.75 = x 
Algebra Lesson 3 – Solving for X
Outline:
A quick review:
1. When you write algebraic expressions, use +, , and = signs. For division, use / , the same way you know that when you see a fraction, it means to divide the top number by the bottom number.
2. For multiplication, write the expression with no symbol or sign between them as the X
(multiplication) symbol can be confused with the variable x. For example 3 times the variable y should be written 3y. You can also use parentheses to indicate multiplication. This is especially useful in longer problems such as (3y)(42x).
3. When you want to multiply something AFTER another expression has been done first, use
parenthesis. For example, if you want to add x and y and THEN multiply the result by 7, write it this way: 7(x + y).
4. To translate from language to a math expression, read the sentence carefully. Then decide what operations it will take to reach a solution. Write this into an algebraic expression.
Something new:
1. To take something OUT of parenthesis, do the operation one number at a time. For example, 7(x+y). First, multiply 7 times x. Then multiply 7 times y. The result is 7x + 7y.
2. When you solve for x, you want to “isolate” the x on one side of the equal sign. To do this, use the opposite sign of the number you want to move and do the same thing to BOTH sides of the equation.
For example: 8x + 2 = 50
8x + 2 – 2 = 50 – 2 (Subtract 2 from both sides of the = sign)
8x = 48 (Divide by 8 to solve because that is opposite of multiplication)
x = 6
And now a fun problem to make you really think. Solve using algebra. Translate into an expression and solve.
You can do it!
Assignment(s) including Answer key:
George is 4 years older than Jon, who is 4 years older than Jim, who is 4 years older than Sam, who is 1/2 the age of George. How old is each boy? Hint: let x represent George’s age.
Answer Key
If x is George’s age, then x – 4 is Jon’s age because George is 4 years older than Jon. George is 8 years older than Jim, making Jim’s age x – 8. George is 12 years older than Sam, so Sam’s age can be represented as
x – 12.
We also know that Sam is 1/2 the age of George.
So, set up the expression:
(x – 12) = 1/2 x
To solve for x, you first need to put 1/2x into a different form.
Multiply 1/2 by x.
x is the same as x over 1 if you put it into fraction form, or x/1. So, when you multiply 1/2 by x, it would actually look like (1/2) (x/1). When you multiply fractions, you multiply the numerators (1)(x) and then the denominators (2)(1). The result is x/2. Now rewrite the expression using x/2 instead of 1/2x.
(x12) = x/2
We learned that when you solve an expression, you want to “isolate” the variable. To do this, multiply BOTH sides by 2.
(x12) = x/2
2(x12) = 2(x/2)
2x 24 = x
Now, you need to get both variables on the same side of the equal sign, so subtract 2x from BOTH sides:
2x – 2x – 24 = x – 2x
24 = 1x (Divide 24 by 1 to solve for x)
24 = x
therefore, George is 24. Jon’s age is represented by x – 4 or 20.
Jim is x – 8, or 16. And Sam is x – 12, or 12.
You can further check the problem by multiplying Sam’s age by 2, to get George’s age, because the problem told us that Sam is half the age of George.
Basic Algebra – Lesson 4
Outline:
In the first lesson, you learned that numbers and variables form sentences, or algebraic “expressions.” When you take information from a sentence and turn it into a mathematical expression, it is called “translating.” In another lesson, you learned that when you write algebraic expressions, use +, , and = signs; and for division, use / , the same way you know that when you see a fraction, it means to divide the top number by the bottom
number. Then, for multiplication, we learned to write the expression with no symbol or sign between them (such as 3a), with an X , or using parentheses. The parenthesis is especially useful in longer problems such as (3y)(42x).
But what if there are no parentheses or brackets? How do you know which to do first? Add, divide, multiply? Which one?
When there are no other indications as to which computation to do first, mathematicians follow The Order of Operations rule. Multiply, Divide, Add, Subtract.
A good way to remember the Order of Operations is to think about My Dear Aunt Sally. The M is for multiply, the D in Dear is for Divide, the A is for Add, and the S in Sally is for subtract.
HINT: If the problem has only multiplication and division, then work from left to right.
HINT: If the problem has only subtraction and addition, then work from left to right.
Let’s Get Started:
13 – 2 X 5
If I just do the math from left to right, I would say 13 – 2 = 11. Then 11 times 5 = 55.
But the Order of Operations tells me to multiply first! So, 2 X 5 = 10. Then, I subtract 10 from 13 and get 3.
You can see that which order you choose makes a BIG difference in the answer! That’s why it’s important to follow the Order of Operations.
Now it’s time for you to try a few.
Assignment(s) including Answer key:
1. 10 – 4 + 3
2. 10 + 4 X 2
3. (5 X 4) 15 + 2
4. 12 – 2(3 + 1)
5. 18 + 2(3)
6. (12 – 2)(3 + 4)
7. 4 X 3 + 5
8. 24 – 6 +2
9. 24 – 6 X 2
10. 36/9 – 2
ANSWER KEY
1. 9
2. 18
3. 7
4. 4
5. 24
6. 70
7. 17
8. 20
9. 12
10. 2
Basic Algebra – Lesson 5 – Expressions from Sentences
Outline:
Review:
In the first lesson, you learned that numbers and variables form sentences, or algebraic “expressions.” When you take information from a sentence and turn it into a mathematical expression, it is called “translating.” In another lesson, you learned that when you write algebraic expressions, use +, , and = signs; and for division, use / , the same way you know that when you see a fraction, it means to divide the top number by the bottom
number. Then, for multiplication, we learned to write the expression with no symbol or sign between them (such as 3a), with an X , or using parentheses. The parenthesis is especially useful in longer problems such as (3y)(42x).
Then you learned how to work problems where there are no parentheses or brackets, using the Order of Operations rule. Multiply, Divide, Add, Subtract.
Now let’s put all of this to use and Yes, that’s right! We’re going to take English sentences – WORDS – and turn them into algebraic expressions.
Let’s Get Started:
Subtract seven from twentyone, then add three.
The algebraic expression is: 21 – 7 + 3
No parentheses are needed because the Order of Operations tells us that addition and subtraction are done in order from left to right.
Now it’s time for you to try a few. Remember your terms: subtract, sum, product, division, multiply, quotient. You may have to use parenthesis on some of them.
Assignment(s) including Answer key:
1. Subtract 2 from x; then add y.
2. Subtract the sum of 2 and y from x.
3. Divide 10 by 3; then multiply by 5.
4. Divide x by the product of 3 and z.
5. Multiply x by 3; then add y.
6. Add x and 3; then multiply by y.
7. Subtract the product of 5 and x from 7.
8. 5 more than the product of 3 and c.
9. 13 less than the quotient 5 divided by p.
10. 4 times the sum of 10 and x.
Answer Key
1. x – 2 + y
2. x – (2 + y)
3. 10/3 X 5
4. x/32
5. 3x + y
6. (x + 3) y
7. 7 – 5x
8. 3c + 5
9. 5/p 13
10. 4(10 + x)
Basic Algebra – Lesson 6 Balancing Equations to Solve Variables
Outline:
In the last lesson, you learned to write expressions and to find the values of expressions.
When two expressions can be written to balance or equal one another, it is called an
equation. In other words, whatever is expressed on one side of the equal sign is calculated to be the exact same value as whatever is on the other side of the equal sign. IN other words, the left member and the right member of the equation “balance.”
Examples of equations:
5 + 7 – 2 = 2(5) (the answer on the left side is 10 and the answer on the right side is 10)
100/25 = 347 – 343 (the answer is 4 on the left of the equal sign and 4 on the right also)
Equations involving unknown variables are solved by balancing the left member and the right member. For example, in the equation y + 5 = 12, I know that the left member must equal 12 for it to balance with the right member. This means y would be 7 because 7 + 5 = 12. That one is easy enough to do the math without a formal procedure. But because algebraic equations can become much longer and more calculated, a system is necessary to solve for variable and balance equations.
Let’s take the above equation:
y + 5 = 12
The proper procedure in solving for y is to “isolate” y. This means we want y to stand by itself on one side of the equal sign.
Now, keep in mind that we want every thing to “balance” on either side of the equal sign. This means whatever I do to the left member, I must do to the right member. So, to isolate y and solve the equation, I must “move” the 5 to the other side of the equal sign. To do this, I must make it zero on the left side of the equation by subtracting 5:
y + 5 = 12
y + 5 – 5 = 12 – 5
y + 0 = 12 – 5
y = 7
Try solving for the variable by keeping each side equal. Remember, the idea is to “balance”
the equation, so what you do to one side, you must do to the other.
100/5 = y + 2
100/5 2 = y +2 – 2
20 – 2 = y
18 = y
Now it’s time for you to try a few…
Assignment(s) including Answer key:
EXERCISE:
1. y – 10 = 17
2. c + 4 = 29
3. 5y = 90
4. 1/3 t = 29
5. 6y = 72
6. 1/9 g = 58
7. 4y = 100
8. 12a = 132
9. r + 9 = 48
10. x + 79 = 422
ANSWER KEY:
1. 27
2. 25
3. 18
4. 87
5. 12
6. 522
7. 25
8. 11
9. 39
10. 343
Basic Algebra – Lesson 7 Using Equations to Solve Puzzles
Outline:
In the last lesson, you learned to balance an equation, making what is on one side of the equal sign equal to what is on the other side. In other words, the left member and the right member of the equation “balance.”
Examples of equations:
5 + 7 – 2 = 2(5) (the answer on the left side is 10 and the answer on the right side is 10)
100/25 = 347 – 343 (the answer is 4 on the left of the equal sign and 4 on the right also)
You also learned to isolate variables. Equations involving unknown variables are solved by balancing the left member and the right member. In the equation, y + 5 = 12, the proper procedure in solving for y is to “isolate” y. This means we want y to stand by itself on one side of the equal sign.
Now, keep in mind that we want every thing to “balance” on either side of the equal sign. This means whatever I do to the left member, I must do to the right member. So, to isolate y and solve the equation, I must “move” the 5 to the other side of the equal sign. To do this, I must make it zero on the left side of the equation by subtracting 5:
y + 5 = 12
y + 5 – 5 = 12 – 5
y + 0 = 12 – 5
y = 7
Equations can also be used to solve puzzles. You must set up an equation that “matches” in numbers what the puzzle expresses in words. Then, solve for the variable. That will be the answer to the puzzle.
Why don’t you try a few?
Assignment(s) including Answer key:
1. Farmer Brown told Bob and Sue that they could pick apples from his tree, but that neither of them could take more than 20. They worked for a while, and then Bob asked Sue, “Have you picked your limit yet?”
Sue replied, “Not yet. But if I had twice as many as I have now, plus half as many as I have now, I would have my limit.” How many did Sue have?
2. A little boy was told not to eat the grapes from the vine for fear that he would eat too many and get a stomachache. Sneaking out to the grape arbor when his mother wasn’t looking, the little boy ate grapes for five days, each day eating 6 more than the day before. In fact, after five days, the little boy was so sick that he had to confess to his mother that he had eaten 100 grapes. How many grapes did the little boy eat on EACH of the five days?
3. How high is a tree that is 15 feet shorter than a pole three times as tall as the tree?
ANSWER KEY:
1. Let x = the number of apples she had.
2x + 1/2 x = 20
(2x as a fraction with 2 as the denominator would be written 4x/2.)
4x/2 + 1/2 x = 20
5x/2 = 20
5x/2 X 2 = 20 X 2
5x = 40
5x/5 = 40/5
x = 8 apples
2. Let x = number of grapes the little boy ate the first day
x + 6= number of grapes eaten the second day
x + 6 + 6 = number of grapes eaten the third day
x + 6 + 6 + 6 = number of grapes eaten the fourth day
x + 6 + 6 + 6 + 6 = number of grapes eaten the fifth day
Five days’ worth of grapes = 100 in all. Therefore, the equation to set up is:
x + (x + 6) + (x + 6 + 6) + (x + 6 + 6 + 6) + (x + 6 + 6 + 6 + 6) = 100
x + x + 6 + x + 12 + x + 18 + x + 24 = 100
5x + 60 = 100
5x + 60 – 60 = 100 – 60
5x = 40
5x/5 = 40/5
x = 8 grapes eaten on the first day
(Now put 8 in place of x in all the other expressions.)
x + 6 = 14 grapes eaten the second day
x + 6 + 6 = 20 grapes eaten the third day
x + 6 + 6 + 6 = 26 grapes eaten the fourth day
x + 6 + 6 + 6 + 6 = 32 grapes eaten the fifth day
(To check, add 8, 14, 20, 26, ad 32. They equal 100.)
3. Let y = number of feet in height of tree
3y would equal the number of feet in height of the pole
Set up the equation:
y = 3y – 15
y – 3y = 3y – 3y – 15
2y = 15
2y/2 = 15/2
y = 7.5 feet
3y = 22.5 feet
Algebra Lesson 8 – The Equation Behind an Old Trick
Outline:
In the last lesson, you learned that equations can be used to solve puzzles. For years, mathematicians have been solving the following puzzle to the amazement of their students. Not only is it a fun “trick,” but it is based on a sound mathematical principle that makes it work:
By using your house number, a mathematician can calculate your age!
Here’s how it works:
1.Double your house number.
2.Add 5.
3.Multiply by 50.
4.Add your age.
5.Add the number of days in a year (365).
6.Take that number and subtract 615.
7.Pretend it is an amount of money and position the decimal point for dollars and cents.
Answer: the dollars will be your house number and the cents will be your age.
But now the big question is: HOW does it work? Give it a little thought before looking at the answer.
ANSWER:
To understand why the puzzle works, you must solve it algebraically.
Let x = your house number.
Let y = your age.
Let A = answer
Using the numbered list of instructions, set up the problem as an algebraic equation. It would look like this:
A = 50(2x + 5) + y + 365 – 615
100
A = 100x + 250 + y +365 615
100
A = 100x + y
100
A = x + y/100
Therefore,
x = the house number (You started with this number.)
y/100 = your age (When you divide by 100, you move the decimal point two places to the left.)
Algebra Lesson 9 – Positives and Negatives on the Number Line
Outline:
Different kinds of numbers are used to stand for different things. There are positive numbers and there are negative numbers. Positive numbers are more than zero and negative numbers are less than zero.
The best way to “see” negative and positive numbers is to look at a number line:
/_____/_____/_____/_____/_____/_____/_____/_____/_____/_____/_____/_____/_____/_____/
7 … 6 ….. 5 …. 4 …… 3 …… 2 ….. 1 ….. 0 …… 1 …… 2 …… 3 …… 4 …… 5 …… 6 …… 7
When you use a number line, you add positives as you go to the right and you add negatives as you move to the left. For example, 3 + 2 means that you start at 3 and move two places to the left. The answer is 1. To add a positive to a negative, you work from left to right. For example, 4 + 3 is 1. You get that by starting at 4 and moving three places to the right. To add two negatives, move to the left. For example, 1 + 3 = 4. You arrive at that answer by starting at 1 and counting three places to the left.
Assignment(s) including Answer key:
Now you try:
1 . 1 + 1
2. 4 + 2
3. 2 + 4
4. 2 + 5
5. 5 + 4
6. – 3 + 5
7. 2 + 2 + 1
8. 5 + 6
9. 7 +2
10 4 + 2 + 2
Now try a few that are beyond the scope of the number line shown above. You can do it! Just use the principles you have learned.
 15 + 13
 39 + 40
 23 + 20
 2 + 11
 13 + 4
CUT HERE_________________________________________________________________
ANSWERS:
1. 0
2. 2
3. 6
4. 7
5. 1
6. 2
7. 5
8. 1
9. 5
10 4
********

 2
 79
 3
 9
17
Algebra Lesson #10 – Absolute Values
Outline:
In lesson 9, we learned about negative and positive numbers by using a number line. Negatives go left from the zero, and positives are counted off to the right. To add a negative and a positive, you start at one of the numbers and then count either to the left or the right depending on the sign. For example, to add 2 and 3, you can start at 2 and count to the right three places. You would end up at 1. 2 + 3 = 1
There is a term that is used in math to represent the distance that a number is from zero. That term is absolute value. The absolute value of 3 is three. The absolute value of 6 is six. This means that in terms of absolute value, 6 is larger than 3. In other words, 6 takes up more places on the number line. Look at the number line below. The red numbers represent the numbers of places it takes to reach 6 from zero. The blue color represents three. You can see that the red line is much longer than the blue one.
/_____/ _____/_____/_____/_____/_____/_____/_____/_____/_____/_____ /_____/
6 … 5 …… 4 ….. 3 ….. 2 ….. 1 ….. 0 .…. 1 …… 2 …… 3 …….4 …… 5 ..…. 6
The concept of absolute value is important when adding positive and negative numbers. For example, if you have 7 + 4, you may wonder which number to use to begin the addition process. The number four is positive, so you might think that it represents “more.” Anyone would rather have $4 in their checking account that be in the hole $7; right?! But in order to do the math, we must concern ourselves only with absolute value. The absolute value of seven is greater than the absolute value of four. Therefore, we begin with 7 and treat it as the larger number.
Here’s the problem:
7 + 4 =
Subtract the values. 7 – 4 = 3. To determine which sign to use, you must consider absolute value. Since 7 is larger in absolute value, its sign will dominate. The answer will be 3.
7 + 4 = 3
Suppose you have two negative numbers:
7 + 2
Add the values. 7 + 2 = 9. Since you added absolute values, the answer is – 9.
7 + 2 = 9
Assignment(s) including Answer key:
Now you try:
 7 + 9
 7 + 9
 6 + 5
 2 + 5
 5 + 4
 – 3 + 5
 5 + 2
 15 + 6
 17 +22
 4 + 21
ANSWERS:

 16
 2
 1
 7
 1
 8
 3
 9
 5
25
Algebra Lesson 11 – Additive Inverses
Outline:
Additive Inverses are opposites. Two numbers are opposites if their sum equals zero. For example, 8 and 8 are additive inverses because their sums total zero. This makes them opposites.
Another way to write the opposite of –8 is to write –(8). To subtract a negative 8 is the same as making it a positive 8, or finding its opposite.
You can do the same thing with variables. For example, (x) = x.
Using this principle, let’s add a negative number to a positive one: 8 + (3) = 8 – 3 = 5. This could be read as 8 plus a negative 3 OR 8 minus 3 OR 8 plus the opposite of 3.
In algebra, subtracting a number can also be described as adding its opposite.
For example, x – y = x + (y).
OR
8 – (5) = 8 + 5 = 13
Now, let’s turn things around a bit. Try this one:
14 – 28 = x
How can you rewrite that to use what you’ve learned about opposites?
14 + (28) = x
–14 = x
Let’s try one more:
11 – (3) – 4 = x
11 + 3 – 4 = x
14 – 4 = x
10 = x
Assignment(s) including Answer key:
1. 5 – 7
2. 8 – (10)
3. – 3 – 7
4. – 7 – 9
5. 7 + (3)
6. 25 – 250
7. 5.8 – 2.3
8. 2.3 – 5.8
9. (34 – 13) – (15 – 17)
10. 11 – 5 – [6 + (13)]
Answer key:
1. –2
2. 18
3. –10
4. –16
5. 4
6. –225
7. 3.5
8. –3.5
9. 23
10 .13
Algebra Lesson 12: Multiplying Positives and Negatives
As we learned last lesson: Additive Inverses are opposites. Two numbers are opposites if their sum equals zero. For example, 8 and 8 are additive inverses because their sums total zero. This makes them opposites. You can do the same thing with variables. For example, (x) = x.
When this principle is used in multiplication, these rules of thumb emerge:
1. A negative times a negative gives a positive answer.
2. A negative times a positive is negative.
3. A positive times a positive is a positive result.
Examples
(6)(8) = 48
(3)(12) = 36
(12)(11) = 132
Now, what happens when there are three numbers to multiply? Simply work in sequence, following the rules you have learned.
Example: (5) (4) (20)
(20) (20) [multiply –5 times –4 to get positive 20]
400 [multiply positive 20 from last step by –20 to get –400]
And what about exponents? Just write them out and follow the rules.
Example: 45 = (4) (4) (4) (4) (4)
(16) (4) (4) (4)
64 (4) (4)
256 (4)
1024
Exercises:
1. (10) (8)
2. (5) (26)
3. 53
4. (2) (3)2
5. (1.5) (4)
6. (5) (22) (2)
7. (3) (5) (4)
8. (1) (3)4
9. (1)15
10. (2)2 (3)4
Answer key:
1. –80
2. 130
3. –125
4. –18
5. –6
6. 220
7. –60
8. –81
9. –1 *
10. 324
**Note: Did you learn something about exponents? If the exponent is even, the answer is positive. If the exponent is odd, the answer is negative.