class01 math grade10 algebra notes

Math Grade 10 notes on class 1

1

Algebra

►Basic Knowledge

1. Laws of Exponents

The exponent of a number says how many times to use the number in a multiplication.

In this example: 82 = 8 × 8 = 64

So an Exponent just saves you writing out lots of multiplies!

Example: 7a

7a a a a a a a a aaaaaaa

Notice how I just wrote the letters together to mean multiply? We will do that a lot here.

Example: 6y y y y y y y yyyyyy

The Key to the Laws

Writing all the letters down is the key to understanding the Laws

Example: 2 3 5( )( )x x xx xxx xxxxx x

Which shows that 2 3 2 3 5x x x x , but more on that later!

A fractional exponent like 1/n means to take the nth root: 1

nnx x

2. Using Exponents in Algebra

You might like to read the page on Exponents first.

1) Whole Number Exponents

The exponent "n" in an says how many times to use a in a multiplication:

...na a a a n a

2) Negative Exponents

Example: 5-3

= 1/53 = 1/125 = 0.008

A positive exponent an is equal to 1/a

-n (1 divided by the negative exponent)

1n

na

a

3) Variables with Exponents

What is a Variable with an Exponent?

A Variable is a symbol for a number we don't know yet. It is usually a letter like x or y.


2

An exponent (such as the 2 in x2) says how many times to use the variable in a multiplication.

Example: y2 = yy

(yy means y multipled by y, because in Algebra putting two letters next to each other means to

multiply them)

Likewise z3 = zzz and x

5 = xxxxx

4) Exponents of 1 and 0

a) Exponent of 1

If the exponent is 1, then you just have the variable itself (example x1 = x)

We usually don't write the "1", but it sometimes helps to remember that x is also x1

b) Exponent of 0

If the exponent is 0, then you are not multiplying by anything and the answer is just "1" (example y0 =

1)

5) Multiplying Variables with Exponents

So, how do you multiply this:

(y2)(y

3)

We know that y2 = yy, and y

3 = yyy so let us write out all the multiplies:

y2 y

3 = yyyyy

That is 5 "y"s multiplied together, so the new exponent must be 5:

y2 y

3 = y

5

But why count the "y"s when the exponents already tell us how many?

The exponents tell us that there are two "y"s multiplied by 3 "y"s for a total of 5 "y"s:

y2 y

3 = y

2 + 3 = y

5

So, the simplest method is to just add the exponents! (Note: this is one of the Laws of Exponents)

6) Negative Exponents

Negative Exponents Mean Dividing!

3 1x

x

, 2

2

1x

x

, 3

3

1x

x

Get familiar with this idea, it is very important and useful!

7) Dividing

33 2 1

2

y yyyy y y

y yy

OR, you could have done it like this:


3

33 2 3 2 1

2

yy y y y y

y

So, it just subtract the exponents of the variables you are dividing by!

You can see what is going on if you write down all the multiplies, then "cross out" the variables that

are both top and bottom:

3 2 2

2 2

x yz xxxyzz xx x

xy z xyyzz y y

3. Complex Fractions

A complex fraction is a fraction where the numerator, denominator, or both contain a fraction.

Example 1: 3

1/ 2is a complex fraction. The numerator is 3 and the denominator is 1/2.

Example 2: 3/ 7

9is a complex fraction. The numerator is 3/7 and the denominator is 9.

Example 3: 3/ 4

9 /10is a complex fraction. The numerator is 3/4 and the denominator is 9/10.

Rule:

To multiply two complex fractions, convert the fractions to simple fractions and follow the steps you

use to multiply two simple fractions.

Example:

Calculate 3/ 4 8

1/ 2 3/16 .

Solution 1:

Convert the numerator 3/ 4

1/ 2to a simple fraction.

3/ 4

1/ 2can be written

3/ 4 3 1 3 2 3 2 3 31

1/ 2 4 2 4 1 2 2 2 2

.

Convert the denominator to a simple fraction. 8

3/16can be written

8 8 3 8 16 8 16 128

3/16 1 16 1 3 1 3 3

.


4

The problem 3/ 4 8

1/ 2 3/16 can now be written

3 128

2 3 .

Multiply the numerators and multiply the denominators.3 128

2 6

.

The problem is reduced by 3 2 64 2 3 64

642 3 2 3 1

.

Solution 2:

Convert the 8 to a fraction, multiply the numerators, and multiply the denominators.

3 8 6 323/ 4 8 6 324 1 1 1 64

1 3 3 321/ 2 3/16 3

2 16 32 1

.

4. Remember these Identities

1) abxbaxbxax )())(( 2

2) 222 2)( bababa

3) 32233 33)( babbaaba

4) ))((22 bababa

5) ))(( 2233 babababa

6) 1 2 3 2 2 1( )( ) ( is a positive integer)n n n n n n na b a b a a b a b ab b n

7) cabcabcbacba 222)( 2222

8) ))((3 222333 cabcabcbacbaabccba

9) a

b

c

b

a c

b

10)

bd

bcad

d

c

b

a

11) a

b

c

d

ac

bd 12)

a

b

c

d

ad

bc


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►Examples

Example 1

1. Find the sum of the following: 1 1 1 1 1 1 1 1 1 1 1

(1 ) ( ) ( ) ... ( )2 3 2 3 5 2 5 7 2 51 53

Solution:

This sequence “telescopes” if you multiply it our and rearrange the terms. It will look like:

3

1

2

1

2

1+

5

1

2

1

3

1

2

1+

7

1

2

1

5

1

2

1+···+

51

1

2

1

49

1

2

1+

53

1

2

1

51

1

2

1

=2

1+

3

1

2

1

3

1

2

1+

5

1

2

1

5

1

2

1+

7

1

2

1

7

1

2

1+···+

51

1

2

1

51

1

2

1-

2

1·

53

1

=2

1-

2

1·

53

1=

2

1

53

11 =

2

1

53

52=

53

26

Example 2

1

11

1

a a

c bb b

a b

, where a, b, and c are positive integers, find the number of different ordered triples (a,

b, c) such that a+2b+c ≤ 40.

Solution:

ab +ac+ bc

11bc+ab +ac

bc

ac

11ac

bc , or a=11b

By substitution, the condition a +2b + c = 40 becomes 13b +c = 40.

Since b and c are positive integers, then b can only take on the values 1, 2, or 3. The values of a

correspond directly to the values of b, since a = 11b.

If b=3, there is one corresponding value of c. When b = 2, there are 14 possible values of c. Finally

if b=1, there are 27 possible values of c.

Therefore, the number of different ordered triples satisfying the given conditions is 1+14+27=42.


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Example 3

There are four unequal, positive integers a, b, c, and N such that N=5a+3b+5c. It is also true that

N=4a+5b+4c and N is between 131 and 150. What is the value of a b c?

Solution:

We are told that N=5a+3b+5c (1) and N=4a+5b+4c (2). Multiply equation (1) by 4 to get

4N=20a+12b+20c (3). Similarly, multiply equation (2) by 5 to get 5N=20a+25b+20c (4).

Subtract equation (3) from equation (4) to get N=13b.

Since N and b are both positive integers with 131<N<150, N must be a multiple of 13. The only

possible value for N is 143, when b =11.

Substitute N = 143 and b =11 into equation (1) to get 143 =5a+3(11)+ 5c

110=5a+5c

Thus, the value of a + b + c is 22+11=33.

►Questions in class

1. If ac dbc d42 and cd3, what is the value of a b c d?

2. If 32

+32

+ 32 =3

a, what is the value of a?

3. What is the sum of the digits of the integer equal to 777 777 777 777 7772-222 222 222 222

2232 ?

4. Find the value of 2003 2002

2002 2003

(4 )( 3 )

(6 )( 2 )


7

5

written as the sum of exactly two elements of the set?

6. The time on a digital clock is 5:55. How many minutes will pass before the clock next shows a

time with all digits identical?

7. In an election for class president, 61 votes are cast by students who are voting to choose one of

four candidates. Each student must vote for only one candidate. The candidate with the highest

number of votes is the winner. What is the smallest number of votes the winner can receive?

8. Let N be the smallest positive integer whose digits have a product of 2000. What is the sum of

the digits of N?

9. The 6 members of an executive committee want to call a meeting. Each of them phones 6

different people, who in turn each calls 6 other people. If no one is called more than once, how

many people will know about the meeting?

10. A “double-single” number is a three-digit number made up of two identical digits followed by a

different digit. For example, 553 is a double-single number. How many double-single numbers are

there between 100 and 1000?

11. The product of the digits of a four-digit number is 810. If none of the digits is repeated, what is

the sum of the digits?

12. In 2004, Gerry downloaded 200 songs. In 2005, Gerry downloaded 360 songs at a cost per song

which was 32 cents less than in 2004. Gerry’s total cost each year was the same. What was the cost

of downloading the 360 songs in 2005?

13. A box contains apple and pears. An equal number of apples and pears are rotten.

2

3 of all of the apples are rotten.

3

4 of all of the pears are rotten.

What fraction of the total number of pieces of fruit in the box is rotten?

class01 math grade10 algebra notes

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