Download Student Worksheets for Important Concepts

Dynamic Solution Exercise for Basic Mathematics
Arithmetic to Algebra – For the Student – Level 1
Concept #1 – Division by Zero
Division by zero is always an important concept we work on with students and introducing examples
such as the following is helpful. It is important to start with this concept. Notice that in normal division,
you should always be able to multiply the divisor time the quotient to get back to the dividend.
6
= 3 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 2 ∗ 3 = 6
2
0
= 0 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 2 ∗ 0 = 0
2
6
= ? 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 0 ∗? = 6
0
Since there is no number times zero that will give us 6, we say that division by zero is undefined.
Answer the following problems.
12
=
𝑏𝑒𝑐𝑎𝑢𝑠𝑒 ____ ∗ ___ = ____
3 _____
0
= ____ 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 ____ ∗ ____ = ____
5
9
= ____ 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 ____ ∗ ____ = ____
0
We can include examples such as the following. Use the order of operations to add the numbers in the
numerator and subtract the numbers in the denominator. Once we have a division by zero, we know
the answer is undefined.
12 + 8 20
=
= 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
4−4
0
In the following example, we subtract the numbers in the numerator, add the numbers in the
denominator, and once we see we have zero divided by a number we know the answer is zero.
15 − 15 0
= =0
5+5
5
Find the answer to the following problems.
10 − 10
=
7+2
15 + 3
=
8−8
15 + 7
=
5−5
9−9
=
5+5
Dynamic Solution Exercise for Basic Mathematics
Arithmetic to Algebra - For the Student – Level 1
Concept #2 – Comparing Addition verses Multiplication
Look at the first example. In addition, notice that you can’t divide out/cancel the 3’s, or the 2 into 4. If
you try to divide out/cancel the 3 into the 3 and the 2 into the 4, and then add, you don’t get the correct
answer.
3+2 5
=
3+4 7
In multiplication, you can divide out/cancel the common factors and get the same answer. You can
divide out/cancel the 5 into the 10 and the 7 into the 21.
5 7
1 1 1
∗
= ∗ =
10 21 2 3 6
Notice that you could multiply the numerators and denominators first, you can still divide out/cancel the
common factor of 35 to get the same result.
5 7
35
35 ∗ 1
1 1
∗
=
=
=1∗ =
10 21 210 35 ∗ 6
6 6
Work each of the two following problems using the correct method showing all steps. Simplify as
needed and cancel common factors where possible.
Problem:
5+7
=
5 + 14
Problem:
3 4
∗
=
9 16
The following examples give you some slight variations to the original problems to practice your skills.
3
3
=
3+4 7
Notice we can’t divide out/cancel since the 3 and 4 are added. They are not factors.
3 ∗ 4 12
=
3+4
7
Notice we can’t divide out/ cancel since the 3 and 4 in the denominator are added, not multiplied. The 3
and 4 in the denominator are not factors.
3∗2
1 1
=1∗ =
3∗4
2 2
In the last example, we could divide out/cancel the 3 into the 3 and the 2 into the 4 since it is
multiplication and we are dividing out/canceling factors.
Simplify the following problems showing your steps.
5
=
5+4
5∗7
=
5+7
5∗6
=
5∗8
Dynamic Solution Exercise for Basic Mathematics
Arithmetic to Algebra - For the Student – Level 1
Concept #3 - Location of Negative Sign in a Fraction
We have discussed that a +/- or -/+ equal a negative answer. So the following is true:
3 −3
3
− =
=
4
4
−4
You can see that moving the negative sign from the numerator to the denominator makes an equivalent
expression.
A simple explanation for this might be that
3
−4
3
1
3
−1
= −4 ∗ 1 = −4 ∗ −1 =
−3
4
You are not changing the value of the fraction, just rewriting it by multiplying by a well chosen one, in
this case, -1/-1 = +1. This shows you can move the negative sign from the numerator to the
denominator without changing the value of the fraction.
Given this and thinking about adding fractions, we look at the following examples.
Example:
2 3 5
+ =
7 7 7
Example:
5
3
5 −3 5 + (−3) 2
+
= +
=
=
7 −7 7
7
7
7
Work the following problems showing your work.
Problem:
2 5
+ =
9 9
Problem:
2
5
+
=
9 −9
This example includes finding the common denominator
Example:
2
3
2 −3 2 4 −3 3
8 −9 −1
+
= +
= ∗ +
∗ =
+
=
3 −4 3
4
3 4
4 3 12 12 12
Work the following problem using the same process and show your steps.
Problem:
2
3
+
=
5 −4
Dynamic Solution Exercise for Basic Mathematics
Arithmetic to Algebra – For the Student – Level 1
Concept #4 – The Distributive Property
Here is an interesting property of our number system illustrated by the following example. We get the
same result regardless of which way we perform the process. This property of our number system is
called the Distributive Property.
In the first line we add the numbers in the parenthesis using the order of operations followed by
multiplying the sum by 5.
5(3 + 4) = 5*7 = 35
In the next example we multiply the 5 by each number in the parenthesis and then add. We say that
we have distributed the 5 to each number in the sum and then added the results. We are applying the
Distributive Property.
5(3 + 4) = 5*3 + 5*4 = 15 + 20 = 35
This property of our number system is called the distributive property and it will be used extensively
as you progress in your study of mathematics including algebra.
Notice we can follow the same process regardless of the number of items being added inside the
parenthesis.
5(3 + 4 + 7) = 5*14 = 70
5(3 + 4 + 7) = 5*3 + 5*4 + 5*7 = 15 + 20 + 35 = 70
We can apply the distribute property when subtraction is involved since a subtraction can be written as
an addition problem.
4(7 – 2) = 4(7 + -2) = 4*7 + 4*(-2) = 28 + -8 = 20
Work the following problems both ways showing your work as demonstrated above.
Problem: 5(8 + 7) =
Problem: 6(9 + 4) =
Problem: 5(8 – 3) =
Work the following two examples both ways showing your work as demonstrated above.
Problem: 6(9 + 5 + 4) =
Problem: 4(8 + 3 + 5 + 2) =
Dynamic Solution Exercise for Basic Mathematics
Arithmetic to Algebra – For the Student – Level 1
Concept #5 - Using the Distributive Property in Reverse - Factoring
The next two examples show how to use the distributive property in reverse. We have found the
greatest common factor 7, and written each number as a product of 7 and the other factor that makes it
equivalent to the original number. Notice we can move the 7 to the front, (we say we have factored
out the 7), you can then add the numbers left in the parenthesis and multiply to get the same answer.
Example:
42 + 56 = 98 The greatest common factor is 7
42 + 56 = 7*6 + 7*8 = 7(6+8) = 7(14) = 98
Example:
10 + 15 + 25 = 50 The greatest common factor is 5
10 + 15 + 25 = 5*2 + 5*3 + 5*5 = 5(2 + 3 + 5) = 5*10 = 50
Follow the pattern of these two examples to remove the greatest common factor and simplify. Check
your answer by adding the original numbers.
Problem: 14 + 35 =
Problem:
12 + 18 + 30 =
Dynamic Solution Exercise for Basic Mathematics
Arithmetic to Algebra – For the Student – Level 1
Concept #6 - Reducing Fractional expressions to lowest terms by factoring and reducing.
We will use this concept to simplify a quotient as shown in the following examples.
Keep in mind the idea of reducing fractions to lowest terms.
Example:
12
15
3∗4
3
4
4
4
= 3∗5 = 3 ∗ 5 = 1 ∗ 5 = 5
We are able to divide out/cancel common factors, in this case 3.
Example: We will write each number as a product of 7 and the other factor. We then factor 7 from each
sum. Since 7*(6 + 8) is a product of two factors, and 7*(4 + 9) is a product of two factors, we can divide
out/cancel the common factor of 7 and simplify.
42 + 56
7∗6 + 7∗8
7(6 + 8) (6 + 8) 14
=
=
=
=
28 + 63
7∗4+7∗9
7(4 + 9) (4 + 9) 13
Notice we can’t divide out/cancel the 4 and the 8 or take a 3 from the 6 and 9 since these numbers are
added. We can only divide out/cancel common factors.
Notice we could solve this problem by following the order of operations. We would add the numbers in
the numerator and denominator, and reduce to lowest terms.
42 + 56 98 7 ∗ 14
14 14
=
=
=1∗
=
28 + 63 91 7 ∗ 13
13 13
Problem: Work the following problem both ways showing all steps to prove your answer.
15 + 35
=
20 + 45
In the following example, you can see that both 3 and the quantity (3+4) are common factors that can
be divided out/canceled.
9 + 12
3∗3 + 3∗4
3(3 + 4) 1(3 + 4) 1
=
=
=
=
27 + 36
9∗3+9∗4
9(3 + 4) 3(3 + 4) 3
Notice we could just simplify by adding the numerators and denominators and then reducing to lowest
terms.
9 + 12
21 7 ∗ 3
3 1
=
=
=1∗ =
27 + 36 63 7 ∗ 9
9 3
Work the following problem both ways showing all steps to prove your answer.
Problem:
10 + 14
=
20 + 28
Here is a similar example and problem using subtraction.
30 − 6
6 ∗ 5 − 6 ∗ 1 6(5 − 1) 6(4) 1
=
=
=
= =1
48 − 24 6 ∗ 8 − 6 ∗ 4 6(8 − 4) 6(4) 1
Notice we could follow the order of operations and subtract in the numerator and denominator first.
30 − 6
24
=
=1
48 − 24 24
Work the following problem using the same pattern.
20 − 15
=
40 − 30
Here is an example with a single number in the denominator.
9 + 15 3 ∗ 3 + 3 ∗ 5 3(3 + 5) 3(8) 8
=
=
=
= =2
12
3∗4
3(4)
3(4) 4
We could follow the order o f operations and add the numbers in the numerator to get the same
answer.
9 + 15 24 2
=
= =2
12
12 1
Work the following problem by factoring out the greatest common factor in the numerator as shown
above and check by adding the numbers in the numerator first and then reducing.
55 + 35
=
45
The last example and problem involves having a single term in the numerator. We factor the greatest
common factor 4 from the denominator and then divide out/cancel the factor of 4.
16
4∗4
4∗4
4
4
=
=
=
=
4 + 8 4 ∗ 1 + 4 ∗ 2 4(1 + 2) (1 + 2) 3
Notice we can’t divide out/ cancel the 2 into the 4 on the next to last step since they are not factors.
We could follow the order o f operations and add the numbers in the numerator to get the same
answer.
16
16 4 ∗ 4 4
=
=
=
4 + 8 12 4 ∗ 3 3
Work the following problem by factoring out the greatest common factor in the denominator as shown
above and check by adding the numbers in the denominator first and then reducing.
21
=
3 + 12
Dynamic Solution Exercise for Basic Mathematics
Arithmetic to Algebra - For the Student – Level 1
Concept #7 - Multiplication – Rewriting to use the Distributive Property
An alternative way to perform multiplication is shown in the next example. You are using repeated
distribution to get the final product by multiplying each number in the first quantity times each number
in the second quantity, and adding the results to get the final answer. The 30 is distributed to the 40
and the 5. The 2 is then distributed to the 40 and the 5. The sums are added to get you final product.
(32)(45) = (30+2)(40+5) = 30*40 + 30*5 + 2*40 + 2*5 = 1200 + 150 + 80 + 10 = 1440
There are many ways to write each number as a sum or difference of two numbers as can be seen in the
following example.
(32)(45) = (40 – 8)(40 + 5) = 40*40 + 40*5 - 8*40 - 8*5 = 1600 +200 -320 – 40 = 1440
Perform the following multiplication using the process illustrated above.
(23)(34) = (20 + 3)(30 + 4) =
Find another set of sums or differences that would be equal to (23)(34) and see if you can produce the
same answer. Show your work.
We could use the same concept when multiplying mixed numbers such as the following problem.
1
3
1
3
3 1
1 3
3 ∗ 2 = (3 + ) ∗ (2 + ) = 3 ∗ 2 + 3 ∗ + ∗ 2 + ∗
2
4
2
4
4 2
2 4
9
3 48 18 8 3 77
5
6+ +1+ =
+
+ + =
=9
4
8
8
8 8 8
8
8
Remember, we have taught you to rewrite the problem mixed numbers as improper fractions and then
multiply.
1
3 7 11 77
5
3 ∗2 = ∗
=
=9
2
4 2 4
8
8
Work the following problem as shown above using the distributive property and check by rewriting the
mixed numbers as improper fractions and multiplying.
1
2
2 ∗3 =
4
3
Dynamic Solution Exercise for Basic Mathematics
Arithmetic to Algebra - For the Student – Level 1
Concept #8 - Quantities Equal to -1
Look at the problem of simplifying:
5−8
8−5
Start with the simplification of:
−(5 − 8) = −1(5 − 8) = −1[5 + (−8)] = −1 ∗ 5 + −1 ∗ −8 = −5 + 8 = 8 − 5
Therefore:
5 − 8 −3
5 − 8 −1(−5 + 8) −1(8 − 5)
=
= −1 𝑜𝑟 𝑤𝑒 𝑐𝑜𝑢𝑙𝑑 𝑠𝑎𝑦
=
=
= −1
8−5
3
8−5
8−5
(8 − 5)
We can divide out/cancel the common factor of (8-5) from the numerator and denominator.
Show the steps to simplify:
– (7 – 4) =
Show the steps to simplify:
4−7
=
7−4
Give it another try on this problem:
9−5
=
5−9