Download Faculty Overview of Important Concepts

Dynamic Solution Exercises for Elementary Algebra - Overview
Arithmetic to Algebra – For the Faculty – Level 2
This handout is an overview of the topics that can be demonstrated in class during Math 0920, 0921,
and 0960 to better prepare students for Math 1310. You have many options for using these examples
and worksheets. You may use the examples in class at the appropriate time. You may want to let the
students do the worksheets as homework, or let them work in class in groups to complete them, or
complete them outside class and then work in groups to finalize their answers for presentation to the
class. Grading should be based on the students demonstrating the correct process for solving the
problem. These problems could be used as extra credit on the exams, etc. They don’t need to take a lot
of extra time, nor does a lot of homework need to be assigned. The idea is for the students to see and
try out the processes and to improve conceptual understanding. Tell them that they will be given much
more practice and explanation throughout this class and Math 1310.
Concept #1 – Comparing Addition verses Multiplication
This concept could be covered anytime after reviewing the addition and subtraction of fractions.
It is also good for students to review the difference between addition and 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
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. You don’t get the
same answer if you tried to divide out/cancel the 3’s.
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. You don’t get the same answer if you try to divide out/cancel
the 3’s and 4’s.
3∗2
1 1
3∗2
6
1
= 1 ∗ = 𝐼𝑓 𝑤𝑒 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑓𝑖𝑟𝑠𝑡 𝑤𝑒 𝑔𝑒𝑡
=
=
3∗4
2 2
3 ∗ 4 12 2
In the last example, we could divide out/cancel the 3’s and the 2 into the 4 since it is multiplication and
we are dividing out/canceling factors.
Concept #2 - Location of Negative Sign in a Fraction
This concept could be covered anytime after covering the addition of real numbers.
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
Another reason for the above statement is the fact that we show them that a negative number divided
by a positive number yields a negative answer regardless of the position of the negative sign.
When talking about slope when introducing graphing of linear equations, this is a great place to show
that a move of down 3 and to the right four is the same as a move of up three and to the left four to
illustrate the concept graphically.
In Math 0960, we can begin to show them problems like shown below. Addition and subtraction are
covered in Math 1310, but some simple examples can be presented in Math 0921 and Math 0960. We
don’t do addition of rational expressions in Math 0960, but presenting this example prepares them for
Math 1310 where we do present this problem with variable factors.
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
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
Concept #3 – Division by Zero
This concept could be covered after covering multiplication and division of real numbers.
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.
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
Concept #4 – The Distributive Property
This concept could be covered anytime after covering the Properties of the Real Number System.
The distributive property is an important concept in algebra and can be illustrated in Math 0920 and
Math 0960 with numerical examples as we introduce and reinforce the concepts. These examples can
be presented when working in the first chapter of the text and when beginning to simplify expressions
using the distributive property.
An interesting property of our number system is 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. Here we are adding
the numbers in the parenthesis and multiplying last. 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 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 Following the order of operations and adding in the parenthesis first
5(3 + 4 + 7) = 5*3 + 5*4 + 5*7 = 15 + 20 + 35 = 70 Distributing the 5 to each term and adding the results
We can also use the same property with subtraction by rewriting as addition.
4*(8-3) = 4*5 = 20
4(8 – 3) = 4[8 + (-3)] = 4*8 + 4*(-3) = 32 + (-12) = 20
Notice this is the same as the answer if you follow the order of operations and subtract in the
parenthesis first followed by multiplying by 4, verses distributing the 4 to each term in the parenthesis
and then adding. We can distribute multiplication over addition and subtraction.
The next example has unlike variable terms. Since we can’t combine unlike terms, all we can do is
distribute the number each front to each term in the parenthesis.
7(3x + 4y - 7z + 1) = 7*3x + 7*4y + 7*(-7z) + 7*1 = 21x + 28y - 49z + 7
Unlike terms can’t be combined.
Here is one more example and problem where a negative number is distributed across the parenthesis.
-5(3x – 4y + 7z – 1) = -5[3x + (-4y) + 7z + (–1)] = -15x + 20y – 35z + 5
Concept #5 - Using the Distributive Property in Reverse – Factoring
This concept could be covered anytime after covering the Properties of the Real Number System.
The next two examples show how to use the distributive property in reverse and can be done early in
the course to prepare students for factoring later in the course. We have found the greatest common
factor 5, and written each number as a product of 5 and the other factor that makes it equivalent to the
original number. Notice we can move the 5 to the front of the quantity, (We say we have factored out
the 5 in the 1st example, 7 in the 2nd example, and 5 in the 3rd example.), you can then add the
numbers left in the parenthesis and multiply to get the same answer. This is a good place to describe
the process as factoring as they will see it later when we remove the greatest common factor from a
sum or difference of terms in the chapter on factoring.
The next two examples show how to use the distributive property in reverse. We have found the
greatest common factor 5, and written each number as a product of 5 and the other factor that makes it
equivalent to the original number. Notice we can move the 5 to the front, (We say we have factored
out the 5 in the 1st example, 7 in the 2nd example), you can then add the numbers left in the
parenthesis and multiply to get the same answer. In each case we are finding the greatest common
factor and factoring it from each of the terms of the expression.
Example:
10 + 15 + 25 = 50
10 + 15 + 25 = 5*2 + 5*3 + 5*5 = 5(2 + 3 + 5) = 5*10 = 50
Example:
42 + 56 = 98
42 + 56 = 7*6 + 7*8 = 7(6+8) = 7(14) = 98
The next example has variable terms with a common factor in each term which we will factor out.
Example: 10x + 20y + 35z = 5*2x + 5*4y + 5*7z = 5(2x + 5y + 7z)
Concept #6 - Distributing -1 to a Quantity
This concept could be covered after presenting the properties of the real numbers and after
simplifying expressions using the distributive property.
Distributing a -1 throughout a quantity and can be demonstrated with a few examples.
−(5 − 8) = −1[5 + (−8)] = (−1) ∗ 5 + (−1) ∗ (−8) = −5 + 8 = 8 − 5 = 3
Notice we could perform the parenthesis first and get the same answer.
- (5 – 8) = -[5 + (-8)] = - (-3) = 3
We can introduce the same concept with variable terms.
−(𝑥 − 8) = −1[𝑥 + (−8)] = −1 ∗ 𝑥 + −1 ∗ (−8) = −𝑥 + 8 = 8 − 𝑥
-(x – y + 4z – 4) = -1[x + (–y) + 4z + (–4)] = (-1)*x + (-1)*(-y) + (-1)*4z +( -1)*(-4) = -x + y + -4z + 4
Concept #7 – Interpreting a fractional Coefficient of a Variable Term
This concept could be covered anytime after covering the section on combining like terms with
variable terms.
Students many times struggle with the interpretation of a term written with a fractional coefficient.
Examples like the following will help them to understand how to work with adding and multiplying
terms with fractional coefficients. The first two examples show that the x can be put in the numerator
or multiplied by the fraction. The 3rd example shows how to interpret the factional coefficients when
adding.
3𝑥 3 𝑥 3
= ∗ = 𝑥
4
4 1 4
𝑥 1 𝑥 1
= ∗ = 𝑥
7 7 1 7
𝑥 2𝑥 1 𝑥 2 𝑥 3
4
7
7𝑥
+
= ∗ + ∗ = 𝑥+ 𝑥= 𝑥=
2 3
2 1 3 1 6
6
6
6
𝑁𝑜𝑡𝑖𝑐𝑒 𝑡ℎ𝑎𝑡
3
3
𝑥≠
4
4𝑥
Concept #8 - Clearing Fractions in an Equation
This concept could be covered anytime after covering how to solve linear equations with variables on
one side of the equation.
For the concept of clearing fractions when solving linear equations in Math 0960, presenting these
examples may help the student to understand why we are able to clear the fractions in an equation. We
begin by showing you a true equation which you can see is true when you get a common denominator
and compare the left and right sides.
−1 3 5
+ =
28 4 7
−1 21 20
20 20
+
=
𝑠𝑜 𝑤𝑒 𝑐𝑎𝑛 𝑠𝑒𝑒 𝑡ℎ𝑎𝑡
=
28 28 28
28 28
Then put the variable x in for the -1 and solve for x:
𝑥 3 5
+ =
28 4 7
𝑥 21 20
21
+
=
𝑠𝑜 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡
𝑓𝑟𝑜𝑚 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠
28 28 28
28
[
𝑥
−1
=
]
28 28
Multiply both sides by 28 to get:
28 𝑥
−1 28
[ =
]
1 28 28 1
𝑥 = −1
This shows the justification for clearing fractions as we show in Math 0960.
28 𝑥 3 5 28
[ + = ]
1 28 4 7 1
𝑥+7∗3=5∗4
𝑥 + 21 = 20
𝑥 = −1
Concept #9 - Multiplication – Rewriting to use the Distributive Property
This concept could be covered anytime after presenting the section on the multiplication of
polynomials.
The next concept we can introduce at the Math 0921 and Math 0960 level and can be reviewed in Math
1310 is the idea of multiplying two binomials. They will see the idea of how we foil and distribute for
use in Algebra with variables and numbers. This example fits in when you are beginning to multiply
binomials.
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
Some other examples are given.
(32)(45) = [40 +(– 8)](40 + 5) = 40*40 + 40*5 (-8)*40 + (-8)*5 = 1600 +200 -320 – 40 = 1440
Here is an example involving multiplying a negative times a negative. There are many ways we could
perform the multiplication.
(47)(13) = [50+(– 3)][20 +(– 7)] = 50*20 + 50*(-7) + (-3)*20 + (-3)*-7 = 1000 – 350 - 60 + 21 = 611
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.
3
1
2 ∗4 =
4
3
Below is an example where we divide the numbers into three sums and then distribute to multiply.
(125)(213) =
= (100 + 20 + 5)(200 + 10 + 3) = 100*200 + 100*10 + 100*3 + 20*200 + 20*10 + 20*3 + 5*200 + 5*10 + 5*3
= 20000 + 1000 +300 + 4000 +200 + 60 + 1000 + 50 + 15 = 26,625
Concept #10 - Simplifying Rational Expressions – Identifying Restricted Values
This concept could be covered anytime after presenting the section on simplifying rational
expressions.
In working with rational expressions, we emphasize that division by zero is not possible. Remember
that 6/0 is undefined, so it is important that in any rational expression we do not allow the denominator
to become zero.
𝑥−8
𝑥−3
If x = 8, the rational expression would be equivalent to 0/5 = 0. If x = 3, the rational expression would
be equivalent to -5/0 which is undefined. It is important to note that our restrictions on the
acceptable values for the rational expression (domain) are based on the original rational expression(s)
before any dividing out/canceling is done.
In the following example, after you factor the numerator and denominator, it is important to note that x
can’t equal -3 or 8 even though we cancel the factor (x + 3). Our restriction is based on the original
rational expression given. At this point, leave it at that, but later in Math 1310, it may be mentioned
that if this represented a function to be graphed, there would be a hole in the graph at that point.
In the following example, after you factor the numerator and denominator, it is important to note that x
can’t equal -3 or 8 even though we divide out/cancel the factor (x + 3). Our restriction is based on the
original rational expression given. Furthermore,
𝑥 2 − 2𝑥 − 15 (𝑥 + 3)(𝑥 − 5) (𝑥 − 5)
=
=
𝑎𝑠 𝑙𝑜𝑛𝑔 𝑎𝑠 𝑥 ≠ −3𝑜𝑟 8
𝑥 2 − 5𝑥 − 24 (𝑥 + 3)(𝑥 − 8) (𝑥 − 8)
Concept #11 – Simplifying Rational Expressions with Numerical Values
This concept could be covered anytime after presenting the section on simplifying rational
expressions.
Once students understand the distributive property as shown above, we can move to dividing
out/canceling common factors in rational expressions beginning with numerical examples. Here we can
show them that only common factors can be canceled, in this case the 7’s. These examples come in
when presenting the multiplication and division of rational expressions, but could be done early in the
course as a preparation for what is coming later. They can be used again when you get to the chapter
on multiplying and dividing rational expressions.
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
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
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
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
Concept #12 - Simplifying Rational Expressions Equal to One
Another example to be presented in Math 0960 and Math 1310 would be this one dealing with dealing
with a quotient equal to -1. Begin with the examples below and progress to the fraction.
−(5 − 8) = −1(5 − 8) = −1 ∗ 5 + −1 ∗ −8 = −5 + 8 = 8 − 5
−(𝑥 − 8) = −1(𝑥 − 8) = −1 ∗ 𝑥 + −1 ∗ −8 = −𝑥 + 8 = 8 − 𝑥
The first example shows that we can make the quotient equal to -1 after we have a common factor. The
2nd example shows the same process with a variable expression.
5 − 8 −3
−1(−5 + 8) −1(8 − 5)
=
= −1 =
=
= −1
8−5
3
8−5
(8 − 5)
𝑥 − 8 −1(−𝑥 + 8) −1(8 − 𝑥)
=
=
= −1
8−𝑥
8−𝑥
(8 − 𝑥)
This example shows us to factor out the -1 to divide out/cancel the quantity (5 - 8) and simplify the
numerator and denominator since nothing more can be divided out/canceled. The 2nd example is similar
with variable factors, the factor (x – 8) can be divided out/canceled.
(8 − 5)(7 − 3)
−1(−8 + 5)(7 − 3)
−1(5 − 8)(7 − 3)
−1(7 − 3) −4
=
=
=
=
(5
(5
(5 − 8)(6 + 9)
− 8)(6 + 9)
− 8)(6 + 9)
(6 + 9)
15
(8 − 𝑥)(𝑥 − 3)
−1(−8 + x)(x − 3)
−1(x − 8)(x − 3)
−1(𝑥 − 3) −𝑥 + 3
=
=
=
=
(𝑥
(𝑥
(𝑥 − 8)(𝑥 + 9)
− 8)(𝑥 + 9)
− 8)(𝑥 + 9)
(𝑥 + 9)
𝑥+9
Concept #13 – Simplifying Rational Expressions with Variable Terms
This concept could be covered anytime after presenting the section on simplifying rational
expressions.
The first example requires us to factor -1 from both factors. You can also divide out/cancel the common
quantity (3+4). The second example involves variable terms and illustrates taking out a common factor
of 3 and dividing out/canceling the common quantity which is the factor (3x + 4). In both examples we
are dividing out/canceling only common factors.
(−9 − 12)
3(−3 − 4)
−3(3 + 4)
−1
=
=
=
(27 + 36)
9(3 + 4)
9(3 + 4)
3
Notice you get the same answer if you simplify the numerator and denominator first, and then reduce.
−9 − 12 −21 7 ∗ −3 −3 −1
=
=
=
=
27 + 36
63
7∗9
9
3
We now introduce variable terms into the rational expressions
−9𝑥 − 12
3(−3x − 4)
−3(3x + 4)
−1
=
=
=
27𝑥 + 36
9(3𝑥 + 4)
9(3𝑥 + 4)
3
This example allows us to factor out a common factor of 2, but we can’t divide out/cancel a common
quantity or factor here.
(−4 + 3)
−8 + 6
2(−4 + 3)
−1
=
=
=
(7
14 − 4
2(7 − 2)
− 2)
5
Notice we get the same answer if we simplify the numerator and denominator first and then reduce.
−8 + 6
−2
−1
=
=
14 − 4
10
5
The second example is similar, but has variable factors and ends up with unlike linear factors that can’t
be simplified any further.
(−4x + 3)
−8𝑥 + 6
2(−4x + 3)
=
=
(7𝑥 − 2)
14𝑥 − 4
2(7𝑥 − 2)
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 of operations and add the numbers in the numerator to get the same answer.
16
16 4 ∗ 4 4
=
=
=
4 + 8 12 4 ∗ 3 3
We could have a similar type of problem with variable terms.
18𝑥
3 ∗ 6𝑥
6𝑥
=
=
6𝑥 + 9 3(2𝑥 + 3) 2𝑥 + 3
Notice we can’t divide out/cancel the 2x into the 6x since the 2x is added in the denominator and not a
factor. We can divide out/cancel the factor of 3 since it is a factor in both the numerator and
denominator.