Chapter 3: Solving Equations and Problem Solving
Index
Section Page
3.1 Simplifying Algebraic Expressions 2 – 8
3.2 Solving Equations: The Addition Property 9 – 13
3.3 Solving Equations: The Mult. Property 14 – 21
Ch. 3 Integrated Review 22 – 26
3.4 Solving Linear Equations in 1 Variable 27 – 36
3.5 Linear Equations in 1 Var. & Problem Solving 37 – 42
Review 43 – 54
Practice Test 55 – 59
§3.1 Simplifying Algebraic Expressions
First, let’s review some vocabulary that we encountered in section 8 of chapter 1. This is vocabulary that we need in order to talk about simplification of algebraic expressions.
Variable – Any unknown value denoted by a letter of the alphabet (most commonly “x”).
Example: a) x
b) z
Term – Number, variable, product of a number and a variable or a variable raised to a power.
Example: a) 5
b) 5x
c) xy
d) x2
Numeric Coefficient – The numeric portion of a term with a variable.
Example: What is the numeric coefficient?
a) 3x2
b) x/2
c) - 5x/2
d) – z
Constant – Any term that is a single number, which is not multiplied by a variable.
Example: What is the constant in each expression?
a) 3x + 5
b) 27y - 1
c) –72 + y
This next one is a new concept, but it follows directly from the information that we already have.
Like Term – Terms that have a variable, or combination of variables, that are raised to the exact same power.
Example: Are the following like terms?
a) 7x 10xy
b) - 15z 23z
c) t 15tv
d) 5 5w
e) xy 6xy
f) xy - 2xyz
Simplifying an algebraic expression by combining like terms means adding or subtracting terms in an algebraic expression that are alike.
Simplifying An Algebraic Expression
Step 1: Change all subtraction to addition
Step 2: Use the distributive property wherever necessary
Step 3: Group like terms
Step 4: Add numeric coefficients of like terms
Step 5: Don’t forget to separate each term in your simplified
expression by an addition symbol!!
Example: Simplify each
a) 15x + 13x
b) 8z - 9z - 35z
c) 2x + 4y + 5x - 2y + 5
d) 5xy + 2 + 7 - 2xy
e) 6(9 + 2xy)
f) 6(x + 7) - 2x + 5
g) -9(2 + y) + 4y - y
h) x + 6(6 - x) - 4x
j) 5(x + 7) + 2(x - 4)
k) –(x + y) + 2y
Note: This problem shows an important concept of how to distribute the negative sign when it is in front of parentheses.
l) 7 - (x + 2)
Note: This problem shows another important concept – dealing with the subtraction when there are parentheses following it.
The last thing that we need to discuss in this section is the use of algebraic expressions in defining areas and perimeters of figures. We will be using the area formulas that we have already discussed in chapter one and we will be using the knowledge that we have just acquired in simplifying algebraic expressions.
Example: Find the perimeter of the following figures.
a) The following is a regular octagon.
b) Ignore the fact that this figure looks regular.
Example: Find the area of the following figures.
a)
b) The following is a compound shape. Find its area.
HW §3.1
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§ 3.2 The Addition Property of Equality
Linear Equations in One Variable are equations that can be simplified to an equation with one term involving a variable raised to the first power which is added to a number and equivalent to a constant. Such an equation can be written as follows:
ax + b = c
a, b & c are constants
a ¹ 0
x is a variable
Our goal in this and the next section will be to rewrite a linear equation into an equivalent equation (an equation with the same value) in order to arrive at a solution. We will do this by forming the equivalent equation:
x = # or # = x
x is a variable
# is any constant
Remember that only one value will make a linear equation true and that is the value that we desire. This also means that we can check our value and make sure that the original equation, when evaluated at the constant, makes a true statement.
The Addition Property of Equality says that given an equation, if you add (subtract) the same thing to both sides of the equal sign, then the new equation is equivalent to the original.
a = b and a + c = b + c
are equivalent equations
Example: Form an equivalent equation by adding the opposite of the
constant term on the left.
9 + x = 12
We’ll use the additive property of equality to move all variables to one side of the equation (for simplicity, in the beginning, we will move variables to the left) and all numbers to the other side (in the beginning to the right). This is known as isolating the variable.
Isolating the variable is done by adding the opposite of the constant term to both sides of the equation (the expression on each side of the equal sign must first be simplified of course.) We add the opposite of the constant term in order to yield zero! This makes the constant disappear from the side of the equal sign with the variable! Don’t get to carried away and forget that in order to make it disappear, you must add the opposite to both sides of the equal sign!
Example: Solve by using the addition property of equality and the
additive inverse. In order to solve these, you must look at
the equation and ask yourself, “How will I make x stand
alone?” This is where the additive inverse comes in --
You must add the opposite of the constant term in order to
get the variable to “stand alone.”
a) x + 9 = 11
b) 15 = x + 2
c) x - 3 = 10
d) 8 = x - 3
e) x + 3 = 0
Yes, you can probably look at each of the above equations and tell me what the variable should be equivalent to, but the point here is to develop a method that will work when the equation is more complicated!!
How do we know that we got the right number for the above answers? We check them using evaluation!
Example: Check to problems from the previous example
a) x = 2 so we replace x with 2 and get the following
(2) + 9 = 11
11 = 11 true statement, therefore(\)this checks
b)
c)
d)
e)
Not only can we move the constant from one side of the equation to the other, but we can also move the variable term from one side of the equation to the other.
Example: Solve each equation using the addition property of equality.
a) 8 + 7y = 8y
b) 19z = 9 + 18z
Of course, every problem will not be as simplistic as the ones used as examples so far, and sometimes we will have to simplify a problem before we can solve it. We will do this by treating each side of the equation as an algebraic or numeric expression (recall algebraic expressions involve terms with variables, but no equal sign) and use the skills developed in section 1 of chapter 3.
Example: Solve the following problems.
a) 6a - 2 - 5a = -9 + 1
b) 5a + 6 - 4a = 7a + 8 - 7a
c) 3(a + 7) - 2a = 1
d) - 2(2a - 12) - (-5a) = 4
e) 3(a + 6) = 4a + 1
Note: When our equations get complicated we must always remember that once we have simplified both sides of the equation that we must remember to keep the variable positive because at this time we do not know what to do if the variable is not positive!
HW §3.2
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§ 3.3 The Multiplication Property of Equality
Let’s first introduce a concept that your book will not discuss until chapter 4, but one that I prefer to discuss when introducing the multiplication property of equality. The concept is that of a reciprocal. A reciprocal is any number which when multiplied by the number at hand yields the identity element of multiplication (recall this is one). The reciprocal can be thought of in several other ways also. The easiest way at this time will be to think of it as one over the number, so that it will look like a fraction, 1/x. Remember that whenever you multiply any number by its reciprocal, the product is one!! Let’s practice the reciprocal and then we will talk about one other important concept.
Example: Find the reciprocal of each number.
a) 5
b) -7
c) 3
d) 0
Division is multiplication by the reciprocal of the divisor!!! I have mentioned this in passing before, but now I am stating it as a new way of defining division, and we will be using the new definition so we should get used to it!
Example: Divide
a) 27(1/3)
b) 8(1/2)
c) 105(1/5)
Note: I prefer to introduce these concepts and teach the multiplication property of equality using them rather than learning it one way (the way the book shows, using division) and then having to go back and seemingly re-learn the whole process when we introduce numeric coefficients that are fractions. By learning the concept this way we can apply it to either numeric coefficients that are whole numbers or that are fractions!
The Mulitiplication Property of Equality says that two equations are equivalent if the second is the same as the first, where both sides have been multiplied by a nonzero constant.
a = b is equivalent to ac = bc
We will be using this to solve such problems as:
Example: 4x = 24
Ø By inspection we see that if x = 6, both sides are equivalent!
Ø However, if we want to isolate x what must 4x be multiplied by? Think about it’s reciprocal! (Remember that we just learned that if we multiply a number by its reciprocal we will always get 1!)
Ø Using the idea of isolating the variable and the multiplication property of equality, we can arrive at a solution for x!
Example: Solve using the multiplicative property of equality
a) 16x = 48
b) 14x = 28
c) 2x = 24
Now let’s combine what we have learned in the last section (3.2) with what we have learned in this section. We will use the addition property first to move all variables to one side and number to the other and then we will use the multiplication property to allow the variable to stand alone, thus arriving at our solution. Here are the steps that we’ll follow:
Solving Algebraic Equations
Step 1: Simplify the expressions on either side of the equal sign as much as possible
by combining all like terms in each expression.
Step 2: Move all variables to one side of the equal sign using the addition property of
equality. Try to keep the variable positive.
Step 3: Move all the numbers to the opposite side of the equal sign from the variables
using the addition property of equality.
Step 4: Remove the numeric coefficient by multiplying both sides of the equation by
its reciprocal (divide by numeric coefficient; multiplication property of
equality).
Step 5: Check your solution by evaluating the original equation at the answer.
Example: Solve the following and check
a) 2x + 3 = 13 - 4
b) 7x - 2 + 2x = 25
c) 2 - 17 - 8 = -2 + 5x - 2x
The answer to an equation can be zero. (We will find as we progress that some equations also have no solution, but at this point we will have no problems that have no solution.) The answer will be zero because we will have multiplication by zero. Do not stop solving an equation, until you have the variable standing alone, even if the equation is something equals zero, be sure that you multiply both sides by the reciprocal of the numeric coefficient and obtain the answer.
Example: Solve 7x + 2x - 9 = -9
The next concept that has great importance in this chapter is the translation of algebraic expressions. This is just like the translation of algebraic expressions that we covered in section 1.8, but they may be a little more complicated. Here are the phrasings again.
Words and Phrasing for Translation Problems, by Operation
Note: Let any unknown be the variable x.
Addition
Word / Phrasing / Algebraic Expressionsum / The sum of a number and 2 / x + 2
more than / 5 more than some number / x + 5
added to / Some number added to 10 / 10 + x
greater than / 7 greater than some number / x + 7
increased by / Some number increased by 20 / x + 20
years older than / 15 years older than John / x + 15
Note: Because addition is commutative, each expression can be written equivalently in reverse, i.e. x + 2 = 2 + x
Subtraction
Word / Phrasing / Algebraic Expressiondifference of / The difference of some number and 2
The difference of 2 and some number / x - 2
2 - x
years younger than / Sam's age if he is 3 years younger than John / x - 3
diminished by / 15 diminished by some number
Some number diminished by 15 / 15 - x
x - 15
less than / 17 less than some number
Some number less than 17 / x - 17
17 - x
decreased by / Some number decreased by 15
15 decreased by some number / x - 15
15 - x
subtract from / Subtract some number from 51
Subtract 51 from some number / 51 - x
x - 51
Note: Because subtraction is not commutative, x - 2 ¹ 2 - x