Download A Review of Linear Eq. in 1 Var.

Note to interview committee: These notes represent a typical 1 hour 10min. lecture on linear equations. It would
follow a previous 1 hour 10 min. on the addition and multiplication properties of equality
Linear Equations in 1 Variable & The Addition & Multiplication Prop. of Equality
Recall:
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0
x is a variable
Our goal is to rewrite a linear equation into an equivalent equation (an equation with the same
solution set) in order to derive a solution set (the answer; note that I will simply call it a solution at this
point). We will do this by forming the equivalent equation:
x=#
or
#=x
x is a variable
# is any constant
Recall:
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 is the same as A + C = B + C (since C is added to both A & B)
We use the addition property to move terms across the equal sign, so that we can get all variable
terms on one side and all constant terms on the other side. The addition property works on terms
that have been added(subtracted) to(from) other terms.
Example:
5x + 3 = 4x  4
Recall:
The Mulitiplication Property of Equality says that two equations are equivalent if both sides
have been multiplied by a nonzero constant.
a = b
is equivalent to
ac = bc
a, b & c are real number
c  0
We use the multiplication property to “remove” the numeric coefficient (a number that is multiplied
by a variable). It should only be applied only once, after the addition property has been applied.
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Example:
/3 x = 14
Y. Butterworth
Linear Equations in 1 Variable
1
Today we will cover solving algebraic equations that will require simplification, at least one application of the
addition property as well as the multiplication property. The following will be our goals:
1)
Review the process for solving an algebraic equation in one variable
2)
Solve problems with ascending degree of difficulty using the step-through process
3)
Discuss clearing an equation of fractions/decimals, and solve problems that involve clearing
4)
Discuss the 3 types of equations and their solutions
Solving Linear Equations in One Variable
Step 1:
Simplify both sides of the equation
a) Use distributive property to simplify complex terms
b) Clear the equation of fraction/decimals
c) Combine like terms on left and like terms on right
Step 2:
Step 3:
Move all variables to one side of the equation using addition prop.
Move all constants to the opposite side of the equation from variables
using addition prop.
Remove numeric coefficient of variable by multiplying by its reciprocal in order to
solve the equation (multiplication prop.)
Check your solution
Step 4:
Step 5:
Example:
a)
Solve each of the following linear equations in one variable. We will
check the last one only as a review for checking.
12x  5 = 23  2x
Note: At this time I would allow the students time to work the following two examples on their own while walking
around the classroom to answer questions. I would allow 3-5 minutes for this process, after which time I would
quickly solve the examples as you will see me do.
b)
2x + 10 + 14x = 2x  18
c)
5(2x  1)  6x = 4
Now we need to spend a little extra time working on one particular step in the simplification
process. Although the steps in simplifying can be done in any order, and some people prefer to
clear before distributing, I find that it is better to distribute before clearing for the beginning
student. Fewer errors are made in doing the steps in the following manner.
Simplifying: Getting Rid of Fractions (Clearing the Equation of Fractions)
Step 1: Distribute (although this is not necessary, I think less errors are made this way)
Step 2: Find LCD of all fractions (on both left and right)
Step 3: Multiply all terms by LCD
a)
b)
c)
Multiply symbolically
Cancel
Multiply out
Step 4: Simplify left & right sides of the equation
Y. Butterworth
Linear Equations in 1 Variable
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Example:
a)
Solve by clearing the equation of fractions before using the addition and
multiplication properties.
1
/2 x  6/7 = 1
b)
4
c)
x
/3 (5  w) = - w
/2  1 = x/5 + 2
Simplifying: Getting the Decimals Out
Step 1: Distribute (although this is not necessary, I think less errors are made this way)
Step 2: Examine all of the decimals in the equation, and determine which has the most
decimal places. Count those decimal places.
Step 3: Multiply every term in the entire equation by a factor of 10 with the same
number of zeros as the number counted in step 1.
Step 4: Simplify both sides of the equation
Example:
a)
b)
Solve by clearing each equation of decimals first.
2.7  x = 5.4
4(y + 2.81) = 2.81
Note: In a problem involving the distributive property, it may be less confusing for you to simplify the distributive
property before doing the first step.
c)
Y. Butterworth
0.60(z  300) + 0.05z = 0.70z  0.41(500)
Linear Equations in 1 Variable
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Finally, we need to discuss equations that have no solution or infinitely many solutions. An
equation that has no solution yields an untrue statement. In other words, when you go through
the process of solving such an equation you will get a false statement such as 0 = 5. This is a
contradiction (an equation with no solution) and its solution set is written as a null set (symbolized with
 or { }). An equation with infinitely many solutions yields a true statement. In other words,
when you go through the process of solving such an equation you will get a true statement such
as 5 = 5. This is called an identity (an equation with infinite solutions) and its solution set is all real
numbers. I will allow you to write . The following is a little table that shows all the
possibilities.
Type
Conditional
Contradiction
Identity
Example
2x + 5 = 5x  5
2x + 5 = 2(x  5)
2x + 5 = 2(x  5) + 15
Example:
a)
When Solved Looks Like Solution
x = 0
x = 0 or {0}
0 = -5
 or { }
0=0

Solve the following equations
2x + 5 = 8x  2(3x + 5)
b)
2(x  5) = 8x  2(3x + 5)
c)
2x + 5x  10 = 8x  2(3x + 5)
Note: Watch the conditional equations that have a solution of zero! If you move the constants before you move the
variables, it is easy to think that you have a solution that yields a false statement! If you do the problem by moving
the constants first you will get a variable expression that equals a different variable expression. This can be
mistaken as a false statement but you can not yet determine this because all the variables are not on one side and
the constants on the other! Before you can see the solution, the variable must be on one side and the constants on
the other!! For part c), you would have gotten 7x = 6x which does not yield a false statement! 
Note: I will make copies of my notes to hand out to my students for sections that require a lot of writing, such as
word problems, but the notes are always available to students on my web-site. You can find notes similar to these at
on my Algebra page under Ch. 2 notes either for Lial or Martin-Gay at
http://hhh.gavilan.edu/ybutterworth/algebra/algebraindex.htm.
Y. Butterworth
Linear Equations in 1 Variable
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