Download Section 1.1: Solving Linear Equations in One Variable

Remember to read the textbook before attempting to do your homework.
Section 1.1: Solving Linear Equations in One Variable
Warm-up: Panchito needs your help. He has been assigned a set of homework problems. Write down
every single problem that he will have to do for that assignment. Note: eoo means “every other odd”
a. 7 – 11 odd, 16 – 20 ALL, 29 – 53 eoo
b. 29 – 53 eoo
Definitions:
A set is a collection of objects. We use braces { } to enclose its elements.
A linear equation is an equation that can be written in the form:
ax + b = c, where a, b, and c are real numbers, and a ≠ 0.
The solution set of an equation is the set of all numbers that make the equation a true statement
Note: We can use set notation to list all the solutions of an equation.
e.g. The solution set of x + 1 = 9 is {8}. Why??? Go ahead and solve the equation.
Types of equations . . .
An identity is an equation that is satisfied by every real number. As a result, the solution set is
given by the set of all real numbers, ℝ.
A conditional equation is an equation that has at least ONE solution (but not an infinite number
of solutions).
An inconsistent equation is an equation that has no solution. As a result, the solution set is given
by the empty set, ∅ .
Question: When solving an equation, how will I be able to tell if the equation is inconsistent
or an identity?
Answer:
When you’re solving an equation, if the variables cancel out and you get . . .
a TRUE statement, then the equation is an identity. Therefore, the solution
set to this type of equation is given by the set of all real numbers, ℝ.
a FALSE statement, then the equation is inconsistent. Therefore, this
type of equation has no solution, ∅.
Properties of Equality
Let
a=b
If a, b, and c represent real numbers, where c ≠ 0, then we have the following properties:
Addition Property of Equality The addition property of equality states that adding the same number to
both sides of an equation does not change its solution.
If a = b, then a + c = b + c
The subtraction property of equality states that subtracting the same
Subtraction Prop. of Equality
number from both sides of an equation does not change its solution.
If a = b, then a – c = b – c
Multiplication Prop. of Equality The multiplication property of equality states that multiplying the same
number to both sides of an equation does not change its solution.
If a = b, then a • c = b • c
The division property of equality states that dividing both sides of an
Division Property of Equality
equation by a nonzero number does not change its solution.
If a = b, then a ÷ c = b ÷ c
Math 80 || Intermediate Algebra || Cerritos College
– Pg 1 –
Chapter 2 Lecture Notes by Maria Torres
Strategies for solving linear equations in one variable:
– Note: Some equations may not require all these steps –
1. Clear fractions (or decimals): ← this part is optional (but you’ll be glad you did it)
Eliminate fractions by multiplying each side by the least common denominator.
Eliminate decimals by multiplying each side by the appropriate power of 10.
1,
e.g. 0.3x + 5 = 7
← to get rid of the tenths place value, 10
multiply both sides of the equation by 10
1 ,
e.g. 0.3x + 5 = 0.27 ← to get rid of the hundredths place value, 100
multiply both sides of the equation by 100.
2. Simplify each side separately:
Clear grouping symbols, and then combine like terms on each side of the equation.
3. Isolate the variable terms on one side of the equation:
Get all the variables on one side, and get all the numbers on the other side.
4. Isolate the variable:
Get the equation with just the variable (with coefficient 1) by itself on one side.
5. Check. Substitute your answer into the original equation
(and make sure you get a “true statement” – if not, find your mistake and fix it ☺).
Example 1: Solve. You may not use a calculator to answer.
a. 3x – 5(6 – 2x) = 4(x – 8) + 3
b. 0.05x – 0.1x + 0.6 =0.04x + 2.22
x − 1 3x − 4
1
d. 4 [ 6 − (1 + 2 x)] + 10 x = 2(10 − 3 x) + 8 x
c.
−
=
2
6
3
Note-taking tip: It is recommended, not required, that you write the problems with a pen and
the solutions with a pencil . Why?? So that way the solutions stand out from the actual problems.
And as added bonus, you get a nice visual break. ☺
(rewrite each problem below and show all work)
Ran out of space???  Don’t worry! Just continue writing on a separate sheet of paper.
Don’t forget to buy your textbook soon so that way you can start working on today’s homework in a
timely manner. You can check out the textbook for FREE through the Library’s Reserve System.
Instructor also has extra copies of the textbook for you to use during her office times.
Math 80 || Intermediate Algebra || Cerritos College
– Pg 2 –
Chapter 2 Lecture Notes by Maria Torres
Remember to read the textbook before attempting to do your homework.
Section 1.2: Formulas
A literal equation is an equation which consists primarily of letters. Formulas are an example of
literal equations. A formula is an equation involving two or more variables.
Solving Literal Equations and formulas
To solve a literal equation or to solve a formula means to rewrite the equation so a different
variable stands alone on one side of the equals sign. We have to be told for which variable we
want to solve.
Step #1:
Step #2:
Step #3:
Step #4:
Identify the variable that you want to have alone.
Treat all other letters as if they were numbers.
You can add, subtract, or multiply by a letter.
You can also divide by a variable as long as it is never zero.
Use all of the rules of algebra that you learned to solve equations in the past.
You may need to add the opposite of a variable to both sides of an equation.
You might have to multiply both sides by the reciprocal of the variable.
Get the variable you want by itself on one side of the equation.
Example 1: Solve for the specified variable.
(end of Section 1.2)
Ran out of space???  Don’t panic! Just continue writing on the back of this page. ☺
Looking for more practice on literal equations? Then work on the attached worksheet, Pizzazz #72
Math 80 || Intermediate Algebra || Cerritos College
– Pg 3 –
Chapter 2 Lecture Notes by Maria Torres
Remember to read the textbook before attempting to do your homework.
Section 1.3: Applications (aka Word Problems ← your favorite!)
Tips for solving application problems, a.k.a. word problems
1. Read the entire problem: What are you trying to find? What information is given?
2. Plan your approach to the problem.
Make a drawing, list operations you will use, and/or create a table if needed.
3. Solve the problem. Identify the unknown value, call it a variable, and then define the variable.
Translate the problem into algebra to form an equation, and then solve it.
Question: What does it mean to define the variable?
Answer: This means that you are to tell the reader what the variable represents.
It’s kind of like when you are going to use a dictionary to look up the definition of a word, you
expect the dictionary to define, or tell you, what that word means.
▪ If you define your variable at any time before your write the equation, then please use a
“Let statement”, e.g. Let n = the number of nickels.
▪ But if you define your variable RIGHT AFTER you write your equation, then please use a
“where statement”, e.g. where n = the number of nickels
4. Write your answer in a complete sentence. The reader has to have a clear idea of what you are
talking about.
5. Check your answer: Does your answer make sense? Is it reasonable?
The perimeter (P) of a figure is given by the sum of all its sides – it measures the distance all around.
Rectangle:
W
A = L• W
P = 2L + 2W, where L = length, W = width
L
Ex 1: If the perimeter of a rectangular courtyard is 34 yards and the length is 2 yards longer than twice
the width, then what are the dimensions of the courtyard?
Note: Your lecture notes were copied single-sided so you may continue writing on the back, if needed.
Math 80 || Intermediate Algebra || Cerritos College
– Pg 4 –
Chapter 2 Lecture Notes by Maria Torres
Bonus Problems: Feel free to answer these problems right now. You do not have to write your
answers in complete sentences. Please show how you got your answers.
a. The sum of Panchito’s and Conchita’s savings b. The sum of Panchito’s and Conchita’s savings
is $5000. If Panchito saved $2000, then how
is $5000. If Panchito saved d dollars, then write
an expression for how much Conchita saved.
much did Conchita save?
Simple Interest Formula:
Interest = principal • rate • time
I = Prt , where
or just
I = Interest (money earned)
P = Principal (amount invested/deposited/borrowed)
r = interest rate (this is the percent) – the rate is used to calculate the amount of interest to be paid
t = time that the money is invested/deposited/borrowed, usually in years
Memory aid:
sounds like
I am pretty
I am Pre-tee
I = Prt
Example 2: Panchito had $3000 to invest. He invested part of it in an investment paying 4% annual
interest, and the rest in an investment paying 9% annual interest. If the total annual income from these
investments was $170, then how much did he invest at each rate?
Math 80 || Intermediate Algebra || Cerritos College
– Pg 5 –
Chapter 2 Lecture Notes by Maria Torres
Example 3: Panchito received a year-end bonus of $17,000 from his company and invested the money
in an account paying 6.5% annual interest. How much additional money should he deposit in an account
paying 5% so that the return on the two investments will be 6%?
Mixture Problems
Example 4: How many ounces of real fruit juice are in each container? You do not have to write your
answers in complete sentences.
10% real
30% real
orange
apple
juice (50 oz.)
juice (40 oz.)
100% real
cranberry
fruit juice (50 oz.)
Math 80 || Intermediate Algebra || Cerritos College
Refreshing
Water
– Pg 6 –
(40 oz.)
Chapter 2 Lecture Notes by Maria Torres
Example 5: A 90-ounce juice mix is made by mixing 50 ounces of a 10% orange juice and 40 ounces
of a 30% apple juice. How many ounces of real fruit juice are in the juice mix?
Solution:
orange
apple
mix
amt. of
drink →
+
% juice →
10%
amt. of
→ (amt. of pure juice)
juice
There are
=
30%
+
(amt. of pure juice)
=
(amt. of pure juice)
ounces of real fruit juice in the 90-ounce juice mix. ☺
Example 6: How many liters of a 10% alcohol solution should be mixed with 12 liters of a 20% alcohol
solution to obtain a 14% alcohol solution?
Math 80 || Intermediate Algebra || Cerritos College
– Pg 7 –
Chapter 2 Lecture Notes by Maria Torres
Ex 7: Panchito has two solutions available in the lab, one with 6% alcohol and another with 14% alcohol.
How much of each should she mix together to obtain 10 gallons of a solution that contains 12% alcohol?
Example 8: How much pure antifreeze must be mixed with 3 gal of 30% antifreeze solution to obtain
40% antifreeze solution?
(end of Section 1.3)
Math 80 || Intermediate Algebra || Cerritos College
– Pg 8 –
Chapter 2 Lecture Notes by Maria Torres
Bonus Problems for Section 1.3
BP1: How much water must be added to 20 liters of 50% antifreeze to reduce it to 40% antifreeze?
Answer: 5 liters
BP2: Panchito has two solutions available in the lab, one with 5% alcohol and another with 13% alcohol.
How much of each should he mix together to obtain 6 gallons of a solution that contains 8% alcohol?
Answer: 3.75 gal of 5% alcohol, 2.25 gal of 13% alcohol
BP3: Panchito invested some money at 3.5% simple interest, and $5000 more than three times this
amount at 4%. He earned $1440 in annual interest. How much did he invest at each rate?
Answer: $8,000 at 3.5%, $29,000 at 4%
Math 80 || Intermediate Algebra || Cerritos College
– Pg 9 –
Chapter 2 Lecture Notes by Maria Torres