Transcript Chapter_02 - USF Math Center
Section 2.1
The Addition & Subtraction Properties of Equality
OBJECTIVES
A Determine whether a number satisfies an equation.
OBJECTIVES
B Use the addition and subtraction properties of equality to solve equations.
OBJECTIVES
C Use both properties together to solve an equation.
SOLUTIONS The replacements of the variable that make the equation a
true
statement.
“ Solved the equation.”
SOLUTIONS A number that is a solution of an equation satisfies the equation.
SOLUTIONS For example 3 is a solution of 2x = 6 because: 2 3 = 6 3 satisfies the equation.
PROCEDURE: Solving Equations by Adding or Subtracting 1. Simplify both sides if necessary.
PROCEDURE: Solving Equations by Adding or Subtracting 2. Add or subtract the same numbers on both sides so that one side contains only variables.
PROCEDURE: Solving Equations by Adding or Subtracting 3. Add or subtract the same expressions on both sides so that the other side contains only numbers.
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.1
Exercise #1
Does the number 3 satisfy the equation 6 = 9 – x ?
6 = 9 – x
Does the number 3 satisfy the equation 6 = 9 – x ?
6 = 9 – 3 Yes.
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.1
Exercise #2
Solve x – 2 7 = 3 7 .
Add 2 7 : x – 2 7 + 2 7 = 3 7 + 2 7 Simplify: x = 5 7
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.1
Exercise #3
Solve – 2x + 7 8 + 3x – 5 8 = 5 8 .
Simplify: x + 2 8 = 5 8 Subtract 2 8 : x + 2 8 – 2 8 = 5 8 – 2 8 x = 3 8
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.1
Exercise #4
+ 5 – 2x.
Use Distributive Property: 2 = 3x – 3 + 5 – 2x Simplify: 2 = x + 2 Subtract 2: 2 – 2 = x + 2 – 2 0 = x
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.1
Exercise #5
= 8 + 5x.
Use Distributive Property: 2 + 5x + 5 = 8 + 5x Simplify: 7 + 5x = 8 + 5x Subtract 5x: 7 + 5x – 5x = 8 + 5x – 5x 7 = 8 Impossible. No solution.
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.1
Exercise #6
= – 1 – 2x.
Use Distributive Property: – 3 – 2x + 2 = – 1 – 2x Simplify: Identity: – 1 – 2x = – 1 – 2x All real numbers.
Section 2.2
The Multiplication and Division Properties of Equality
OBJECTIVES
A Use the multiplication and division properties of equality to solve equations.
OBJECTIVES
B Multiply by reciprocals to solve equations.
OBJECTIVES
C Multiply by LCMs to solve equations.
OBJECTIVES
D Solve applications involving percents.
PROCEDURE: Clearing Fractions Multiply both sides of the equation by the LCM of the denominators. Or, multiply each term by the LCM.
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.2
Exercise #7
Solve 2 3 x = – 4 3 • 2 3 x = – 4 • 3 2x = – 12 2 2 x = – 6
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.2
Exercise #8
Solve – 2 3 x = – 6 – 3 2 • – 2 3 x = – 6 • – 3 2 3 • – 3 1 2 x = 9
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.2
Exercise #9
Solve x 4 + 2x 3 = 11 LCD = 12 3 12 x 4 1
4 + 12 2x 3 1
= 11 • 12 3x + 8x = 132 11x = 132 x = 12
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.2
Exercise #10
Solve x 3 – x 5 = 2 LCD = 15 5 15
1 1 5x – 3x = 30 2x = 30 x = 15
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.2
Exercise #11
Solve. 8 40
x
x – 2 5 – 2 – x + 1 8
x
= 0 + 1 5 1 5 – 40 8 1 8
x
x
LCD = 40 8x – 16 – 5x – 5 = 0 3x – 21 = 0 3x = 21 x = 7
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.2
Exercise #12
What percent of 55 is 11?
x • 55 = 11 55x = 11 x = 0.20
x = 20%
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.2
Exercise #13
Nine is 36% percent of what number?
9 = 0.36 • x 9 = 0.36x 25 = x The number is 25.
Section 2.3
Linear Equations
OBJECTIVES
A Solve linear equations in one variable.
OBJECTIVES
B Solve a literal equation for one of the unknowns.
PROCEDURE: Solving Linear Equations
x
4 1 6 7 12
x
– 2
1. Clear fractions (Multiply by the LCM of the denominator).
PROCEDURE: Solving Linear Equations
x
4 1 6 7 12
x
– 2
12
x
4 – 12 1 6 1 2
7 12
x
– 2
PROCEDURE: Solving Linear Equations 2. Remove parentheses and simplify.
3x – 2 = 7(x – 2) 3x – 2 = 7x – 14
PROCEDURE: Solving Linear Equations 3. Isolate the variable.
3x – 2 = 7x – 14 3x – 2 + 2 = 7x – 14 + 2 3x = 7x – 12
PROCEDURE: Solving Linear Equations 3x = 7x – 12 3x – 7x = 7x – 7x – 12 – 4x = – 12
PROCEDURE: Solving Linear Equations 4. If coefficient is NOT 1, divide both sides by coefficient – 4x = – 12 – 4x – 4
PROCEDURE: Solving Linear Equations – 4x – 4
x = 3
PROCEDURE: Solving Linear Equations 5. Check your answer in the original equation.
3 4 6 7 12
3 – 2
PROCEDURE: Solving Linear Equations 3 4 6 7 12 7 12 9 12 – 2 = 7
3 – 2
Literal Equation An equation that contains several variables.
D = RT and I = Prt are examples.
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.3
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.3
Exercise #14
Solve. 1 5 23
x
+ 5
3 – 5x = 23x + 115 – 5x = 23x + 112 – 28x = 112 x = – 4
x
+ 3
15 LCD = 15
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.3
Exercise #15
Solve for h in S = 1 3
r
2
h
3S = 1
r
2
h
3S 1
r
2 = h h = 3S
r
2
Section 2.4
Problem Solving: Integer, General, and Geometry Problems
OBJECTIVES
Use RSTUV method to solve:
A Integer problems B General word problems C Geometry word problems
PROCEDURE: RSTUV Method for Solving Word Problems 1. Read the problem and decide what is asked for (the unknown).
PROCEDURE: RSTUV Method for Solving Word Problems 2. Select a variable to represent this unknown.
PROCEDURE: RSTUV Method for Solving Word Problems 3. Think of a plan to help you write an equation.
PROCEDURE: RSTUV Method for Solving Word Problems 4. Use algebra to solve the resulting equation.
PROCEDURE: RSTUV Method for Solving Word Problems 5. Verify the answer.
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.4
Exercise #16
The sum of two numbers is 75. If one of the numbers is 15 more than the other, what are the numbers?
Let x = smaller number x + 15 = larger number
x
+
x
x + x + 15 = 75 2x + 15 = 75 2x = 60
The sum of two numbers is 75. If one of the numbers is 15 more than the other, what are the numbers?
Let x = smaller number x + 15 = larger number 2x = 60 x = 30 x + 15 = 45 The numbers are 30 and 45.
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.4
Exercise #17
A man invested a certain amount of money in stocks and bonds. His annual return from these investments is $840. If the stocks produce $230 more in returns than the bonds, how much money does he receive annually from each investment?
Let x = return from stocks 840 – x = return from bonds
x
x
x = 840 – x + 230 x = 1070 – x 2x = 1070 x = 535 840 – x = 305 $535 from stocks and $305 from bonds.
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.4
Exercise #18
Find the measure of an angle whose supplement is 50° less than 3 times its complement.
Let x = measure of the angle 90 – x = complement of the angle 180 – x = supplement of the angle
Find the measure of an angle whose supplement is 50° less than 3 times its complement.
180 – x = 270 – 3x – 50 180 – x = 220 – 3x 180 + 2x = 220 2x = 40 x = 20 The angle measures 20°.
Section 2.5
Problem Solving: Motion, Mixture, and Investment Problems
OBJECTIVES
Use the RSTUV method to solve:
A Motion problems B Mixture problems C Investment problems
PROCEDURE: RSTUV Method for Solving Word Problems 1. Read the problem and decide what is asked for (the unknown).
PROCEDURE: RSTUV Method for Solving Word Problems 2. Select a variable to represent this unknown.
PROCEDURE: RSTUV Method for Solving Word Problems 3. Think of a plan to help you write an equation.
PROCEDURE: RSTUV Method for Solving Word Problems 4. Use algebra to solve the resulting equation.
PROCEDURE: RSTUV Method for Solving Word Problems 5. Verify the answer.
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.5
Exercise #19
A freight train leaves a station traveling at 30 miles per hour. Two hours later, a passenger train leaves the same station traveling in the same direction at 42 miles per hour. How long does it take for the passenger train to catch the freight train?
Let x = time of the passenger train x + 2 = time of the freight train r = 42 r = 30 D = r t D = 42x
D
= 30
x
+ 2
D = 42x
D
= 30
x
+ 2
30
x x
30x + 60 = 42x 30x = 42x – 60 – 12x = – 60 x = 5 The passenger train catches the freight train in 5 hours.
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.5
Exercise #21
An investor bought some municipal bonds yielding 5% annually and some certificates of deposit yielding 7% annually. If her total investment amounts to $20,000 and her annual return is $1160, how much money is invested in bonds and how much in certificates of deposit?
Let x = amount invested in bonds at 5% 20,000 – x = amount invested in CD's at 7% I = RT interest = 0.05x
Total interest = 1160
Let x = amount invested in bonds at 5% 20,000 – x = amount invested in CD's at 7% Total interest = 1160
5x + 140,000 – 7x = 116,000 140,000 – 2x = 116,000
Let x = amount invested in bonds at 5% 20,000 – x = amount invested in CD's at 7% 140,000 – 2x = 116,000 I = RT – 2x = – 24,000 x = 12,000 20,000 – x = 8,000 $12,000 is invested in bonds at 5%, $8000 is invested in CD's at 7%
Section 2.6
Formulas and Geometry Applications
OBJECTIVES
A Solve a formula for one variable and use the result to solve a problem.
OBJECTIVES
B Solve problems involving geometric formulas.
OBJECTIVES
C Solve geometric problems involving angle measurement.
OBJECTIVES
D Solve an application.
DEFINITION Angle of measure 1
One complete revolution around a circle is 360
°,
and 1 360 of a complete revolution is 1
°
DEFINITION Complementary Angles
Two angles whose sum is 90°
are called complementary angles.
DEFINITION Complementary Angles
90
– x
30° + 60 ° = 90 °
x
DEFINITION Supplementary Angles
Two angles whose sum is 180°
are called supplementary angles.
DEFINITION Supplementary Angles
180
– x
x
50° + 130° = 180°
DEFINITION Vertical Angles Are formed in opposite sides of two intersecting lines and have equal measures.
DEFINITION Vertical Angles 1 4 2 3
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.6
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.6
Exercise #22
The cost C of riding a taxi is C = 1.95 + 0.85m, where m is the number of miles (or fraction) you travel.
a. Solve for m.
b. How many miles did you travel if the cost of the ride was $20.65?
The cost C of riding a taxi is C = 1.95 + 0.85m, where m is the number of miles (or fraction) you travel.
a. Solve for m.
C = 1.95 + 0.85m C – 1.95 = 0.85m C – 1.95
0.85
= m m = C – 1.95
0.85
The cost C of riding a taxi is C = 1.95 + 0.85m, where m is the number of miles (or fraction) you travel.
b. How many miles did you travel if the cost of the ride was $20.65?
m = C – 1.95
; C = 20.65
0.85
m = 20.65 – 1.95
0.85
= 18.70
= 22 0.85
22 miles were traveled.
Section 2.7
Properties of Inequalities
OBJECTIVES
A Determine which of two numbers is greater.
OBJECTIVES
B Solve and graph linear inequalities.
OBJECTIVES
C Write, solve, and graph a compound inequality.
OBJECTIVES
D Solve an application.
DEFINITIONS On the number line, greater numbers are always to the right of smaller ones. –3 –2 –1 0 1 3 > 1 or 1 < 3 2 3
DEFINITIONS Addition and Subtraction Properties of Inequalities.
You can add or subtract the same number on both sides of an inequality and obtain an equivalent inequality.
DEFINITIONS Addition and Subtraction Properties of Inequalities.
If x <y then x + a<y + a or x – b<y – b
DEFINITIONS Addition and Subtraction Properties of Inequalities.
If x >y then x + a>y + a or x – b>y – b
DEFINITIONS Multiplication and Division Properties of Inequalities for Positive Numbers You can multiply or divide both sides of an inequality by any positive number and obtain an equivalent inequality.
DEFINITIONS Multiplication and Division Properties of Inequalities for Positive Numbers If x <y then ax <ay or xa <
y a
DEFINITIONS Multiplication and Division Properties of Inequalities for Positive Numbers If x >y then ax >ay or xa >
y a
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.7
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.7
Exercise #24
Fill in the blank with < or > to make the resulting statement true.
a. – 3 – 5 – 5 b. – 1 3 – 3 0 3 – 1 3 0 3
Fill in the blank with < or > to make the resulting statement true.
> – 5 b. – 1 3 – 3 < – 1 3 0 0 3
Chapter 2 Equations, Problem Solving, and Inequalities Section 2.7
Exercise #25
Solve and graph the inequality.
2 4 –
x
2 .
x
2 + 4
x
4
x x
+ 2
x
4 LCD = 4
x
Solve and graph the inequality.
.
x
2
x
x
+ 2 4
x
– 1 0
Solve and graph the inequality.
b.
x + 1 Š 3 and – 2x < 6 x Š 2 and – 2x < 6 x Š 2 and x > – 3 –3 < x Š 2 – 3 0 2