chapter · equations have expressions on each side of the equals sign. the sentence “14 – 10 =...

CHAPTER

2 Solving Equations and Inequalities

Slide 2 Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

2.1 Solving Equations: The Addition Principle 2.2 Solving Equations: The Multiplication Principle 2.3 Using the Principles Together 2.4 Formulas 2.5 Applications of Percent 2.6 Applications and Problem Solving 2.7 Solving Inequalities 2.8 Applications and Problem Solving with Inequalities

OBJECTIVES

2.1 Solving Equations: The Addition Principle


a Determine whether a given number is a solution of a given equation.

b Solve equations using the addition principle.


Equation


An equation is a number sentence that says that the expressions on either side of the equals sign, =, represent the same number.




Here are some examples of equations: Equations have expressions on each side of the equals sign. The sentence “14 – 10 = 1 + 3” asserts that the expressions 14 – 10 and 1 + 3 name the same number. Some equations are true. Some are false. Some are neither true nor false.

EXAMPLE



Determine whether each equation is true, false, or neither.


EXAMPLE Solution





Solution of an Equation


Any replacement for the variable that makes an equation true is called a solution of the equation. To solve an equation means to find all of its solutions. One way to determine whether a number is a solution of an equation is to evaluate the expression on each side of the equals sign by substitution. If the values are the same, then the number is a solution.

EXAMPLE



5 Determine whether 19 is a solution of 7x = 141.


Since the left-hand and the right-hand sides are not the same, 19 is not a solution of the equation.




We now begin to consider principles that allow us to start with an equation like x + 6 = 13 and end up with an equivalent equation, like x = 7, in which the variable is alone on one side and for which the solution is easier to find.


Equivalent Equations


Equations with the same solutions are called equivalent equations.


The Addition Principle for Equations


For any real numbers a, b, and c, a = b is equivalent to a + c = b + c.




Let’s solve the equation x + 6 = 13 using the addition principle. We want to get x alone on one side. To do so, we use the addition principle, choosing to add –6 because 6 + (–6) = 0.




When we use the addition principle, we sometimes say that we “add the same number on both sides of the equation.” This is also true for subtraction, since we can express every subtraction as an addition. That is, since a – c = b – c is equivalent a + (–c) = b + (–c), the addition principle tells us that we can “subtract the same number on both sides of the equation.”

EXAMPLE



6 Solve:


EXAMPLE Solution



6


EXAMPLE



9 Solve:


EXAMPLE Solution



9


CHAPTER




OBJECTIVES

2.2 Solving Equations: The Multiplication Principle


a Solve equations using the multiplication principle.


The Multiplication Principle for Equations


EXAMPLE



1 Solve: 5x = 70.


To get x alone, we multiply by the multiplicative inverse, or reciprocal, of 5. Then we get the multiplicative identity 1 times x, or which simplifies to x. This allows us to eliminate 5 on the left.

EXAMPLE Solution



1





The multiplication principle also tells us that we can “divide on both sides of the equation by the same nonzero number.” This is because dividing is the same as multiplying by a reciprocal. That is,

EXAMPLE



3 Solve:


EXAMPLE



4 Solve:


EXAMPLE Solution



4


EXAMPLE Solution



4


Another approach is to multiply both sides by –1.

EXAMPLE



6 Solve:


In practice, it is generally more convenient to divide on both sides of the equation if the coefficient of the variable is in decimal notation or is an integer. If the coefficient is in fraction notation, it is usually more convenient to multiply by a reciprocal.

EXAMPLE Solution



6


EXAMPLE



7 Solve:


EXAMPLE Solution



7


EXAMPLE



8 Solve:


(Use the multiplication principle to solve an equation that involves division.)

EXAMPLE Solution



8


EXAMPLE Solution



8


There are other ways to solve the equation in Example 8. One is by multiplying by on both sides as follows:

CHAPTER




OBJECTIVES

2.3 Using the Principles Together


a Solve equations using both the addition principle and the multiplication principle.

b Solve equations in which like terms may need to be collected.

c Solve equations by first removing parentheses and collecting like terms; solve equations with an infinite number of solutions and equations with no solutions.




Consider the equation 3x + 4 = 13.

It is more complicated than those we discussed in the preceding two sections.

In order to solve such an equation, we first isolate the x-term, 3x, using the addition principle.

Then we apply the multiplication principle to get x by itself.

EXAMPLE



1 Solve:


EXAMPLE



2 Solve:


EXAMPLE Solution



2


EXAMPLE



5 Solve:


If there are like terms on one side of the equation, we collect them before using the addition principle or the multiplication principle.

EXAMPLE Solution



5


EXAMPLE



6 Solve:


If there are like terms on opposite sides of the equation, we get them on the same side by using the addition principle. Then we collect them. In other words, we get all the terms with a variable on one side of the equation and all the terms without a variable on the other side.

EXAMPLE Solution



6


EXAMPLE



7 Solve:


If there are like terms on one side at the outset, they should be collected first.

EXAMPLE Solution



7





In general, equations are easier to solve if they do not contain fractions or decimals. Consider, for example, the equations




If we multiply by 4 on both sides of the first equation and by 10 on both sides of the second equation, we have

The first equation has been “cleared of fractions” and the second equation has been “cleared of decimals.”

EXAMPLE



8 Solve:


The easiest way to clear an equation of fractions is to multiply every term on both sides by the least common multiple of all the denominators.

EXAMPLE Solution



8


EXAMPLE



9 Solve:


The greatest number of decimal places in any one number is two. Multiplying by 100, which has two 0’s, will clear all decimals.

EXAMPLE Solution



9


EXAMPLE


c Solve equations by first removing parentheses and collecting like terms.

10 Solve:


To solve certain kinds of equations that contain parentheses, we first use the distributive laws to remove the parentheses. Then we proceed as before.

EXAMPLE Solution


c Solve equations by first removing parentheses and collecting like terms.

10



An Equation-Solving Procedure


EXAMPLE


c Solve equations with an infinite number of solutions and equations with no solutions.

12 Solve:


EXAMPLE


c Solve equations with an infinite number of solutions and equations with no solutions.

13 Solve:


CHAPTER




OBJECTIVES

2.4 Formulas


a Evaluate a formula. b Solve a formula for a specified letter.

2.4 Formulas

a Evaluate a formula.


A formula is a “recipe” for doing a certain type of calculation. Formulas are often given as equations.

EXAMPLE

2.4 Formulas


1 Distance from Lightning


Consider the formula This formula gives the distance M (miles) of a lightning strike that has a thunder delay of t seconds.

Suppose it takes 10 sec for the sound of thunder to reach you after you have seen a flash of lightning. How far away did the lightning strike?

EXAMPLE Solution

2.4 Formulas


1


EXAMPLE

2.4 Formulas


3 Distance, Rate, and Time


The distance d that a car will travel at a rate, or speed, r in time t is given by

2.4 Formulas


EXAMPLE

2.4 Formulas

b Solve a formula for a specified letter.

11 Circumference


Solve for r : This is a formula for the circumference C of a circle of radius r.

EXAMPLE Solution

2.4 Formulas


11

EXAMPLE

2.4 Formulas


12 Averages


Solve for a: This is a formula for the average A of three numbers a, b, and c.

EXAMPLE Solution

2.4 Formulas


12


CHAPTER




OBJECTIVES

2.5 Applications of Percent


a Solve applied problems involving percent.




In solving percent problems, we first translate the problem to an equation. Then we solve the equation using the techniques discussed in Sections 2.1–2.3. The key words in the translation are as follows.


Key Words in Percent Traslations


EXAMPLE



1 Translate:


EXAMPLE



2 Translate:


EXAMPLE



3 Translate:


EXAMPLE



4 What number is 11% of 49?


EXAMPLE Solution



4


EXAMPLE



5 3 is 16% of what number?


EXAMPLE Solution



5


EXAMPLE



6 $32 is what percent of $50?


EXAMPLE Solution



6


EXAMPLE



7 Foreign Visitors to China


About 22 million foreign travelers visited China in 2006. Of this number, 9% were from the United States. How many Americans visited China in 2006?

EXAMPLE Solution



7


Thus, 1.98 million is 9% of 22 million, so 1.98 million Americans visited China in 2006.

EXAMPLE



8 Public School Enrollment


In the fall of 2008, 14.9 million students enrolled in grades 9–12 in U.S. public schools. This was 30% of the total enrollment in public schools. What was the total enrollment?

EXAMPLE Solution



8


EXAMPLE



9 Employment Outlook


There were 280 thousand dental assistants in 2006. This number is expected to grow to 362 thousand in 2016. What is the percent of increase?

EXAMPLE Solution



9


CHAPTER




OBJECTIVES

2.6 Applications and Problem Solving


a Solve applied problems by translating to equations.


Five Steps for Problem Solving in Algebra



To Familiarize Yourself with a Problem


EXAMPLE



1 Knitted Scarf


Lily knitted a scarf in three shades of blue, starting with a light-blue section, then a medium-blue section, and finally a dark-blue section. The medium-blue section is one-half the length of the light-blue section. The dark-blue section is one-fourth the length of the light-blue section. The scarf is 7 ft long. Find the length of each section of the scarf.

EXAMPLE Solution



1


1. Familiarize. Because the lengths of the medium-blue section and the dark-blue section are expressed in terms of the length of the light-blue section, we let

EXAMPLE Solution



1


EXAMPLE Solution



1


2. Translate. From the statement of the problem and the drawing, we know that the lengths add up to 7 ft. This gives us our translation:

EXAMPLE Solution



1


3. Solve. First, we clear fractions and then carry out the solution as follows:

EXAMPLE Solution



1


4. Check. Do we have an answer to the original problem? If the length of the light-blue section is 4 ft, then the length of the medium-blue section is , or 2 ft, and the length of the dark-blue section is , or 1 ft. The sum of these lengths is 7 ft, so the answer checks.

EXAMPLE Solution



1


5. State. The length of the light-blue section is 4 ft, the length of the medium-blue section is 2 ft, and the length of the dark-blue section is 1 ft. (Note that we must include the unit, feet, in the answer.)

EXAMPLE



3 Interstate Mile Markers


The sum of two consecutive mile markers on I-70 in Kansas is 559. Find the numbers on the markers.

EXAMPLE Solution



3


1. Familiarize. The numbers on the mile markers are consecutive positive integers. Thus if we let x = the smaller number, then x + 1 = the larger number. 2. Translate. We reword the problem and translate as follows.

EXAMPLE Solution



3


3. Solve. We solve the equation:

EXAMPLE Solution



3


4. Check. Our possible answers are 279 and 280. These are consecutive positive integers and 279 + 280 = 559, so the answers check. 5. State. The mile markers are 279 and 280.

EXAMPLE



5 Perimeter of NBA Court


The perimeter of an NBA basketball court is 288 ft. The length is 44 ft longer than the width. Find the dimensions of the court.

EXAMPLE Solution



5


1. Familiarize. We first make a drawing.

EXAMPLE Solution



5


2. Translate. To translate the problem, we substitute w + 44 for l and 288 for P:

EXAMPLE Solution



5


3. Solve. We solve the equation:

EXAMPLE Solution



5


4. Check. If the width is 50 ft and the length is 94 ft, then the perimeter is 2(50) + 2 (94) or 288 ft. This checks. 5. State. The width is 50 ft and the length is 94 ft.

EXAMPLE



6 Roof Gable


In a triangular gable end of a roof, the angle of the peak is twice as large as the angle of the back side of the house. The measure of the angle on the front side is 20° greater than the angle on the back side. How large are the angles?

EXAMPLE Solution



6


1. Familiarize. We first make a drawing as shown above. We let:

EXAMPLE Solution



6


2. Translate.

EXAMPLE Solution



6


3. Solve.

EXAMPLE Solution



6


3. Solve.

4. Check. The peak is twice the back and the front is 20° greater than the back. The sum is 180°. The angles check.

EXAMPLE Solution



6


5. State. The measures of the angles are 40°, 80°, and 60°.

EXAMPLE



8 Simple Interest


An investment is made at 3% simple interest for 1 year. It grows to $746.75. How much was originally invested (the principal)?

EXAMPLE Solution



8


1. Familiarize.

2. Translate.

EXAMPLE Solution



8


3. Solve.

EXAMPLE Solution



8


4. Check. 5. State.

CHAPTER




OBJECTIVES

2.7 Solving Inequalities


a Determine whether a given number is a solution of an inequality.

b Graph an inequality on the number line. c Solve inequalities using the addition principle. d Solve inequalities using the multiplication principle. e Solve inequalities using the addition principle and the

multiplication principle together.




An inequality is a number sentence with >, <, ≥, or ≤ as its verb—for example,

Some replacements for a variable in an inequality make it true and some make it false. (There are some exceptions to this statement, but we will not consider them here.)


Solution


A replacement that makes an inequality true is called a solution. The set of all solutions is called the solution set. When we have found the set of all solutions of an inequality, we say that we have solved the inequality.

EXAMPLE



Determine whether each number is a solution of x < 2.



b Graph an inequality on the number line.


Inequalities frequently have infinitely many solutions. Because we cannot list them all individually, it is helpful to make a drawing that represents all the solutions. A graph of an inequality is a drawing that represents its solutions. An inequality in one variable can be graphed on the number line. An inequality in two variables can be graphed on the coordinate plane. We will study such graphs in Chapter 9.

EXAMPLE



5 Graph:


EXAMPLE



7 Graph:


In order to be a solution of this inequality, a number must be a solution of both –3 ≤ x and x < 2.


c Solve inequalities using the addition principle.


Consider the true inequality 3 < 7. If we add 2 on both sides, we get another true inequality:

Similarly, if we add –4 on both sides of x + 4 < 10, we get an equivalent inequality:

To say that x + 4 < 10 and x < 6 are equivalent is to say that they have the same solution set.


The Addition Principle for Inequalities


EXAMPLE



9 Solve: Then graph.


EXAMPLE Solution



9


EXAMPLE Solution



9


The solution may also be stated using set-builder notation:


d Solve inequalities using the multiplication principle.


Consider the true inequality 3 < 7. If we multiply on both sides by a positive number, like 2, we get another true inequality: If we multiply on both sides by a negative number, like –2, and we do not change the direction of the inequality symbol, we get a false inequality:




if we reverse (change the direction of) the inequality symbol, we get a true inequality:


The Multiplication Principle for Inequalities





In other words, when we multiply or divide by a positive number on both sides of an inequality, the direction of the inequality symbol stays the same. When we multiply or divide by a negative number on both sides of an inequality, the direction of the inequality symbol is reversed.

EXAMPLE





EXAMPLE Solution



12


EXAMPLE


e Solve inequalities using the addition principle and the multiplication principle together.

13 Solve:


EXAMPLE Solution



13


EXAMPLE



16 Solve:


First, use the distributive law to remove parentheses. Next, collect like terms and then use the addition and multiplication principles for inequalities to get an equivalent inequality with x alone on one side.

EXAMPLE Solution



16


CHAPTER




OBJECTIVES

2.8 Applications and Problem Solving with Inequalities


a Translate number sentences to inequalities. b Solve applied problems using inequalities.


a Translate number sentences to inequalities.



Translating “at least” and “at most”


EXAMPLE


b Solve applied problems using inequalities.

1 Catering Costs


To cater a party, Curtis’ Barbeque charges a $150 setup fee plus $15.50 per person. The cost of Berry Manufacturing’s annual picnic cannot exceed $2100. How many people can attend the picnic?

EXAMPLE Solution



1


1. Familiarize. Suppose that 110 people were to attend the picnic. The cost would then be $150 + $15.50(110) or $1855. This shows that more than 110 people could attend the picnic without exceeding $2100. Instead of making another guess, we let n = the number of people in attendance.

EXAMPLE Solution



1


2. Translate.

EXAMPLE Solution



1


3. Solve.

EXAMPLE Solution



1


4. Check. Although the solution set of the inequality is all numbers less than or equal to about 125.8, since n = the number of people in attendance, we round down to 125 people. If 125 people attend, the cost will be $150 + $15.50(125), or $2087.50. If 126 attend, the cost will exceed $2100.

EXAMPLE Solution



1


5. State. At most, 125 people can attend the picnic.

EXAMPLE



2 Nutrition


The U.S. Department of Agriculture recommends that for a typical 2000-calorie daily diet, no more than 20 g of saturated fat be consumed. In the first three days of a four-day vacation, Anthony consumed 26 g, 17 g, and 22 g of saturated fat. Determine (in terms of an inequality) how many grams of saturated fat Anthony can consume in the fourth day if he is to average no more than 20 g of saturated fat per day.

EXAMPLE Solution



2


1. Familiarize. Suppose Anthony consumed 19 g of saturated fat on the fourth day. His daily average for the vacation would then be

EXAMPLE Solution



2


2. Translate.

EXAMPLE Solution



2


3. Solve.

EXAMPLE Solution



2


4. Check. 5. State. Anthony’s average intake of saturated fat for the vacation will not exceed 20 g per day if he consumes no more than 15 g of saturated fat on the fourth day.

chapter · equations have expressions on each side of the equals sign. the sentence “14 – 10 =...

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