Equations, Inequalities, and Problem Solving Chapter 2 Equations, Inequalities, and Problem Solving 2.1 2.2 2.3 2.4 2.5 2.6 2.7 1

Linear Equations in One Variable 2.1 Linear Equations in One Variable Objectives: Solve linear equations using properties of equality Solve linear equations that can be simplified by combining like terms Solve linear equations containing fractions or decimals Recognize identities and equations with no solutions Key Vocabulary: Solution, equivalent equation, conditional equation, contradiction, identity

Linear Equations An algebraic equation is a statement in which two expressions have equal value. Solving algebraic equations involves finding values for a variable that make the equation true. Equivalent equations are equations with the same solution set.

Example 1 Solve for x: 3x + 7 = 16. Check: 3x + 7 = 16 - 7 - 7 The solution or the solution set is {3}.

Example 3 Solve: 4p – 11 – p = 2 + 2p – 20. 3p – 11 = 2p – 18 Combine like terms. – 2p – 2p . Subtract 2p from both sides. p – 11 = – 18 Simplify. + 11 + 11 Add 11 to both sides. p = – 7 Simplify. Of course you should check the solution. Let’s check this one using the sto→ button on the calculator

Example 3 Solve: 12x + 30 + 8x – 6 = 10. 20x + 24 = 10 Simplify left side. – 24 – 24 Subtract 24 from both sides. 20x = –14 Simplify both sides. Divide both sides by 20. Simplify both sides. Check using the sto→ button on the calculator

Example 4 Solve: 5(3 + z) – (8z + 9) = – 4z. 15 + 5z – 8z – 9 = – 4z Use distributive property. 6 – 3z = – 4z Simplify left side. + 4z + 4z Add 4z to both sides. 6 + z = 0 Simplify both sides. – 6 – 6 Subtract 6 from both sides. z = – 6 Simplify both sides. Check using the sto→ button on the calculator

Solving a Linear Equation in One Variable Clear the equation of fractions by multiplying both sides of the equation by the least common denominator (LCD) of all denominators in the equation. Use the distributive property to remove grouping symbols such as parentheses. Combine like terms on each side of the equation. Use the addition property of equality to rewrite the equation as an equivalent equation with variable terms on one side and numbers on the other side. Use the multiplication property of equality to isolate the variable. Check the proposed solution in the original equation.

Example 6 Solve for y: -3y -3y 9 = 7y + 30 -30 -30 -30 -30 Check using the sto→ button on the calculator

A linear equation in one variable that has exactly one solution is called a conditional equation. An equation in one variable that has no solution is called a contradiction. An equation in one variable that has every number (for which the equation is defined) as a solution is called an identity. (infinite solutions)

Example 8 Solve for x: 3x – 7 = 3(x + 1). 3x – 7 = 3x + 3 Use distributive property. –3x –3x Subtract 3x from both sides. –7 = 3 Simplify both sides. The equation 7 = 3 is false no matter what value the variable x might have. Thus, the original equation has no solution. This equation is a contradiction.

Example 9 Solve for x: 5x – 5 = 2(x + 1) + 3x – 7. 5x – 5 = 2x + 2 + 3x – 7 Use distributive property. 5x – 5 = 5x – 5 Simplify the right side. – 5x – 5x Subtract 5x from both sides. – 5 = – 5 Simplify. Since – 5 = – 5 is a true statement for every value of x, all real numbers are solutions. The solution set is the set of all real numbers. The equation is called an identity.

Solve: additional ex 5 4

Linear Equations in One Variable 2.1 summary Linear Equations in One Variable Objectives: Solve linear equations using properties of equality Solve linear equations that can be simplified by combining like terms Solve linear equations containing fractions or decimals Recognize identities and equations with no solutions Key Vocabulary: Solution, equivalent equation, conditional equation, contradiction, identity

An Introduction to Problem Solving 2.2 An Introduction to Problem Solving Objectives: Write algebraic expressions that can be simplified Apply the steps for problem solving Key Vocabulary: Consecutive integers, complementary, supplementary

Consecutive, even, and odd integers and their representations. + 2 + 1 Consecutive Integers: x x + 1 x + 2 + 4 Consecutive Even Integers: + 2 x x + 2 x + 4 Consecutive Odd Integers: + 4 + 2 x x + 2 x + 4

Example 1 Write the following as an algebraic expression. Then simplify. The sum of three consecutive odd integers, if x is the first consecutive integer. In words: Translate: first integer plus next odd integer plus next odd integer x + (x + 2) + (x + 4) = x + x + 2 + x + 4 = 3x + 6

General Strategy for Problem Solving UNDERSTAND the problem. During this step, become comfortable with the problem. Some way of doing this are: Read and reread the problem. Propose a solution and check. Construct a drawing. Choose a variable to represent the unknown. TRANSLATE the problem into an equation. SOLVE the equation. INTERPRET the result. Check the proposed solution in the stated problem and state your conclusion.

Example 3 Find three numbers such that the second is 3 more than twice the first number, and the third number is four times the first number. The sum of the three numbers is 164. Let x = the first number, then 2x + 3 = the second number 4x = the third number 7x + 3 = 164 7x = 161 x = 23 In words: Translate: x + (2x + 3) + 4x = 164 first number added to second number added to third number is 164 1st # = 23 2nd # = 49 3rd # = 92

Example 6 Kelsey Ohleger was helping her friend Benji Burnstine study for an algebra exam. Kelsey told Benji that her last three latest art history quiz scores are three consecutive even integers whose sum is 264. Help Benji find the scores. x = the first integer. Then x + 2 = the second consecutive even integer x + 4 = the third consecutive even integer. 1st # = 86 3x + 6 = 264 3x = 258 x = 86 2nd # = 88 3rd # = 90

Add on: find the perimeter The problem is miss leading on the system because they put a question mark on the picture which isn’t necessary y + 1 2 10 20 + ? Peri = 20 + 2(7y – 3) 7y – 3 = 20 + 14y – 6 Peri = 14 + 14y

Add on: In 2010, the population of the country was 34. 6 million Add on: In 2010, the population of the country was 34.6 million. This represented an increase in population of 8.1% since 2001. What was the population of the country in 2001? Round to the nearest hundredth of a million. 2001population + 8.1%of(2001population) = 2010population p + 0.081p = 34.6 1.081p = 34.6 p = 32.01 If the total bill for a TV is $456.72 and the tax rate is 8.5%, find the original price of the TV. Round to the nearest cent. TV + 0.085TV = 456.72 1.085TV = 456.72 $420.94

An Introduction to Problem Solving 2.2 summary An Introduction to Problem Solving Objectives: Write algebraic expressions that can be simplified Apply the steps for problem solving Key Vocabulary: Consecutive integers, complementary, supplementary

Formulas and Problem Solving 2.3 Formulas and Problem Solving Objectives: Solve a formula for a specified variable Use formulas to solve problems

Formulas A formula is an equation that states a known relationship among multiple quantities (has more than one variable in it). A = lw Area of a rectangle = length · width I = PRT Simple Interest = Principal · Rate · Time P = a + b + c Perimeter of a triangle = side a + side b + side c d = rt Distance = rate · time V = lwh Volume of a rectangular solid = length · width · height

Example 1 & 2 Solve: V = lwh for w. Solve 4y – 5x = 9 for y.

Solving Equations for a Specific Variable Clear the equations of fractions by multiplying each side of the equation by the least common denominator. Use the distributive property to remove grouping symbols such as parentheses. Combine like terms on each side of the equation. Use the addition property of equality to rewrite the equation as an equivalent equation with terms containing the specified variable on one side and all other terms on the other side. Use the distributive property and the multiplication property of equality to isolate the specified variable.

Example 3 Example 4 A = P + PRT solve for T. A = P + PRT solve for P.

Compound interest formula Compound interest formula. A = amount in account after t years P = principal or amount invested t = time in years r = annual rate of interest n = number of times compounded per year For these questions you need a scientific calculator

Example 5 P = r = t = n = Formula: Substitute $15,000 0.03 5 years June Myers just received an inheritance of $15,000 and plans to place all the money in a savings account that pays 3% compounded quarterly to help her son go to college in 5 years. How much money will be in the account in 5 years? P = r = t = n = Formula: Substitute $15,000 0.03 5 years 4 times/yr Answer: In 5 years, the account will contain $17,417.76.

Example 6 The fastest average speed by a cyclist across the continental United States is 15.4 mph, by Pete Penseyres. If he traveled a total distance of about 3107.5 miles at this speed, find his time cycling. Write the time in days, hours, and minutes. (Source: The Guinness Book of World Records) distance formula d = rt 201.7857 ÷ 24 = 8.4077 days d = r = 3107.5 miles 15.4 mph 0.4077(24) = 9.7857 hours 0.7857(60) = 47.1429 minutes Answer: Petes cycling time was approximately 8 days, 9 hours, 47 minutes. 201.7857 hours

Formulas and Problem Solving 2.3 summary Formulas and Problem Solving Objectives: Solve a formula for a specified variable Use formulas to solve problems

Linear Inequalities and Problem Solving 2.4 Linear Inequalities and Problem Solving Objectives: Graph inequalities Solve linear inequalities Solve problems that can be modeled by linear inequalities

The solution set of an inequality is the set of all solutions. A solution of an inequality is a value of the variable that makes the inequality a true statement. The solution set of an inequality is the set of all solutions. The inequality x > 2 has a solution set of {x | x > 2|, read as The set of all numbers x such that x is greater than 2

Example 1 Graph each on a number line. a. x ≥ 3 b. x < –3 c. 0.5 < x ≤ 4 2 – 2 1 3 4 5 – 1 – 3 – 4 – 5 2 – 2 1 3 4 5 – 1 – 3 – 4 – 5 2 – 2 1 3 4 5 – 1 – 3 – 4 – 5

Addition Property of Inequality If a, b, and c are real numbers, then a < b and a + c < b + c are equivalent inequalities. Multiplication Property of Inequality If a, b, and c are real numbers and c is positive, then a < b and ac < bc are equivalent inequalities. If a, b, and c are real numbers and c is negative, then a < b and ac > bc

Example 2 Solve each and graph the solutions on a number line. + 4 + 4 x < 13 4x + 5 ≥ 3x – 7 – 3x – 3x x + 5 ≥ –7 – 5 – 5 x ≥ –12 12 10 11 13 14 15 – 12 – 11 – 13 – 14 – 15

Example 3 Solve and graph each on a number line. 2 – 2 1 3 4 5 – 1 – 3 1 3 4 5 – 1 – 3 – 4 – 5 2 – 2 1 3 4 5 – 1 – 3 – 4 – 5

Example 4 Solve and graph on a number line: 3x + 9  5(x – 1) 7(x – 2) + x > – 4(5 – x) – 12 3x + 9  5x – 5 7x – 14 + x > – 20 + 4x – 12 8x – 14 > 4x – 32 – 4x – 4x 4x – 14 > –32 + 14 + 14 4x > –18 – 3x – 3x 9  2x – 5 + 5 + 5 14  2x 7  x x ≤ 7 5 4 6 7 8 2 – 2 1 3 4 5 – 1 – 3 – 4 – 5

Example 7 Solve: 3(x + 4) > 3x + 5. 3(x + 4) > 3x + 5 3x + 12 > 3x + 5 Distribute on the left side. – 3x – 3x Subtract 3x from both sides. 12 > 5 Simplify. 12 > 5 is a true statement for all values of x, so this inequality and the original inequality are true for all numbers. All real numbers are solutions.

Example 8 A salesperson earns $800 per month plus a commission of 25% of sales. Find the minimum amount of sales needed to receive a total income of at least $1800 per month. Let x = amount of sales. 800 + 25% commission of sales ≥ 1800 Answer: The minimum amount of sales needed for the salesperson to earn at least $1800 per month is $4000 per month.

Example 9 In the United States, the annual consumption of cigarettes is declining. The consumption c in billions of cigarettes per year since the year 1990 can be approximated by the formula c = –9.2t + 527.33 where t is the number of years after 1990. Use this formula to predict the years that the consumption of cigarettes will be less than 200 billion per year. Answer: The annual consumption of cigarettes will be less than 200 billion more than 35.58 years after 1990, or in approximately 2026.

Linear Inequalities and Problem Solving 2.4 Summary Linear Inequalities and Problem Solving Objectives: Graph inequalities Solve linear inequalities Solve problems that can be modeled by linear inequalities

Compound Inequalities 2.5 Compound Inequalities Objectives: Find the intersection and union of two sets Solve compound inequalities containing and and or

The solution set of a compound inequality formed by the word and is the intersection of the solution sets of the two inequalities. We use the symbol  to represent “intersection.” Intersection of Two Sets The intersection of two sets, A and B, is the set of all elements common to both sets. A intersect B is denoted by A B A  B

The solution set of a compound inequality formed by the word or is the union of the solution sets of the two inequalities. We use the symbol  to denote “union.” Union of Two Sets The union of two sets, A and B, is the set of elements that belong to either of the sets. A union B is denoted by A A  B B

Graphs Or (υ) Union And (∩) Intersection x > 2 x > 8 All real #’s 2 < x < 8 2 8 x < 2 or x > 8 No solution 2 8

Example 1 If A = {x | x is an odd number greater than 0 but less than 9} and B = {4, 5, 6, 7, 8}, find A  B. List the elements of A. A = {1, 3, 5, 7} List the elements of B. A = {4, 5, 6, 7, 8} The numbers 5 and 7 are in sets A and B. The intersection is {5, 7}.

Example 2 Solve: x + 4 > 0 and 4x > 0. Solve each inequality separately. x + 4 > 0 and 4x > 0 x > 4 and x > 0 Graph the two inequalities on a number line and find their intersection. x > 4 x > 0 x > 4 and x > 0 Solution is x > 0 2 – 2 1 3 4 5 – 1 – 3 – 4 – 5 2 – 2 1 3 4 5 – 1 – 3 – 4 – 5

Example 3 Solve: 5x > 0 and 3x  4 ≤ 13. Solve each inequality separately. 5x > 0 and 3x – 4 ≤ 13 x > 0 and 3x ≤ 9 x ≤ 3 Graph the two inequalities and find their intersection. x > 0 x > 0 and x ≤ 3 There is no number that is greater than 0 and less than or equal to 3. The answer is no solution. 2 – 2 1 3 4 5 – 1 – 3 – 4 – 5 2 – 2 1 3 4 5 – 1 – 3 – 4 – 5

To solve a compound inequality written in compact form, such as 3 < 5 – x < 9, we get x alone in the “middle part.” We must perform the same operations on all three parts of the inequality.

Example 4 Solve and graph on a number line: 3 < 5 – x < 9. 3 < 5 – x < 9 –5 –5 –5 2 – 2 1 3 4 5 – 1 – 3 – 4 – 5

Example 6 If A = {x | x is an odd number greater than 0 but less than 9} and B = {4, 5, 6, 7, 8}, find A  B. List the elements of A. A = {1, 3, 5, 7} B = {4, 5, 6, 7, 8} The numbers that are in either set or both sets are {1, 3, 4, 5, 6, 7, 8} This set is the union.

Example 7 Solve: 6x – 4 ≤ 12 or x + 2 ≥ 8. Solve each inequality separately. Find the union of the graphs. 4 2 3 5 6 7 1 – 1 – 2 – 3 4 2 3 5 6 7 1 – 1 – 2 – 3 The solutions are x ≤ 8/3 or x ≥ 6.

Example 8 Solve: 2x – 6 < 2 or 8x < 0. Find the union. Solve each inequality separately. Find the union. 2 – 2 1 3 4 5 – 1 – 3 – 4 – 5 2 – 2 1 3 4 5 – 1 – 3 – 4 – 5 The solutions are all real numbers.

Compound Inequalities 2.5 summary Compound Inequalities Objectives: Find the intersection and union of two sets Solve compound inequalities containing and and or

Absolute Value Equations 2.6 Absolute Value Equations Objectives: Solve absolute value equations

Example 1 Solve: 6 + 2n = 4. 6 + 2n = 4 or 6 + 2n = 4 Check: To check, let n = 1 and then n = 5 in the original equation. |6 + 2(1)| = 4 |6 + 2(5)| = 4 |6 – 2| = 4 |6 – 10| = 4 |4| = 4 | –4| = 4 4 = 4 True 4 = 4 True The solutions are 1 and 5.

Example 2 Solve: 2x 6 = 4 2x 6 = 4 2x= 10 2x = 10 or 2x = 10 We want the absolute value expression alone on one side of the equation, so begin by adding 6 to both sides. Then apply the absolute value property. 2x 6 = 4 2x= 10 2x = 10 or 2x = 10 x = 5 or x = 5 The solutions are 5 and 5.

Example 3 Example 4 Solve: 7x+2 = 0. 7x +2= 0 x = –2/7 Solve: 3z  2+ 8 = 1. First, isolate the absolute value. 3z  2+ 8 = 1 3z  2 = 7 The absolute value of a number is NEVER negative, so this equation has no solution.

Example 5 Solve: The absolute value of any expression is NEVER negative, so no solution exists. This equation has no solution.

Example 6 Solve: The solutions are 1/3 and 1. This equation is true if the expressions inside the absolute value bars are equal to or are opposites of each other. The solutions are 1/3 and 1.

Example 7 Solve: Solution: x = 5

Absolute Value Equations 2.6 summary Absolute Value Equations Objectives: Solve absolute value equations

Absolute Value Inequalities 2.7 Absolute Value Inequalities Objectives: Solve absolute value inequalities

Example 1 Solve: |x| ≤ 7. The solution set of this inequality contains all numbers whose distance from 0 is less than or equal to 7. Thus 7, 7, and all numbers between 7 and 7 are in the solution set. The solutions are 7 ≤ x ≤ 7.

Example 2 Solve for x: x + 4 < 6. |x + 4| < 6 is equivalent to 6 < x + 4 < 6 6 < x + 4 < 6 –4 –4 –4 10 < x < 2

Example 3 Solve for x: |x – 3| + 6 ≤ 7. +3 +3 +3 First, isolate the absolute value expression by subtracting 6 from both sides. Solve for x: |x – 3| + 6 ≤ 7. +3 +3 +3

Example 4 Solve for x: 8x  3 < 2. The absolute value of a number is always nonnegative and can never be less than 2. Thus this absolute value inequality has no solution.

Example 5 Solve for y: |y – 5| > 8. Solve the compound inequality. y – 5 < – 8 or y – 5 > 8 +5 +5 +5 +5 y < –3 or y > 13 The solutions are all numbers y such that y < 3 or y > 13.

Example 6 Solve for x: 4x + 5 + 6 > 2. 4x + 5 > –4 Isolate the absolute value expression by subtracting 6 from both sides. 4x + 5 > –4 The absolute value of any number is always nonnegative and thus is always greater than 4. This inequality and the original inequality are true for all values of x. The solutions are all real numbers.

Example 7 Solve 10 + 3x + 1 > 2. 10 + 3x > 1 Isolate the absolute value expression by subtracting 1 from both sides. 10 + 3x > 1 Write the absolute value inequality as an equivalent compound inequality and solve. -10 -10 -10 -10 3x < -11 or 3x > -9 x < 11/3 or x > 3

Example 7 Solve The solutions are x ≤ 17 or x ≥ 3. Isolate the absolute value expression by adding 1 to both sides. Write the absolute value inequality as an equivalent compound inequality and solve. The solutions are x ≤ 17 or x ≥ 3.

Example 8 Solve The solution is –2. Recall that “≤” means “less than or equal to.” The absolute value of any expression will never be less than 0, but it may be equal to 0. Thus, to solve set the expression equal to 0. Solve The solution is –2.

Absolute Value Inequalities 2.7 summary Absolute Value Inequalities Objectives: Solve absolute value inequalities