1 Equations and Inequalities © 2008 Pearson Addison-Wesley. All rights reserved Sections 1.1–1.4

1.1 Linear Equations 1.2 Applications and Modeling with Linear Equations Equations and Inequalities 1

Linear Equations 1.1 Basic Terminology of Equations ▪ Solving Linear Equations ▪ Identities, Conditional Equations, and Contradictions ▪ Solving for a Specified Variable (Literal Equations)

Basic Terminology of Equations An equation is a statement that two expressions are equal. For example: To solve an equation means to find all numbers that make the equation a true statement. These numbers are called the solutions or roots of the equation. These solutions make up the solution set. Equations with the same solution set are called equivalent equations. We will focus on solving linear equations. A linear equation in one variable is an equation that can be written in the form ax + b = 0 where a and b are real numbers and a ≠ 0. We will focus on solving linear equations. A linear equation in one variable is an equation that can be written in the form ax + b = 0 where a and b are real numbers and a ≠ 0.

Addition and Multiplication Properties of Equality Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-5 For real numbers a, b, and c: If a = b, then a + c = b + c. That is, the same number or expression may be added to (or subtracted from) both sides of an equation without changing the solution set. If a = b and c ≠ 0, then ac = bc. That is, both sides of an equation may be multiplied (or divided) by the same nonzero number without changing the solution set.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-6 Solve. 1.1 Example 1 Solving a Linear Equation (page 85) Solution set: {6} Distributive property Combine terms. Add 4 to both sides. Add 12x to both sides. Combine terms. Divide both sides by 4.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-7 Solve. 1.1 Example 2 Clearing Fractions Before Solving a Linear Equation (page 85) Solution set: {–10} Multiply by 10, the LCD of all the fractions. Distributive property Combine terms. Add –4s and –6 to both sides. Combine terms. Divide both sides by –6.

Types of Equations 1.Identity – is an equation satisfied by every number. 2.Conditional Equations – is an equation satisfied by some numbers but not others. 3.Contradiction – is an equation that has no solution. Indentifying the type of linear equation comes from whether the resulting statement is true (1), unique (2) or false (3). Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-8

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-9 Decide whether the equation is an identity, a conditional equation, or a contradiction. Give the solution set. 1.1 Example 3(a) Identifying Types of Equations (page 86) This is a conditional equation. Solution set: {11} Add –4x and 9 to both sides. Combine terms. Divide both sides by 2.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Decide whether the equation is an identity, a conditional equation, or a contradiction. Give the solution set. 1.1 Example 3(b) Identifying Types of Equations (page 86) This is a contradiction. Solution set: ø Distributive property Subtract 14x from both sides.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Decide whether the equation is an identity, a conditional equation, or a contradiction. Give the solution set. 1.1 Example 3(c) Identifying Types of Equations (page 86) This is an identity. Solution set: {all real numbers} Distributive property Combine terms. Add x and –3 to both sides.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Solve for the specified variable. 1.1 Example 4 Solving for a Specified Variable (page 87) Solve for the specified variable. (a) d = rt, for t (b), for k Divide both sides by r. Factor out k. Divide both sides by

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Caden borrowed $2580 to buy new kitchen appliances for his home. He will pay off the loan in 9 months at an annual simple interest rate of 6.0%. How much interest will he pay? 1.1 Example 5 Applying the Simple Interest Formula (page 88) Caden will pay $ in interest.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Solving Applied Problems ▪ Geometry Problems ▪ Motion Problems ▪ Mixture Problems ▪ Modeling with Linear Equations Applications and Modeling with Linear Equations

Step for Solving an Applied Problem 1.Read the problem carefully until you understand what is given and what is to be found. 2.Assign a variable to represent the unknown value, using diagrams or tables as needed. 3.Write an equation using the variable expressions. 4.Solve the equation. 5.State the answer to the problem. 6.Check the answer in the words of the original problem. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 1-15

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 1 Find the Dimensions of a Square (page 91) The length of a rectangle is 2 in. more than the width. If the length and width are each increased by 3 in., the perimeter of the new rectangle will be 4 in. less than 8 times the width of the original rectangle. Find the dimensions of the original rectangle. Assign variables: Let x = the length of the original rectangle. Then, x − 2 = the width of the original rectangle. x + 3 = the length of the new rectangle (x − 2) + 3 = x + 1 = the width of the new rectangle.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 1 Find the Dimensions of a Square (cont.) The perimeter of the new rectangle is The perimeter of the new rectangle is 4 in. less than 8 times the width of the original rectangle, so we have

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 1 Find the Dimensions of a Square (cont.) Distributive property. Combine terms. Add –4x and 20 to both sides. Divide both sides by 4. The length of the original rectangle is 7 in. The width of the original rectangle is 7 – 2 = 5 in.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 2 Solving a Motion Problem (page 92) Krissa drove to her grandmother’s house. She averaged 40 mph driving there. She was able to average 48 mph returning, and her driving time was 1 hr less. What is the distance between Krissa’s house and her grandmother’s house? Let x = the distance between Krissa’s house and her grandmother’s house Use d = rt

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 2 Solving a Motion Problem (cont.) The driving time returning was 1 hour less than the driving time going, so Multiply by 40·48. Distributive property. Subtract 48x. Divide by –8. The distance from Krissa’s home to her grandmother’s home is 240 mi.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 3 Solving a Mixture Problem (page 93) How many liters of a 25% anti-freeze solution should be added to 5 L of a 10% solution to obtain a 15% solution? Let x = the amount of 25% solution The number of liters of pure antifreeze in the 25% solution plus the number of liters of pure antifreeze in the 10% solution must equal the number of liters of pure antifreeze in the 15% solution.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 3 Solving a Mixture Problem (cont.) Create a table to show the relationships in the problem. Write an equation:

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 3 Solving a Mixture Problem (cont.) 2.5 liters of the 25% solution should be added. Distributive property. Subtract.15x and.5. Divide by.1.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 4 Solving an Investment Problem (page 94) Last year, Owen earned a total of $1456 in interest from two investments. He invested a total of $28,000, part at 4.8% and the rest at 5.5%. How much did he invest at each rate? Let x = amount invested at 4.8%. Then 28,000 − x = amount invested at 5.5%.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 4 Solving an Investment Problem (cont.) Create a table to show the relationships in the problem. The amount of interest from the 4.8% account plus the amount of interest from the 5.5% account must equal the total amount of interest.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 4 Solving an Investment Problem (cont.) Owen invested $12,000 at 4.8% and $28,000 − $12,000 = $16,000 at 5.5%. Distributive property Combine terms. Subtract Divide by –.007.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 5 Modeling the Prevention of Indoor Pollutants (page 95) A range hood removes contaminants at a rate of F liters of air per second. The percent P of contaminants that are also removed from the surrounding air can be modeled by the linear equation where 10 ≤ F ≤ 75. What flow F must a range hood have to remove 70% of the contaminants from the air?

Copyright © 2008 Pearson Addison-Wesley. All rights reserved Example 5 Modeling the Prevention of Indoor Pollutants (cont.) The flow rate must be approximately L per second to remove 70% of the contaminants. Substitute P = 70 into and solve for F. Subtract Divide by 1.06.