19 22 Solving Equations by Multiplying and Dividing Objectives 221 Use the

19 22 solving equations by multiplying and dividing

This preview shows page 19 - 31 out of 42 pages.

19
Image of page 19
2.2: Solving Equations by Multiplying and Dividing Objectives: 2.2.1: Use the multiplication property to solve equations 2.2.2: Use the multiplication property to solve an application 20
Image of page 20
The Multiplication Property of Equality Suppose we want to solve this equation: 6x = 18. The addition property doesn’t help to solve the equation. 21
Image of page 21
2.2.1: Solving Equations by Using the Multiplication Property Examples: Solve the following equations 1. 6x = 18 2. 5x = 35 3. 9x = 54 4. = 6 5. = 9 3 x 5 x 22
Image of page 22
2.2.1: Solving Equations by Using the Multiplication Property Examples: Solve the following equations Solving Equations by Using Reciprocals 1. x = 9 2. x = 18 Solving Equations by Combining Like Terms 1. 3x + 5x = 40 2. 7x + 4x = 66 5 3 3 2 23
Image of page 23
2.2.2: Use the Multiplication Property to Solve an Application Example: On her first day on the job in a photography lab, Nancy processed all of the film given to her. The following day, her boss gave her four times as much film to process. Over the two days, she processed 60 rolls of film. How many rolls did she process on the first day? 24
Image of page 24
2.3: Combining the Rules to Solve Equations Objectives: 2.3.1: Use both addition and multiplication to solve equations 2.3.2: Solve equations involving fractions 2.3.3: Solve applications 25
Image of page 25
2.3.1: Use Both Addition and Multiplication to Solve Equations Examples: Solve the following equations 1. 3x 5 = 4 2. 4x 7 = 17 3. 5x 11 = 2x 7 4. 7x 12 = 2x 9 26
Image of page 26
2.3.1: Use Both Addition and Multiplication to Solve Equations Examples: Solve the following equations Applying the Properties of Equality with Like Terms 1. 8x + 2 3x = 8 + 3x + 2 2. 7x 3 5x = 10 + 4x + 3 Applying the Properties of Equality with Parentheses 1. x + 3(3x 1) = 4(x + 2) + 4 2. x + 5(x + 2) = 3(3x 2) + 18 27
Image of page 27
2.3.2: Solve Equations Involving Fractions To solve an equation involving fractions, the first step is to multiply both sides of the equation by the least common multiple (LCM) of all denominators in the equation. This clears the equation of fractions, and we can proceed as before. The LCM of a set of denominators is also called the least common denominator (LCD). Examples: Solve the following equations 1. = 2. + 1 = 2 x 3 2 6 5 5 1 2 x 2 x 28
Image of page 28
Conditional Equations, Identities, and Contradictions An equation that is true for only particular values of the variable is called a conditional equation . Here the equation can be written in the form ax + b = 0 in which a ≠ 0 An equation that is true for all possible values of the variable is called an identity . In this case, both a and b are 0, so we get the equation 0 = 0. This is the case, if both sides of the equation reduce to the same expression (a true statement). An equation that is never true , no matter what the value of the variable, is called a contradiction . For example, if a is 0 but b is 4, a contradiction results. This is the case if the equation simplifies as a false statement 29
Image of page 29
Identities and Contradictions Examples: Solve the following equations Solving Identity Equation 1.
Image of page 30
Image of page 31

You've reached the end of your free preview.

Want to read all 42 pages?

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes