Copyright © Cengage Learning. All rights reserved. Equations and Inequalities 2.

Slides:



Advertisements
Similar presentations
SECTION 2-3 APPLICATIONS OF LINEAR EQUATIONS Investment Problems Example 1 An investment counselor invested 75% of a client’s money into a 9% annual simple.
Advertisements

Notes Over 4.5 Solving a Linear System Use an Inverse Matrix to solve the linear system.
OBJECTIVES © 2010 Pearson Education, Inc. All rights reserved 1 Applications of Linear Equations Learn procedures for solving applied problems. Use linear.
Motion, Money, and Mixture Problems
Applications of Systems of Linear Equations Example 1: Steve invested $12,000 for one year in two different accounts, one at 3.5% and the other at 4%.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Systems of Linear Equations in Three Variables Solve linear Systems with.
Applications of Systems of Linear Equations
UNIFORM MOTION PROBLEMS Jo & Jim are 240 miles apart. Jo starts toward Jim at 55 mph at the same time Jim starts toward Jo at 65 mph. How long until they.
Copyright © Cengage Learning. All rights reserved.
CHAPTER 7 Systems of Equations Slide 2Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 7.1Systems of Equations in Two Variables 7.2The Substitution.
2.4 Formulas and Applications BobsMathClass.Com Copyright © 2010 All Rights Reserved. 1 A formula is a general statement expressed in equation form that.
Copyright © 2013 Pearson Education, Inc. Section 2.3 Introduction to Problem Solving.
Lesson 3-9 Weighted Averages.
Holt Algebra Using Algebraic Methods to Solve Linear Systems To solve a system by substitution, you solve one equation for one variable and then.
Chapter 3 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-1 Applications of Algebra.
1 Equations and Inequalities © 2008 Pearson Addison-Wesley. All rights reserved Sections 1.1–1.4.
Copyright © 2010 Pearson Education, Inc. All rights reserved Sec
When solving an application that involves two unknowns, sometimes it is convenient to use a system of linear equations in two variables.
Objective - To solve various problems using systems of linear equations. We will be studying 3 types of problems 1) Number and Value Problems 2) Mixture.
Over Lesson 2–8. Splash Screen Weighted Averages Lesson 2-9A Mixture Problems.
Section 4Chapter 2. 1 Copyright © 2012, 2008, 2004 Pearson Education, Inc. Objectives 2 3 Further Applications of Linear Equations Solve problems about.
SECTION 2-3 APPLICATIONS OF LINEAR EQUATIONS Investment Problems Example 1 An investment counselor invested 75% of a client’s money into a 9% annual simple.
Copyright © Cengage Learning. All rights reserved. Equations and Inequalities 2.
Copyright © 2010 Pearson Education, Inc. All rights reserved Sec
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 2–8) CCSS Then/Now New Vocabulary Example 1:Real-World Example: Mixture Problem Example 2:Real-World.
Section 2Chapter 2. 1 Copyright © 2012, 2008, 2004 Pearson Education, Inc. Objectives Formulas and Percent Solve a formula for a specified variable.
Weighted Averages. An average is a direct measurement of the sum of the measurements divided by the number of things being measured. A weighted average.
Copyright © 2010 Pearson Education, Inc. All rights reserved Sec
Copyright © 2015, 2011, 2007 Pearson Education, Inc. 1 1 Chapter 4 Systems of Linear Equations and Inequalities.
Applications: Interest, Mixture, Uniform Motion, Constant Rate Jobs
Copyright © Cengage Learning. All rights reserved.
Chapter 2 Section 7. Objectives 1 Copyright © 2012, 2008, 2004 Pearson Education, Inc. Further Applications of Linear Equations Use percent in solving.
ACTIVITY 20: Systems of Linear Equations (Section 6.2, pp ) in Two Variables.
Copyright © Cengage Learning. All rights reserved. Fundamentals.
When solving an application that involves two unknowns, sometimes it is convenient to use a system of linear equations in two variables.
Section 2.6 More about Problem Solving. Objectives Solve investment problems Solve uniform motion problems Solve liquid mixture problems Solve dry mixture.
Copyright © 2014, 2010, 2006 Pearson Education, Inc. 1 Chapter 2 Linear Functions and Equations.
 You can use weighted averages to solve uniform motion problems when the objects you are considering are moving at constant rates or speeds.
Splash Screen. Then/Now You translated sentences into equations. Solve mixture problems. Solve uniform motion problems.
Copyright © Cengage Learning. All rights reserved. Systems of Linear Equations and Inequalities in Two Variables 7.
1 Copyright © Cengage Learning. All rights reserved. 2. Equations and Inequalities 2.1 Equations.
1 Copyright © Cengage Learning. All rights reserved. 2. Equations and Inequalities 2.2 Applied Problems.
Solving Equations Containing First, we will look at solving these problems algebraically. Here is an example that we will do together using two different.
10-8 Mixture Problems Standard 15.0: Apply algebraic techniques to percent mixture problems. Standard 15.0: Apply algebraic techniques to percent mixture.
Lesson 2-6 Warm-Up.
Chapter 8 Section 4 Solving System of Equations Applications and Problem Solving.
Solving Application Problems Using System of Equations Section 4.3.
Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1 1 Equations and Inequalities Copyright © 2013, 2009, 2005 Pearson Education, Inc.
1 Equations and Inequalities. 1.2 Applications of Linear Equations.
Applications of Systems of Linear Equations
Equations and Inequalities
Copyright © Cengage Learning. All rights reserved.
Splash Screen.
10.8 Mixture Problems Goal: To solve problems involving the mixture of substances.
HW: Worksheet Aim: How do we solve fractional equation?
Solving Rational Equations
Weighted Averages.
Chapter 2 Section 3.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
3.2 - Solving Systems through Substitution
Solving Equations Containing
Splash Screen.
Chapter 2 Section 2.
2-9 Weighted Averages Weighted Average – the sum of the product of the number of units and the value per unit divided by the sum of the number of units.
Splash Screen.
Algebra 1 Section 3.8.
Algebra 1 Section 7.7.
Splash Screen.
2-9 Weighted Averages Weighted Average – the sum of the product of the number of units and the value per unit divided by the sum of the number of units.
Presentation transcript:

Copyright © Cengage Learning. All rights reserved. Equations and Inequalities 2

Copyright © Cengage Learning. All rights reserved. Section 2.6 Motion and Mixture Applications

3 Objectives 1. Solve a motion application using a linear equation in one variable. 2. Solve a liquid mixture application using a linear equation in one variable. 3. Solve a dry mixture application using a linear equation in one variable

4 Motion and Mixture Applications In this section, we consider uniform motion and mixture applications. In these problems, we will use the following three formulas: r  t = d r  b = a v = p  n The rate multiplied by the time equals the distance. The rate multiplied by the base equals the amount. The value equals the price multiplied by the number.

5 Solve a motion application using a linear equation in one variable 1.

6 Example – Traveling Chicago and Green Bay are about 200 miles apart. If a car leaves Chicago traveling toward Green Bay at 55 mph at the same time as a truck leaves Green Bay bound for Chicago at 45 mph, how long will it take them to meet? 1.What am asked to find? Time it takes for the two vehicles to meet: t.

7 Example – Traveling 2.Form an equation Motion problems are based on the formula d = rt, where d is the distance traveled, r is the rate, and t is the time. cont’d Figure 2-11

8 Example – Traveling We know that the two vehicles travel for the same amount of time, t hours. When the they meet, the total distance traveled is 200 miles. 55t + 45t = 200 cont’d

9 Example – Traveling 3.Solve the equation 55t + 45t = t = 200 t = 2 cont’d

10 Example – Traveling 4.Check the result In 2 hours, the car will travel 55  2 = 110 miles, while the truck will travel 45  2 = 90 miles. The total distance traveled will be = 200 miles. cont’d

11 Solve a liquid mixture application using a linear equation in one variable 2.

12 Example – Mixing Acid A chemist has one solution that is 50% sulfuric acid and another that is 20% sulfuric acid. How much of each should she use to make 12 liters of a solution that is 30% sulfuric acid? 1.What am asked to find? Number of liters for the 50% sulfuric acid solution: x Number of liters for the 20% sulfuric acid soluton: 12 - x

13 Example – Mixing Acid 2.Form an equation Liquid mixture applications are based on the relationship rb = a, where b is the base, r is the rate, and a is the amount. If x represents the number of liters of 50% solution to use, the amount of sulfuric acid in the solution will be 0.50x liters. The amount of sulfuric acid in the 20% solution will be 0.20(12 – x) liters. The amount of sulfuric acid in the final mixture will be 0.30(12) liters. cont’d

14 Example – Mixing Acid We can organize information in a chart or a diagram, as shown in Figure (a) (b) Figure 2-14 cont’d

15 Example – Mixing Acid Since the number of liters of sulfuric acid in the 50% solution plus the number of liters of sulfuric acid in the 20% solution will equal the number of liters of sulfuric acid in the mixture, we can form the equation: cont’d

16 Example Mixing Acid 3.Solve the equation 0.5x + 0.2(12 – x) = 0.3(12) 5x + 2(12 – x) = 3(12) 5x + 24 – 2x = 36 3x + 24 = 36 3x = 12 x = 4 (liters of 50% acid) 12 – 4 = 8 (liters of 20% acid) cont’d

17 Example – Mixing Acid State the conclusion Recall that we let x represent the number of liters of the 50% sulfuric acid solution. The chemist must mix 4 liters of the 50% solution and 12 – 4 = 8 liters of the 20% solution. Check the result The amount of acid in 4 liters of 50% solution is 4(0.50) = 2 liters. The amount of acid in 8 liters of 20% solution is 8(0.20) = 1.6 liters. cont’d

18 Example 4 – Mixing Acid The amount of acid in 12 liters of 30% solution is 12(0.30) = 3.6 liters. Since = 3.6, the results check. cont’d

19 Solve a dry mixture application using a linear equation in one variable 3.

20 Example – Mixing nuts Fancy cashews are not selling at $9 per pound, because they are too expensive. However, filberts are selling well at $6 per pound. How many pounds of filberts should be combined with 50 pounds of cashews to obtain a mixture that can be sold at $7 per pound? 1.What am I asked to find? Number of pounds of filberts in the mixture: x Total number of pounds of mixture: 50 + x

21 Example – Mixing nuts 2.Form an equation Dry mixture problems are based on the formula v = pn, where v is the value of the mixture, p is the price per pound, and n is the number of pounds. Value of filberts: $6x Value of cashews: $9(50) = $45 Value of mixture: $7(50 + x) cont’d

22 Example – Mixing nuts The value of the filberts (in dollars) 6x plus the value of the cashews (in dollars) 450, is equal to the value of the mixture (in dollars) 7(50 + x). We can organize this information in a table or a diagram, as shown in Figure (a) (b) Figure 2-15 cont’d

23 Example – Mixing nuts We can form the equation: 4.Solve the equation. 6x + 9(50) = 7(50 + x) 6x = x 100 = x (lb of filberts) cont’d

24 Example – Mixing nuts 4.Check the result Filberts:100 $6/lb: $600 Cashews: 50 $9/lb: $450 Total: $1050 Mixture: 150 $7/lb: $1,050. cont’d