1 / 41

Business Math

Business Math. Chapter 5: Equations. 5.1 Equations. Solve equations using multiplication or division. Solve equations using addition or subtraction. Solve equations using more than one operation. Solve equations containing multiple unknown terms. Solve equations containing parentheses.

ethelc
Download Presentation

Business Math

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Business Math Chapter 5: Equations

  2. 5.1 Equations • Solve equations using multiplication or division. • Solve equations using addition or subtraction. • Solve equations using more than one operation. • Solve equations containing multiple unknown terms. • Solve equations containing parentheses. • Solve equations that are proportions.

  3. 5.1.1 Solve equations using multiplication or division • An equation is a mathematical statement in which two quantities are equal. • Solving an equation means finding the value of an unknown. • For example: 10x = 30 • To solve this equation, the value of x must be discovered. • Division is used to solve this equation.

  4. Key Terms • The letters represent unknown amounts and are called unknowns or variables. • The numbers are called known or givenamounts. 3x = 27

  5. Remember! • Any operation performed on one side of the equation must be performed on the other side of the equation as well. • If you “multiply by 2” on one side, you must “multiply by 2” on the other side. • If you “divide by 3” on one side, you must “divide by 3” on the other side and so on.

  6. How to solve an equation with multiplication and division 10x = 30 Step one: Isolate the unknown value. Determine if multiplication or division is needed. Step two: Use division to divide both sides by “10.” Step three: Simplify: x = 3

  7. Find the value of an unknown using multiplication. • Find the value of “a” in the following equation. • a/3 = 6 • Multiply both sides by 3 to isolate “a.” • The left side becomes 1a or “a.” • The right side becomes the product of 6 x 3 or “18.” • a = 18

  8. Solve the following: 2b = 40 Determine which operation is needed. Division Perform the same operation to both sides. Divide both sides by “2.” Isolatethevariable and solve the equation b = 40/2 = 20 Do this example.

  9. 5.1.2.Solve an equation with addition or subtraction. 10 + x = 30 Step one: Isolate the unknown value. Determine if addition or subtraction is needed. Step two: Use subtraction to isolate “x.” Step three: Simplify: x = 20

  10. Don’t forget! • Adding or subtracting any number from one side must be carried out on the other side as well. • Subtract “the given amount” from both sides. • Would solving 5 + x = 20 require addition or subtraction of “5” from each side? • Subtraction

  11. Solve the following: b - 12 = 8 Determine which operation is needed. Addition Perform the same operation to both sides. Add “12” to both sides. Isolate the variable and solve the equation. b = 8 + 12 = 20 Do this example.

  12. 5.1.3.Solve equations using more than one operation. • Isolate the unknown value. • Add or subtract as necessary first. • Multiply or divide as necessary second. • Identify the solution: the number on the side opposite the unknown. • Check the solution by “plugging in” the number using the original equation.

  13. Order of Operations • When two or more calculations are written symbolically, it is agreed to perform the operations according to a specified order of operations. • Perform multiplication and division as they appear from left to right. • Perform addition and subtraction as they appear from left to right.

  14. “Undo the operations” To solve an equation, we undo the operations, so we work in reverse order. 1. Undo the addition or subtraction. 2. Undo multiplication or division. 7x + 4 = 39

  15. 7x + 4 = 39 First, undo the addition by subtracting 4 from each side. And that becomes 7x = 35 Next, divide each side by 7. And that becomes x = 35/7 = 5 Verify the result by “plugging 5 in” the place of “x.” 7 (5) + 4 =39 35 + 4 = 39 Try this example.

  16. 5.1.4. Equations containing multiple unknown terms. • In some equations, the unknown value may occur more than once. • The simplest instance is when the unknown value occurs in two addends. • For example: 3a + 2a = 25 • Add the numbers in each addend (2+3). • Multiply the sum by the unknown (5a= 25). • Solve for “a.” (a = 5)

  17. Try this example. • Find a if: a +4a – 5 = 30 • Combine the unknown value addends. a + 4a = 5a 5a – 5 = 30 • “Undo” the subtraction. 5a = 35 • “Undo” the multiplication. a = 7 • Check by replacing “a” with “7. • It is correct.

  18. 5.1.5. Solve equations containing parentheses. 1. Eliminate the parentheses: a. Multiply the number just outside the parentheses by each addend inside the parentheses. b. Show the resulting products as addition or subtraction as indicated. 2. Solve the resulting equation.

  19. Solve the equation:5(A + 3) = 25 Multiply “5” by each addend. 5 multiplied by A + 5 multiplied by 3 Show the resulting products. 5A + 15 = 25 Solve the equation. 5A = 10 A = 2 Look at this example.

  20. Do me first ! Tip! • Remove the parentheses first. • 5 (x - 2) = 45 • 5x -10 = 45 5x = 55 x = 11

  21. 5.1.6. Solve equations that are proportions. • A proportion is based on two pairs of related quantities. • The most common way to write proportions is to use fraction notation. • A number written in fraction notation is also called a ratio. • When two ratios are equal, they form a proportion.

  22. Cross products • An important property of proportions is that the cross products are equal. • A cross product is the product of the numerator of one fraction times the denominator of another fraction. • Example: 3/6 = 2/4 Multiply 3 x 4 = 6 x 2 12 = 12

  23. Do 4/12 and 6/18 form a proportion? 1. Multiply the numerator from the first fraction by the denominator of the second fraction. 4 x 18 = 72 2. Multiply the denominator of the first fraction by the numerator of the second fraction. 6 x 12 = 72 3. Are they equal? Yes, they form a proportion. Verify that two fractions form a proportion.

  24. 5.2 Using equations to solve problems. • There is a list of key words and what operations they imply in your textbook. Please refer to it. • These words help you interpret the information and begin to set up the equation to solve the problem. • Example: “of” often implies multiplication. “¼ of her salary” means “multiply her salary by ¼”

  25. What you know. Known or given facts. What you are looking for? Unknown or missing amounts. Solution Plan Equation or relationship among known / unknown facts. Solution Solve the equation. Conclusion Solution interpreted within context of problem. Five-step problem solving approach for equations.

  26. Use the solution plan • Full time employees work more hours than part-time employees. If the difference is four per day, and part-time employees work six hours per day, how many hours per day do full-timers work? • What are we looking for?Number of hours that FT work • What do we know?PT work 6 hours; The difference between FT and PT is 4 hours.

  27. Use the solution plan • We also know that “difference” implies subtraction. • Set up a solution plan. FT – PT = 4 FT = n [unknown] PT = 6 hours N – 6 = 4 • Solution plan: N = 4 + 6 = 10 • Conclusion: Full time employees work 10 hours.

  28. Try this example. • Jill has three times as many trading cards as Matt. If the total number that both have is 200, how many cards does Jill have? Use the five-step solution plan to solve this problem: 1. What are you looking for? 2. What do you know? 3. Set up a solution plan. 4. Solve it. 5. Draw the conclusion.

  29. What are you looking for? The number of cards that Jill has. What do you know? The relationship in the number of cards is 3:1; The total is 200. Solution plan x (Matt’s)+ 3x (Jill’s) = 200 Solve x + 3x = 2004x = 200; x = 50 Conclusion Jill has “3x” or 150 cards. Solution Plan

  30. Solving a word problem with a total of two types of items. Diane’s Card Shop spent a total of $950 ordering 600 cards from Wit’s End Co., whose humorous cards cost $1.75 and whose nature cards cost $1.50 each. How many of each style of card did the card shop order? • Use the solution plan to solve this problem.

  31. What are you looking for? • How many humorous cards were ordered and how many nature cards were ordered. • The total of H + N = 600 • Another way to look at this is:N = 600 – H • If we let “H” represent the humorous cards, Nature cards will be 600- H. • This will simplify the solution process by using only one unknown: “H.”

  32. What do you know? A total of $950 was spent. Two types of cards were ordered. The total number of cards ordered was 600. The humorous cards cost $1.75 each/nature cards cost $1.50 each. Organize the information.

  33. Total spent Unit prices Volume “unknowns” Solution plan • Set up the equation by multiplying the unit price of each by the volume, represented by the unknowns equaling the total amount spent. • $1.75(H) + $1.50 (600 – H) = $950.00

  34. Solve the equation. • $1.75H + $1.50(600-H) = $950.00 • $1.75H + $900.00-$1.50H = $950.00 • $0.25H + $900.00 = $950.00 • $0.25H = $50.00 • H = 200

  35. Conclusion • H = 200 • The number of humorous cards ordered is 200. • Since nature cards are 600 – H, we can conclude that 400 nature cards were ordered. • Using “200” and “400” in the original equation proves that the volume amounts are correct.

  36. Try this problem. • Denise ordered 75 dinners for the awards banquet. Fish dinners cost $11.75 and chicken dinners cost $9.25 each. If she spent a total of $756.25, how many of each type of dinner did she order? • Use the solution plan to organize the information and solve the problem.

  37. Denise’s order • $11.75(F) + $9.25(75-F) = $756.25 • $11.75 F + $693.75 - $9.25F = $756.25 • $2.50F + $693.75 = $756.25 • $2.50F = $62.50 • F = 25 • Conclusion: 25 fish dinners and 50 chicken dinners were ordered.

  38. Proportions • The relationship between two factors is often described in proportions. You can use proportions to solve for unknowns. • Example: The label on a container of weed killer gives directions to mix three ounces of weed killer with every two gallons of water. For five gallons of water, how many ounces of weed killer should you use?

  39. What are you looking for? The number of ounces of weed killer needed for 5 gallons of water. What do you know? For every 2 gallons of water, you need 3 oz. of weed killer. Set up solution plan. 2/3 = 5/x Solve the equation. Cross multiply.2x = 15; x = 7.5 Conclude You need 7.5 ounces of weed killer for 5 gallons of water. Use the solution plan.

  40. Proportions • Your car gets 23 miles to the gallon. How far can you go on 16 gallons of gas? • 1 gallon/23 miles = 16 gallons/ x miles • Cross multiply: 1x = 368 miles • Conclusion: You can travel 368 miles on 16 gallons of gas.

  41. Direct Proportions • Many business-related problems that involve pairs of numbers that are proportional involve direct proportions. • An increase (or decrease) in one amount causes an increase (or decrease) in the number that pairs with it. • In the previous example, an increase in the amount of gas would directly and proportionately increase the mileage yielded.

More Related