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Instructor: Dr.Gehan Shanmuganathan

Instructor: Dr.Gehan Shanmuganathan. Learning Outcomes. 5-1. 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.

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Instructor: Dr.Gehan Shanmuganathan

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  1. Instructor: Dr.Gehan Shanmuganathan

  2. Learning Outcomes 5-1 • Solve equations usingmultiplication or division. • Solve equations usingaddition or subtraction. • Solve equations usingmore than one operation. • Solve equations containing multipleunknown terms. • Solve equations containing parentheses. • Solve equations that are proportions.

  3. Solve equations usingmultiplication or division 5-1-1 Section 5-1 Equations • An equation is a mathematical statement in which two quantities are equal. • Solving an equation means finding the valueof an unknown. Example: 8x = 24 To solve this equation, thevalue of x must be discovered. Division is used to solve this equation.

  4. Solve equations usingmultiplication or division Section 5-1 Equations Letters, such as (x,y,z)represent unknown amounts and are called unknowns or variables. 4x = 16 The numbers are calledknown or givenamounts.

  5. Solve equations usingmultiplication or division Section 5-1 Equations • Any operation performed on one side of the equation mustbe performed on the otherside 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 also“divide by 3” on the other side.

  6. Solve equations usingmultiplication or division STEP 1 Isolate the unknown value and determineif multiplication or division is needed. STEP 2 Use division to divide both sides by 8. STEP 3 Simplify: x = 3 HOW TO: Section 5-1 Equations 8x = 24 3 x 8 = 24

  7. Find the value of an unknownusing multiplication HOW TO: Section 5-1 Equations Find the value of a: Multiply both sides by 3 to isolate a. The left side becomes 1a or a. The right side becomes theproduct of 6 x 3, or 18. a = 18

  8. STEP 1 Determine which operation is needed. STEP 2 Perform the same operationto both sides. STEP 3 Isolate the variableand solve. An Example… Section 5-1 Equations 2b = 40 Division Divide both sides by 2.

  9. Solve equations usingaddition or subtraction 5-1-2 Section 5-1 Equations • 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 4 + x = 16 require additionor subtractionof “4” from each side? Subtraction

  10. Solve equations usingaddition or subtraction STEP 1 Isolate the unknown value and determineif addition or subtraction is needed. STEP 2 Use subtraction to isolate x. STEP 3 Simplify: x = 6 HOW TO: Section 5-1 Equations 4 + x = 10

  11. STEP 1 Determine which operation is needed. STEP 2 Perform the same operationto both sides. STEP 3 Isolate the variableand solve. An Example… Section 5-1 Equations b - 12 = 8 Addition Add 12 to both sides b = 8 + 12 = 20

  12. Solve equations usingmore than one operation 5-1-3 Section 5-1 Equations • 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” thenumber using the original equation.

  13. When two or more calculations are written symbolically, the operations are performed according to a specified order of operations. First — perform multiplication and division as theyappear from left to right. Second — perform addition and subtraction as theyappear from left to right. Order of operations Section 5-1 Equations

  14. To solve an equation, undo the operations, working in reverse order First — undo the addition or subtraction. Second — undo multiplication or division. Order of operations Section 5-1 Equations

  15. STEP 1 Undo the addition by subtracting 4 from each side. STEP 2 Divide each side by 7. STEP 3 Verify by plugging in 5 in place of x . An Example… Section 5-1 Equations 7x + 4 = 39 7x = 35 7 (5) + 4 = 39 35 + 4 = 39

  16. Solve equations containingmultiple 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, such as 3a + 2a = 25 Add the numbers in each addend (2+3). Multiply the sum by the unknown (5a= 25). Solve for a (a = 5). 5-1-4 Section 5-1 Equations

  17. STEP 1 Combine the unknown value addends. STEP 2 STEP 3 Undo the subtraction. Undo the multiplication. STEP 4 Check by replacing a with 7. An Example… Section 5-1 Equations Find a if: a + 4a – 5 = 39 a + 4a = 5a 5a – 5 = 30 5a = 35 a = 7 Correct! 7 + 4(7) = 35

  18. Eliminate the parentheses. Multiply the number just outside the parenthesesby each addend inside the parentheses. Show the resulting products as addition orsubtraction, as indicated Solve the resulting equation. Solve equations containing parentheses 5-1-5 Section 5-1 Equations

  19. STEP 1 Multiply 6 by each addend. STEP 2 Show the resulting products. STEP 3 Check by replacing a with 7. An Example… Section 5-1 Equations Solve: 6A + 2 = 24 6 multiplied by A+ 6 multiplied by 2 6A + 12 = 24 7 + 4(7) = 35

  20. TIP: Remove the parentheses first. An Example… Section 5-1 Equations 5 (x- 2) = 45 5x -10 = 45 5x = 55 x= 11

  21. A proportion is based on two pairs of related quantities. The most common way to write proportions isto use fraction notation—also called a ratio. When two ratios are equal, they form a proportion. Solve equations that are proportions 5-1-6 Section 5-1 Equations

  22. A cross product is the product of the numerator of one fraction, times the denominator of another fraction. An important property of proportions is that the cross products are equal. Solve equations that are proportions 5-1-6 Section 5-1 Equations

  23. Do and form a proportion? STEP 1 Multiply the numerator from the first fraction by the denominator of the second fraction. STEP 2 Multiply the denominator of the first fraction by the numerator of the second fraction. Verify that two fractions form a proportion HOW TO: Section 5-1 Equations 4 x 18 = 72 6 x 12 = 72 Yes, they form a proportion. Are they equal?

  24. Learning Outcome 5-2 • Use the problem-solvingapproach to analyze andsolve word problems.

  25. Use the problem-solving approachto analyze and solve word problems. 5-2-1 Section 5-2 Using Equations to Solve Problems • Five step problem solving approach: • 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.

  26. Use the problem-solving approachto analyze and solve word problems. Key words in Table 5-1will guide you in using the problem-solving approach. See page533 “of” often implies multiplication. “¼ of her salary” means “multiply her salary by ¼” Example: 5-2-1 Section 5-2 Using Equations to Solve Problems These words help you interpret the information andbegin to set up the equation to solve the problem.

  27. Full time employees work morehours than part-time employees. Use the solution plan HOW TO: Section 5-2 Using Equations to Solve Problems If the differenceis four per day,and part-time employees worksix hours per day, how manyhours per day do full-timers work? What are we looking for? Number of hours that FT employees work. What do we know? PT employees work 6 hours, and thedifference between FT and PT is 4 hours.

  28. Full time employees work morehours than part-time employees. If the differenceis four per day,and part-time employees worksix hours per day, how manyhours per day do full-timers work? Set up a solution plan. FT – PT = 4 FT = N [unknown] PT = 6 hours N – 6 = 4 Solution plan: N = 4 + 6 = 10 What are we looking for? Number of hours that FT employees work. What do we know? PT employees work 6 hours, and thedifference between FT and PT is 4 hours. We also know that “difference” implies subtraction. Conclusion:Full time employees work 10 hours. Use the solution plan HOW TO: Section 5-2 Using Equations to Solve Problems

  29. 1. What are you looking for? The number of cards that Jill has. 2. What do you know? The relationship in the number of cards is 3:1; total is 200 3. Set up a solution plan. x(Matt’s) + 3x(Jill’s) = 200 4. Solve it. x + 3x = 200; 4x = 200; x = 50 5. Draw the conclusion. Jill has 3x, or 150 cards Use the solution plan HOW TO: Section 5-2 Using Equations to Solve Problems Jill has three times as many trading cards as Matt. If thetotal number both have is 200, how many does Jill have?

  30. MORE Use the solution plan HOW TO: Section 5-2 Using Equations to Solve Problems Diane’s Card Shop spent a total of $950 ordering 600 cards from Wit’s End Co., whose humorous cards cost $1.75 each and whose nature cards cost $1.50 each. How many of each style of card did the card shop order? What are you looking for? • How manyhumorous cards were ordered and how manynature cards were ordered—the total of H + N = 600 or N = 600 – H. • If we let H represent the humorous cards, Nature cards will be600 – H, which will simplify the solution process by using onlyone unknown: H.

  31. MORE Use the solution plan HOW TO: Section 5-2 Using Equations to Solve Problems Diane’s Card Shop spent a total of $950 ordering 600 cards from Wit’s End Co., whose humorous cards cost $1.75 each and whose nature cards cost $1.50 each. How many of each style of card did the card shop order? What do you know? A total of $950 was spent. Two types of cards were ordered. The total number of cards ordered was 600. Humorous cards cost $1.75 each and nature cardscost $1.50 each.

  32. Set up the equation by multiplying the unitprice of each by the volume, represented bythe unknowns equaling the total amount spent. Solution Plan Totalspent Unit prices Volumeunknowns MORE Use the solution plan HOW TO: Section 5-2 Using Equations to Solve Problems Diane’s Card Shop spent a total of $950 ordering 600 cards from Wit’s End Co., whose humorous cards cost $1.75 each and whose nature cards cost $1.50 each. How many of each style of card did the card shop order? $1.75(H) + $1.50 (600 – H) = $950.00

  33. MORE Use the solution plan HOW TO: Section 5-2 Using Equations to Solve Problems Diane’s Card Shop spent a total of $950 ordering 600 cards from Wit’s End Co., whose humorous cards cost $1.75 each and whose nature cards cost $1.50 each. How many of each style of card did the card shop order? Solution $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

  34. Use the solution plan HOW TO: Section 5-2 Using Equations to Solve Problems Diane’s Card Shop spent a total of $950 ordering 600 cards from Wit’s End Co., whose humorous cards cost $1.75 each and whose nature cards cost $1.50 each. How many of each style of card did the card shop order? Conclusion 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.

  35. Use the solution plan HOW TO: Section 5-2 Using Equations to Solve Problems 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? $11.75(F) + $9.25(75 - F) = $756.25 $11.75F + $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.

  36. 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 killershould you use? Proportions Section 5-2 Using Equations to Solve Problems

  37. The relationship between two factors is often described in proportions. You can use proportions to solve for unknowns. Proportions Section 5-2 Using Equations to Solve Problems • 1. What are you looking for? • Numberof ouncesof weed killer needed for 5 gallons of water. • 2. What do you know? • For every 2 gallonsof water, you need 3 oz.of weed killer. 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 killershould you use? • Set up a solution plan. • Conclusion. • You need7.5 oz of weedkillerfor 5 gal of water. • Solve it. Cross multiply: 2x = 15; x = 7.5

  38. 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. Proportions Section 5-2 Using Equations to Solve Problems

  39. An Example… Section 5-2 Using Equations to Solve Problems Your car gets 23 miles to the gallon. How far can you go on 16 gallons of gas? Cross multiply: 1x = 368 miles Conclusion: You can travel 368 miles on 16 gallons of gas. In this example, an increase in the amount of gas would directly and proportionately increase the mileage yielded.

  40. Learning Outcomes 5-3 • Evaluate a formula. • Find an equivalent formulaby rearranging the formula.

  41. Evaluate the formula 5-3-1 Section 5-3 Formulas • Write the formula. • Rewrite the formula substituting known valuesfor the letters of the formula. • Solve the equation for the unknown letter or perform the indicated operations, applying the order of operations. • Interpret the solution within the context of the formula.

  42. An Example… Section 5-3 Formulas A plasma TV that costs $2,145 is marked up $854. What is the selling price of the TV? Use the formula S = C + M where S is the selling price, C is the cost, and M is Markup. S = $2,145 + $854 S or Selling Price = $2,999

  43. Find an Equivalent Formulaby Rearranging the Formula Determine which variable of the formula is to be isolated (solved for). Highlight or mentally locate all instances of the variable to be isolated. Treat all other variables of the formula as you would treat numbers in an equation, and perform normal steps for solving an equation. If the isolated variable is on the right side of the equation, interchange the sides so that it appears on the left side. 5-3-2 Section 5-3 Formulas

  44. An Example… Section 5-3 Formulas The formula forSquare Footage = Length x Width or S = L x W. Solve the formula for W or width.Isolate W by dividing both sides by L. The new formula is:

  45. EXERCISE SET A

  46. EXERCISE SET A

  47. EXERCISE SET A

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