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Algebra

Review and Application. Algebra. Chapter 2. PowerPoint Presentation Prepared by C Quinn, Seneca College. Learning Objectives. LO 1. LO 2. LO 3. LO 4. LO 5. After completing this chapter, you will be able to:. Simplify algebraic expressions. Rearrange a formula or equation

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Algebra

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  1. Review and Application Algebra Chapter 2 PowerPoint Presentation Prepared by C Quinn, Seneca College.

  2. Learning Objectives LO 1. LO 2. LO 3. LO 4. LO 5. After completing this chapter, you will be able to: Simplify algebraic expressions Rearrange a formula or equation to isolate a particular variable Solvelinear equations in one or two variables Solve"word problems" that lead to a linear equation in one unknown, or two equations in two unknowns Solve problems involving a percent change

  3. Algebraic Operations LO 1. Algebraic Expression …indicates the mathematical operations to be carried outon acombination ofNUMBERS andVARIABLES

  4. Algebraic Operations Terms …the components of an Algebraic Expression that are separated by ADDITION or SUBTRACTIONsigns x(2x2 –3x – 1)

  5. Algebraic Operations Monomial Binomial Trinomial Polynomial Terms 1 Term 2 Terms 3 Terms …any more than 1 Term! 3x2 +xy 3x2 3x2 +xy – 6y2

  6. Algebraic Operations 36x2y 60xy2 …assumed when two factors are written beside each other! Term …each one in an Expression consists of one or more FACTORS separated by MULTIPLICATION or DIVISIONsign …assumed when one factor is written under an other! xy = x*y

  7. Algebraic Operations 3 x2 Term FACTOR Numerical Coefficient Literal Coefficient 3x2

  8. Review and Application Algebra 36 x2y 60 xy2 Example FACTORS Step 1 Step 2 3x = 5y Division by aMonomial Identify Factors in the numerator and denominator 36x2y 3(12)(x)(x)(y) = 60xy2 5(12)(x)(y)(y) Cancel Factors in the numerator and denominator

  9. Review and Application Algebra 48a2 – 32ab 8a Example Step 1 Step 2 Division by aMonomial 48a2/8a – 32ab/8a or Divide each TERM in the numerator by the denominator 6 4 = 48(a)(a) 32ab - 8a 8a Cancel Factors in the numerator and denominator = 6a – 4b

  10. Review and Application Algebra -x(2x2 – 3x – 1) Example ( ( ( ( ( -x -x -x -1 -3x ) ) ) ) ( 2x2 ) ) MultiplyingPolynomials Multiplyeach term in the TRINOMIALby (–x) = + + The product oftwo negative quantities is positive. x -2x3 + 3x2 + =

  11. Review and Application Algebra LO 2. Use Calculator Example Evaluate S = P(1 + rt) Substitution P = $100 r = 0.09 t = 7/12 7 Months S = $100 [1 + 0.09*(7/12)] Think BEDMAS 105.25 .09 * 7/12 +1 = * 100 = S = $100 [1.0525] S = $105.25

  12. Review and Application Algebra (1 + i)20 (1 + i)8 ExponentsRule of 34 32 *33 (32)4 3 Base = 32 + 3 = 32*4 =(1+ i)20-8 Exponent 3 4 = 3 5 = 3 8 i.e. 3*3*3*3 = (1+ i) 12 = 243 = 81 = 6561 Power

  13. Review and Application Algebra 3x6y3 2 x2z3 Step 1 Step 3 Step 2 Simplifyinside the brackets first Simplify Square each term 3x6y3 2 x2z3 3x4y3 2 z3 9x8y6 32x4*2y3* 2 = = ExponentsRule of X4 = z6 Z3*2

  14. Review and Application Algebra Use Calculator ExponentsRule of …to (a) evaluate (1.62)5 1.62 5 11.16 …to (b) evaluate (1.62)-5 1.62 5 0.0896

  15. Review and Application Algebra LO 2. Example Evaluate S = P(1 + rt) Formula Manipulation r = 0.09 t = ? S = $105.25 P = $100 We must solve for t OR First substitutethe numerical values given,and then manipulatethe equation to solve fort First rearrange the formula to isolatet,and thensubstitute the given numerical values give.

  16. Review and Application Algebra Formula Manipulation Both sides of the equation must be treated in exactly the same way! You can add the same number to both sides You cansubtract the same number from both sides You can multiply or divideboth sides by the same number or variable You can raiseboth sides to the same exponent

  17. Review and Application Algebra LO 2. Example Evaluate S = P(1 + rt) 2-17 Formula Manipulation First rearrange the formula to isolatet,and thensubstitute the given numerical values give. S = P(1 + rt) Multiply through by P S = P + Prt Subtract P from both sides S - P = Prt (S – P) / Pr = Prt/Pr Divideboth sides by Pr (S – P) / Pr = t

  18. Review and Application Algebra LO 2. Formula Manipulation Example continued…. t = (S-P)/ Pr S = $105.25 P = $100 r = 0.09 substitute Now t =(105.25 -100) / 100*0.09 Multiplyby 12 to change years to months t = .5833 years t = 7 months

  19. Review and Application Algebra If...X+5 = $200 LO 3. What does X =? Algebra R&A 2 Solving Linear Equations

  20. Review and Application Algebra A + 9 ? 137 = Solving Linear Equationsin one Unknown Equality in Equations Expressed as: A + 9 = 137 A = 137 – 9 A = 128

  21. Review and Application Algebra Collect like Terms Step Step 1 – 0.025 0.975x Divideboth sides by 0.975 Solving Linear Equationsin one Unknown Solve forxfrom the following: x = 341.25 + 0.025x x - 0.025x = 341.25 0.975x = 341.25 x = 341.25 0.975 x = 350

  22. Review and Application Algebra LO 4. Solving Word Problems Barbie and Ken sell cars at the Auto World. Barbie sold twice as many cars as Ken. Data In Aprilthey sold 15 cars. How many cars did each sell?

  23. Review and Application Algebra Unknown(s) Cars Variable(s) 2 C Barbie Ken C Barbie sold twice as many cars as Ken. In Aprilthey sold 15 cars. How many cars did each sell? 2C + C = 15 3C = 15 Barbie = 2 C =10 Cars Ken = C = 5 Cars C = 5

  24. Review and Application Algebra Q You pay 38% income tax on any additional earnings. You have an opportunity to work overtime @ 1.5 times your base wage of $23.50 per hour. Rounded to the nearest quarter-hour, how much overtime must you work to earn enough money(after tax) to buy a canoe that costs $2,750, including sales tax?

  25. Data Tax Rate 38% on anyadditional earnings Overtime Rate =$23.50 * 1.5 times = $35.25 To Solve... Total Earnings Less 38% of Total Earnings = After Tax Income 1 - .38 = .62 Overtime Rate = $35.25 * .62= $21.86 NetEarnings If Canoe costs $2,750, then $2,750/ $21.86 = 125.8 Overtime Hours

  26. Subtract Divide by -5 Multiply by 2 Substitute y = 2 Solving Two Equations with Two Unknowns Equations 2x – 3y = – 6 x + y = 2 (A) Solve for y (B) Solve for x (A) Solve for y 2x – 3y = – 6 x + y = 4 2x + 2y = 4 -5y = -10 y = 2 (B) Solve for x 2x – 3(2) = – 6 2x – 3y = – 6 2x – 6 = – 6 2x = + 6 – 6 Check… x = 0

  27. 2x – 3y = – 6 x + y = 2 Equations = LeftSide RightSide Show Note Substituting = 2 = – 6 LS =RS LS =RS Solving Two Equations with Two Unknowns You should always checkyour answer by substituting the valuesinto each of the equations! x = 0y=2 Equation 1 Equation 2 LeftSide RightSide LeftSide RightSide = 2x – 3y = x + y = 2(0) – 3(2) = 0 + 2 = – 6 = 2

  28. Setting up linear equations with Two Variables Q York Daycare purchases the same amount of milkand orange juice each week. After price increases from $1.10 to $1.15 per litre for milk, andfrom $0.98 to$1.14 per canof frozen orange juice, the weekly bill rose from $84.40 to $91.70. How manylitres of milkand cans of orange juiceare purchased each week?

  29. Setting up linear equations with Two Variables Purchases Let x = # litres of milk Let y = # cans of orange juice Equations After price increases from $1.10 to $1.15 perlitre of milk, A. Development of… and from $0.98 to$1.14 per canof frozen orange juice, (1) 1.10x+ 0.98y = 84.40 B. 1.15x + 1.14y = 91.70 (2) the weekly bill rose from $84.40 to $91.70. C. Solving…

  30. Setting up linear equations with Two Variables Equation Equation (1) (2) Eliminate x by Dividing by 1.10 Eliminate x by Dividing by 1.15 Let x = # litres of milk Let y = # cans of orange juice 1.10x+ 0.98y = 84.40 (1.10x+ 0.98y)/1.10 = 84.40/1.10 x+ 0.8909y = 76.73 1.15x + 1.14y = 91.70 (1.15x+ 1.14y)/1.15 = 91.70/1.15 x+ 0.9913y = 79.74 …continue

  31. Subtract Equation Equation Equation (1) (2) (1) 1.10x+ 0.98y = 84.40 Substitute into Proof Solving Two Equations with Two Unknowns x+ 0.8909y = 76.73 x+ 0.9913y = 79.74 .1004y = 3.01 y = 29.98 i.e. 30 cans 1.10x+ 0.98(29.98) = 84.40 1.10x+ 29.38 = 84.40 1.10x= 84.40 -29.38 1.10x= 55.02 x= 50.02 i.e. 50 litres

  32. Setting up linear equations with Two Variables Quantity Price $ = New Weekly Cost to Purchase $91.70 Proof Litres of Milk 50 $1.15 $57.50 Cans of Orange Juice 30 1.14 34.20

  33. LO 5. Algebra R&A 2 % Change

  34. Review and Application Algebra Q $ % (/100) 2500 250 2.5 Change 1500 150 1.5 Cross - multiply x 2500 = 100% 1000 Monday’s Sales were $1000 and Tuesday’s were $2500. Find the percent change. Initial Value 1000 100 1 Final Value 150% Increase X = 2500*100% / 1000 = 250% …Alternate solution

  35. Percent % Change Base Method Formula % change = Final Value – Initial Value Initial Value c = Vf– Vi * 100% Vi

  36. Percent % Q Change 2500 Difference $ 1500 $ 1500 Monday’sSales were $1000 and Tuesday’s were $2500. Find the percent change. Initial(Base)Value $1000 This method is referred to as the Base Method Final Value % change = Difference Base % change = $ 1000 c = Vf– Vi * 100% Vi = 1.5 or 150% Increase

  37. Percent Q % Change Kg. % (/100) 3 20 .2 Difference 12 80 .8 Cross - multiply x 3 In the making of dried fruit, 15kg. of fruit shrinks to 3 kg. Find the percent change. 100 1 15 Initial Value Final Value 80% Decrease = 100% 15 X = 3*100% / 15 = 20% …Alternate solution

  38. Percent % Q Change 12 Difference 12 In the making of dried fruit, 15kg. of fruit shrinks to 3 kg. Find the percent change. 15 Initial(Base)Value 3 Final Value % change = Difference Base c = Vf– Vi * 100% Vi % change = 15 = .8 or 80% Decrease

  39. Percent Q % Change Increase in Interest Rate A chartered bank is loweringthe interest rate on its loansfrom9%to7%. What will be the percent decrease in the interest rate on a given balance? = -.02 .09 % change = Difference Base = .07 - .09 .09 Recall = -.2222 or 22.22% Decrease

  40. Percent Q % Change A chartered bank is increasingthe interest rate on its loansfrom 7%to9% What will be the percentincrease in the interest rate on a given balance? % change = Difference Base = .09 - .07 .07 = .02 .07 =.2857 or28.57%Increase

  41. Percent Q % Change You paid $51.75 for a meal in Ontario which includesthe taxes of 14%.What was the price ofthe meal prior to the addition of the taxes? Initial value = Final value 1 + % change = 51.75 1+ .14 = 51.75 1.14 =$45.39

  42. Percent Q % Change Here the change is negative You paid $75 for a new coat after a 20% discount. What was the original price ofthe coat? Initial value = Final value 1 - % change = 75.00 1 - .20 = 75.00 .80 =$93.75

  43. This completes Chapter 2

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