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§ 3.2

§ 3.2. Problem Solving and Business Applications Using Systems of Equations. Systems of Equations in Application. pp 182-183. Blitzer, Intermediate Algebra , 5e – Slide # 2 Section 3.2. Systems of Equations in Application Investment Problem, p 184. EXAMPLE (number 13 on page 192).

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§ 3.2

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  1. §3.2 Problem Solving and Business Applications Using Systems of Equations

  2. Systems of Equations in Application pp 182-183 Blitzer, Intermediate Algebra, 5e – Slide #2 Section 3.2

  3. Systems of Equations in ApplicationInvestment Problem, p 184 EXAMPLE (number 13 on page 192) You invested $7000 in two accounts paying 6% and 8% annual interest, respectively. If the total interest earned for the year was $520, how much was invested at each rate? SOLUTION 1) Use variables to represent unknown quantities. Let x = the amount invested at 6%. Let y = the amount invested at 8%. Blitzer, Intermediate Algebra, 5e – Slide #3 Section 3.2

  4. Systems of Equations in ApplicationInvestment Problem, p 184 CONTINUED 2) Write a system of equations describing the problem’s conditions. Column One of table x + y = 7000 This tells us the sum of the money invested is $7000. Also, we’re concerned with how much money was invested into each of the accounts. We can use a table to organize the information in the problem and obtain a second equation. Blitzer, Intermediate Algebra, 5e – Slide #4 Section 3.2

  5. Systems of Equations in ApplicationInvestment Problem, p 184 CONTINUED The amount of interest from each account is: 0.06x and 0.08y. And since we are concerned with both amounts of interest adding up to $520, the following equation results: Column Three of table 0.06x + 0.08y = 520. 3) Solve the system and answer the problem’s question. The system x + y = 7000 0.06x + 0.08y = 520 Blitzer, Intermediate Algebra, 5e – Slide #5 Section 3.2

  6. Systems of Equations in ApplicationInvestment Problem, p 184 CONTINUED can be solved by substitution or addition. Substitution will work well since both variables in the first equation have coefficients of 1. So, we’ll use substitution. First we’ll isolate x in the first equation. x + y = 7000 x = 7000 - y Subtract y from both sides Now substitute 7000 – y for x in the second equation. Blitzer, Intermediate Algebra, 5e – Slide #6 Section 3.2

  7. Systems of Equations in ApplicationInvestment Problem, p 184 CONTINUED 0.06x + 0.08y = 520 0.06(7000 - y) + 0.08y = 520 Replace x with 7000 - y 420 – 0.06y + 0.08y = 520 Distribute 420 + 0.02y = 520 Add like terms 0.02y = 100 Subtract 100 from both sides y = 5000 Divide both sides by 0.02 Now substitute 5000 for y in either of the original equations. We’ll use the first. Blitzer, Intermediate Algebra, 5e – Slide #7 Section 3.2

  8. Systems of Equations in ApplicationInvestment Problem, p 184 CONTINUED x + y = 7000 x + (5000) = 7000 Replace y with 7000 x = 2000 Subtract 5000 from both sides Therefore, $2000 will be invested into the 6% account and $5000 will be invested in the 8% account. This is the potential solution. Blitzer, Intermediate Algebra, 5e – Slide #8 Section 3.2

  9. Systems of Equations in ApplicationInvestment Problem, p 184 CONTINUED 4) Check the proposed answer in the original wording of the problem. Do both investments add up to $7000? x + y = 2000 + 5000 = 7000 true Do the two investments yield a combined $520 in interest? The 6% investment yields, in interest, 0.06x = 0.06(2000) = $120. The 8% investment yields, in interest, 0.08y = 0.08(5000) = $400. And since $120 + $400 = $520, the solution has been verified. Blitzer, Intermediate Algebra, 5e – Slide #9 Section 3.2

  10. Systems of Equations in ApplicationMixture Problem, p186 EXAMPLE (number 20 on page 193) A jeweler needs to mix an alloy with a 16% gold content and an alloy with a 28% gold content to obtain 32 ounces of a new alloy with a 25% gold content. How many ounces of each of the original alloys must be used? SOLUTION 1) Use variables to represent unknown quantities. Let x = the number of ounces of the 16% alloy to be used in the mixture. Let y = the number of ounces of the 28% alloy to be used in the mixture. Blitzer, Intermediate Algebra, 5e – Slide #10 Section 3.2

  11. Systems of Equations in ApplicationMixture Problem, p186 CONTINUED 2) Write a system of equations describing the problem’s conditions. We need 32 ounces of an alloy containing 25% gold content. We form a table that shows the amount of gold content in each of the three alloys. Blitzer, Intermediate Algebra, 5e – Slide #11 Section 3.2

  12. Systems of Equations in ApplicationMixture Problem, p186 CONTINUED Since adding the two amounts of alloy (x and y) will yield 32 ounces of 25% alloy, Column One of table x + y = 32. The 32-ounce mixture must have a 25% gold content. The amount of gold content must be 25% of 32 ounces, or (0.25)(32) = 8 ounces. Column Three of table 0.16x + 0.28y = 8 Therefore, the system of equations is x + y = 32 0.16x + 0.28y = 8 Blitzer, Intermediate Algebra, 5e – Slide #12 Section 3.2

  13. Systems of Equations in ApplicationMixture Problem, p 186 CONTINUED NOTE: The first equation summarizes how many total ounces of each of the three alloys. The second equation summarizes how much gold content in each of the three solutions. Each equation summarizes exactly one type of quantity. 3) Solve the system and answer the problem’s question. The system can be solved by substitution or addition. Substitution will work well since both variables in the first equation have coefficients of 1. So, we’ll use substitution. First we’ll isolate y in the first equation. Blitzer, Intermediate Algebra, 5e – Slide #13 Section 3.2

  14. Systems of Equations in ApplicationMixture Problem, p 186 CONTINUED x + y = 32. y = 32 - x Subtract x from both sides Now substitute 32 – xfor y in the second equation. 0.16x + 0.28y = 8 Replace y with 32 - x 0.16x + 0.28(32 - x) = 8 Distribute 0.16x + 8.96 – 0.28x = 8 Add like terms 8.96 – 0.12x = 8 Subtract 8.96 from both sides – 0.12x = -0.96 x = 8 Divide both sides by -0.12 Blitzer, Intermediate Algebra, 5e – Slide #14 Section 3.2

  15. Systems of Equations in ApplicationMixture Problem, p 186 CONTINUED Now substitute 8 for x in either of the original equations. We’ll use the first. x + y = 32 8 + y = 32 Replace x with 8 y = 24 Subtract 8 from both sides Therefore, the potential solution is (8,24). That is, 8 ounces of the 16% alloy and 24 ounces of the 28% alloy. Blitzer, Intermediate Algebra, 5e – Slide #15 Section 3.2

  16. Systems of Equations in ApplicationMixture Problem, p 186 CONTINUED 4) Check the proposed answer in the original wording of the problem. Do the amounts of the two alloys add up to the desired 32 ounces? x + y = 8 + 24 = 32 true Also, the problem states that we need 32 ounces of a 25% gold content alloy. The amount of gold content in this mixture is (32)(0.25) = 8 ounces of gold content. The amount of gold content in 8 ounces of a 16% solution is (8)(0.16) = 1.28 ounces. The amount of gold content in 24 ounces of a 28% alloy is (24)(0.28) = 6.72 ounces. So, 1.28 + 6.72 = 8 ounces of gold content, exactly as it should be. Blitzer, Intermediate Algebra, 5e – Slide #16 Section 3.2

  17. Systems of Equations in ApplicationMotion Problem, p 188 EXAMPLE (number 30 on page 193) A motorboat traveled 36 miles downstream, with the current, in 1.5 hours. The return trip upstream, against the current, covered the same distance, but took 2 hours. Find the boat’s rate in still water and the rate of the current. SOLUTION 1) Use variables to represent unknown quantities. Let x = the rate (speed) of the motorboat. Let y = the rate (speed) of the stream. Blitzer, Intermediate Algebra, 5e – Slide #17 Section 3.2

  18. Systems of Equations in ApplicationMotion Problem, p 188 CONTINUED 2) Write a system of equations describing the problem’s conditions. Blitzer, Intermediate Algebra, 5e – Slide #18 Section 3.2

  19. Systems of Equations in ApplicationMotion Problem, p 188 CONTINUED Therefore, the system of equations is: (x + y)1.5 = 36 (x - y)2 =36 3) Solve the system and answer the problem’s question. Upon distributing, the system simplifies to (x + y)1.5 = 36 (x - y)2 =36 1.5x + 1.5y = 36 2x - 2y =36 I will now solve the second equation for x. x =18 + y 2x - 2y =36 2x =36 + 2y Second equation Add 2y Divide by 2 Blitzer, Intermediate Algebra, 5e – Slide #19 Section 3.2

  20. Systems of Equations in ApplicationMotion Problem, p 188 CONTINUED In the first equation, replace x with 18 + y. 1.5x + 1.5y = 36 First equation 1.5(18 + y)+ 1.5y = 36 Replace x with 18 + y 27 + 1.5y + 1.5y = 36 Distribute 27 + 3y = 36 Combine like terms 3y = 9 Subtract 27 from both sides y = 3 Divide both sides by 3 Now we can use either equation to find x. Let’s use the second equation. Blitzer, Intermediate Algebra, 5e – Slide #20 Section 3.2

  21. Systems of Equations in ApplicationMotion Problem, p 188 CONTINUED 2x - 2y =36 Second equation 2x – 2(3) =36 Replace y with 3 2x – 6 =36 Multiply 2x =42 Add 6 to both sides x = 21 Divide both sides by 2 Therefore, the potential solution is (21,3). Blitzer, Intermediate Algebra, 5e – Slide #21 Section 3.2

  22. Systems of Equations in ApplicationMotion Problem, p 188 CONTINUED 4) Check the proposed answer in the original wording of the problem. I now verify that my solutions satisfy the original equations: 1.5x + 1.5y = 36 2x - 2y =36 ? ? 1.5(21)+ 1.5(3) = 36 2(21) – 2(3) =36 ? ? 31.5+ 4.5 = 36 42 – 6 =36 36 = 36 true 36 =36 true So, the motorboat travels 21 miles per hour in still water and the speed of the current is 3 miles per hour. Blitzer, Intermediate Algebra, 5e – Slide #22 Section 3.2

  23. Systems of Equations in ApplicationInvestment Problem, p 185 Check Point 2 You inherited $5,000 with the stipulation that for the first year the money had to be invested in two funds paying 9% and 11% annual interest. How much did you invest at each rate if the total interest earned for the year was $487? Column One of table Column Three of table SOLUTION .09 x .09x .11y y .11 $487 $5,000 Blitzer, Intermediate Algebra, 5e – Slide #23 Section 3.2

  24. Systems of Equations in ApplicationInvestment Problem, p 184 CONTINUED 2) Write a system of equations describing the problem’s conditions. Column One of table x + y = 5000 .09x +.11y = 487 Column Three of table x + y = 5000 x = 5000 - y Subtract y from both sides 0.09x + 0.11y = 487 0.09(5000 - y) + 0.11y = 487 Replace x with 5000 - y 450 – 0.09y + 0.11y = 487 Distribute Blitzer, Intermediate Algebra, 5e – Slide #24 Section 3.2

  25. Systems of Equations in ApplicationInvestment Problem, p 184 CONTINUED 450 + 0.02y = 487 Add like terms 0.02y = 37 Subtract 450 from both sides y = 1850 Divide both sides by 0.02 x + y = 5000 x + (1850) = 5000 Replace y with 1850 x = 3150 Subtract 1850 from both sides Therefore, $3150 will be invested into the 9% account and $1850 will be invested in the 11% account. This is the potential solution. Blitzer, Intermediate Algebra, 5e – Slide #25 Section 3.2

  26. Systems of Equations in ApplicationMixture Problem, p 187 Check Point 3 A chemist needs to mix a 12% acid solution with a 20% acid solution to obtain 160 ounces of a 15% acid solution. How many ounces of each of the acid solutions must be used? Column One of table Column Three of table SOLUTION .12 x .12x .20y y .20 .15 24 160 Blitzer, Intermediate Algebra, 5e – Slide #26 Section 3.2

  27. Systems of Equations in ApplicationMixture Problem, p 187 CONTINUED 2) Write a system of equations describing the problem’s conditions. Column One of table x + y = 160 .12x +.20y = 24 Column Three of table x + y = 160 x = 160 - y Subtract y from both sides 0.12x + 0.20y = 24 0.12(160 - y) + 0.20y = 24 Replace x with 160 - y 19.2 – 0.12y + 0.20y = 24 Distribute Blitzer, Intermediate Algebra, 5e – Slide #27 Section 3.2

  28. Systems of Equations in ApplicationMixture Problem, p 187 CONTINUED 19.2 + 0.08y = 24 Add like terms 0.08y = 4.8 Subtract 19.2 from both sides y = 60 Divide both sides by 0.08 x + y = 160 x + (60) = 160 Replace y with 60 x = 100 Subtract 60 from both sides Therefore, the potential solution is (100,60). That is, 100 ounces of the 12% solution and 60 ounces of the 20% alloy. Blitzer, Intermediate Algebra, 5e – Slide #28 Section 3.2

  29. Systems of Equations in ApplicationMotion Problem, p 189 Check Point 4 With the current, a motorboat can travel 84 miles in 2 hours. Against the current, the same trip takes 3 hours. Find the average rate of the boat in still water and the average rate of the current. Both distances equals 84 SOLUTION 2 x+y 2(x+y) 3(x-y) X-y 3 2(x+y) = 84 3(x-y) = 84 Blitzer, Intermediate Algebra, 5e – Slide #29 Section 3.2

  30. Systems of Equations in ApplicationMotion Problem, p 189 Check Point 4,continued With the current, a motorboat can travel 84 miles in 2 hours. Against the current, the same trip takes 3 hours. Find the average rate of the boat in still water and the average rate of the current. Both distances equals 84 SOLUTION 2 x+y 2(x+y) 3(x-y) X-y 3 2(x+y) = 84 6x+6y = 252 x = 35 3(x-y) = 84 6x-6y = 168 Boat rate is 35 mph, current is 7 mph 12x = 420 Elimination Blitzer, Intermediate Algebra, 5e – Slide #30 Section 3.2

  31. Revenue, Cost and Profit The point of intersection of the revenue and cost functions is called the break-even point. The x-coordinate of the point reveals the number of units that a company must produce and sell so that the money coming in is equal to the money going out. That is, the break-even point is where revenue = cost. Blitzer, Intermediate Algebra, 5e – Slide #31 Section 3.2

  32. Systems of Equations in Application EXAMPLE You invested $30,000 and started a business writing greeting cards. Supplies cost 2 cents per card and you are selling each card for 50 cents. (In solving this exercise, let x represent the number of cards produced and sold.) (a) Write the cost function, C. (b) Write the revenue function, R. (c) Determine the break-even point. Describe what this means. SOLUTION (a) Write the cost function, C. C(x) = 30,000 + 0.02x Blitzer, Intermediate Algebra, 5e – Slide #32 Section 3.2

  33. Systems of Equations in Application CONTINUED (b) Write the revenue function, R. R(x) = 0.50x (c) Determine the break-even point. Describe what this means. The break-even point occurs when revenue and cost are equal. That is, Blitzer, Intermediate Algebra, 5e – Slide #33 Section 3.2

  34. Systems of Equations in Application CONTINUED R(x) = C(x) 0.5x = 30,000 + 0.02x 0.48x = 30,000 Subtract 0.02x from both sides x = 62,500 Divide both sides by 0.48 Therefore, the break-even point will occur when 62,500 cards are sold. When that happens, revenue and cost will be equal. For any number of units over 62,500 sold, the company will make a profit. Blitzer, Intermediate Algebra, 5e – Slide #34 Section 3.2

  35. Systems of Equations in ApplicationMixture Problem on p 193 Number 24 on page 187 A grocer needs to mix raisins at $2.00 per pound with granola at $3.25 per pound to obtain 10 pounds of a mixture that costs $2.50 per pound. How many pounds of raisins and how many pounds of granola must be used? Column One of table Column Three of table SOLUTION 2.00 x 2.00x 3.25y y 3.25 2.50 $25 10 Blitzer, Intermediate Algebra, 5e – Slide #35 Section 3.2

  36. Systems of Equations in ApplicationMixture Problem on p 187 CONTINUED 2) Write a system of equations describing the problem’s conditions. Column One of table x + y = 10 2x +3.25y = 25 Column Three of table x + y = 10 x = 10 - y Subtract y from both sides 2x +3.25y = 25 2(10 - y) + 3.25y = 25 Replace x with 10 - y 20 – 2y + 3.25y = 25 Distribute Blitzer, Intermediate Algebra, 5e – Slide #36 Section 3.2

  37. Systems of Equations in ApplicationMixture Problem on p 187 CONTINUED Add like terms 20 + 1.25y = 25 1.25y = 5 Subtract 20 from both sides y = 4 Divide both sides by 1.25 x + y = 10 x + (4) = 10 Replace y with 4 x = 6 Subtract 4 from both sides Therefore, the potential solution is (6,4). That is, 6 pounds of the raisins and 4 pounds of granola. Blitzer, Intermediate Algebra, 5e – Slide #37 Section 3.2

  38. DONE

  39. DONE

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