Chapter 8: Equatios ad Relatioships Gettig Started, p. 1 1. a) 3 + = 3(6) + = 3 b) + 4 6 = (6) + 4(6) 6 = 1 + 4 6 = 3 c) 6 + 8 = 6(6) (6) + 8 = 36 1 + 8 = 3 d). + 1. =.(6) + 1.(6) = 3 + 9 = 1. a) 4a + = 4(.) + = 1 + = 1 b) 3a. = 3(.). = 7.. = 7 c) 1a = 1(.) = = d) 4a = 4(.) = 1 = 3. a) + 3: Two couters are added each time, so the patter rule must iclude, where represets the figure umber. Figure 1 has couters. accouts for of them, which leaves 3 couters to be expressed. Therefore, add 3 to to complete the algebraic expressio for this patter rule. b) m + 1: Sice oe orage square is added each time, the patter rule must iclude 1m, or just m, where m represets the figure umber. There is always oe gree triagle for all three of the figures, so add 1 to m, which gives m + 1 for this patter rule. 4. a) The patter rule is 4 1, where represets the figure umber. Figure umber Number of couters 1 3 7 3 11 4 1 19 6 3 b) 4 3 3 1 1 Number of Couters Compared to Figure Number 1 3 4 6 7 Figure umber c) Draw a lie up from 1 o the horizotal axis util it touches the exteded lie. Draw a horizotal lie from the itersectio to the vertical axis. It touches at 39, so there are 39 couters i the 1th figure.. a) Fid 1.6 o the horizotal axis. Place your ruler vertically at 1.6 ad see where the ruler touches the graph. The place the ruler horizotally from the itersectio to the vertical axis. This is just below $1 o the vertical axis, so the cost of 1.6 kg of ails is a little less tha $1, about $9.6. b) Fid 4 o the vertical axis. Place your ruler horizotally from 4 across to the graph. The place the ruler vertically from the itersectio to the horizotal axis. It touches at 4, so you could buy 4 kg of ails for $4. 6. You eed to make 3 + equal to 41. Try = 11. 3(11) + = 33 + = 38 too low Try a greater value, such = 1. 3(1) + = 36 + = 41 Figure 1 cotais 41 couters. 7. a) If t = 49, the 7t = 7(49) 8 9 If t = 8, the 7t = 7(8) = 6 = 343 too high 6 is the desired value, so the solutio is t = 8. b) If w = 1, the 3w + 3 = 3(1) + 3 = 36 + 3 = 39 The solutio is w = 1, so there is o eed to evaluate w = 13. c) If a = 7, the If a = 8, the a = (7) a = (8) = 3 = 4 = 3 too low = 3 The solutio is a = 8. 1 11 Nelso Mathematics 8 Solutios 8-1
d) If y = 9, the y 7 = (9) 7 = 18 7 = 11 The solutio is y = 9, so there is o eed to evaluate y =. 8. a) Solve by ispectio. m + = 1 meas somethig + = 1 m = 19 The solutio is m = 19. b) Solve by ispectio. s = 1 meas times somethig = 1 s = The solutio is s =. c) Try differet values for w. For example, w = 3. 4w + = 4(3) + = 1 + = 14 too low Try a greater value, such as w = 4. 4w + = 4(4) + = 18 The solutio is w = 4. d) Try differet values for y. For example, y =. 4y = 4() = = 18 The solutio is y =. 9. Try differet values for. a) For example, if = 6, the 3 = 3(6) = 18 = 16 too high Try a lower value, such as =. 3 = 3() = 1 = 13 This is the desired value, so the solutio is =. b) For example, if =, the 3 = 3() = 6 = 8 too low Try a greater value, such as =. = 3() = 7 = 73 too high Try a lower value, such as = 3. 3 = 3(3) = 69 = 67 This is the desired value, so the solutio is = 3. c) 3 = 31 = 34 = 17 d) 8 7 = 9 8 = 16 = 8.1 Solvig Equatios by Graphig, pp. 4 3. Place a ruler horizotally from 17 o the vertical axis to touch the graph of + 3. The place the ruler vertically from the itersectio to 7 o the horizotal axis. The solutio to the equatio + 3 = 17 is = 7. 4. a) Figure umber Number of tiles 1 3 4 3 b) Each time, 1 tile is added to the previous figure, so 1 is i the patter rule, where represets the figure umber. Each umber of tiles is more tha, so + is the algebraic expressio of this patter rule. c) A algebraic expressio for the patter rule is +, so a equatio to determie which figure umber has tiles is + =. d) 4 18 16 14 1 1 8 6 4 Number of Tiles Compared to Figure Number 4 6 8 1 1 14 16 18 Figure Number Draw a scatter plot usig the data i a). The coect the poits to form a lie. Exted this lie. Draw a horizotal lie from o the vertical axis to touch the graph. They itersect at Figure umber, so Figure will be made with tiles. Check: + =. a) Figure umber Number of couters 1 8 3 11 b) Each time, 3 couters are added, so 3 is i the patter rule, where represets the figure umber. Figure 1 has couters, of which 3 are represeted by 3, so there are couters left to be expressed. Add to 3 to get 3 + for a algebraic expressio of this patter rule. 8- Chapter 8: Equatios ad Relatioships
c) A algebraic expressio for the patter rule is 3 +, so a equatio to determie which figure has 3 couters is 3 + = 3. d) Number of Blue Couters Compared to Figure Number x x 4 18 16 14 1 1 8 6 4 1 3 4 Figure umber Draw a scatter plot usig the data i a). The coect the poits to form a lie. Exted this lie. Draw a horizotal lie from 3 o the vertical axis to touch the graph. They itersect at Figure umber 7, so Figure umber 7 will be made with 3 couters. Check the solutio: 3(7) + = 1 + = 3 6. a) Week umber Amout ($) 1 1 1 3 4 3 b) Each week, the bak balace icreases by $, so w is i the patter rule. I the first week David has $1 i the accout. Of that $1, $ is represeted by w, so the other $ still eeds to be expressed. So, add to w. A algebraic expressio of David s balace after w weeks is w +. c) A algebraic expressio for the bak balace is w +, so a equatio to determie whe the bak balace is 6 is w + = 6. For d), e). ad f) see the followig graph. d) Place a ruler horizotally from $6 o the vertical axis to the graph. They itersect at week 11. David s bak balace will be $6 after 11 weeks. Check the solutio: (11) + = + = 6 e) Place a ruler horizotally from 1 o the vertical axis to the graph. The graph shows that if the patter cotiues, his bak balace will reach $1 after 19 weeks. Check the solutio: (19) + = 9 + = 1 6 7 8 f) Draw a lie up from o the horizotal axis to touch the graph. The graph shows that if the patter cotiues, his bak balace will reach about $1 after weeks. Check the solutio: () + = 1 + = 1 7. a) Make a table of values. Figure umber (term umber) Number of toothpicks (term value) 1 4 7 3 1 The patter rule is 3 + 1, where is the figure umber. 1 8 6 4 1 Square Patter 1 Figure umber (term umber) From the graph, Figure 3 will have 97 toothpicks. b) Make a table of values. Figure umber Number of toothpicks (term umber) (term value) 1 3 3 7 4 9 + 1 1 8 6 4 Triagle Patter 1 1 3 3 Figure umber (term umber) From the graph, Figure 48 will have 97 toothpicks. 4 3 4 3 Nelso Mathematics 8 Solutios 8-3
8. Create a table of values for each equatio. a) + value b) + 7 value c) + 9 value 1 7 1 9 1 11 9 11 13 3 11 3 13 3 1 Use these values to graph the equatios. V a l u e 19 18 17 16 1 14 13 1 11 1 9 8 7 6 4 3 1 Graphs of + 9, + 7, ad + + 7 1 + 9 3 4 + 6 From the graph, the solutios are: + = 19 whe = 7, + 7 = 19 whe = 6, ad + 9 = 19 whe =. The equatios are the same, except that the costat value (, 7, ad 9) icreases by each time. As the costat icreases, the solutio decreases by 1. 9. Create a table of values for each equatio. a) + 3 b) 3 + 3 c) 4 + 3 value value value 1 7 3 9 7 8 1 6 9 3 1 Use these values to graph the equatios. V a l u e 16 1 14 13 1 11 1 9 8 7 6 4 3 1 Graphs of + 3, 3 + 3, ad 4 + 3 4 + 3 3 + 3 + 3 1 7 11 3 1 From the graph, the solutios are: 4 + 3 = 1 whe = 3, 3 + 3 = 1 whe = 4, ad + 3 = 1 whe = 6. Each equatio lie begis at the same spot. But whe the value of the graph is 1, the lies are far apart. This is because each lie is icreasig by a differet value. 1. a) Figure umber Area of pod Number of border tiles 1 1 8 4 1 3 9 16 4 16 b) Let represet the figure umber. The algebraic patter rule for the area of the pod is. For the umber of border tiles, 4 tiles are added each time. So 4 must be icluded i the patter rule. Figure 1 has 8 border tiles, of which 4 are represeted by 4. Therefore, you eed to add 4 to 4 to complete the patter rule for the umber of border tiles, which is 4 + 4. c) 18 16 14 1 1 8 6 4 Border Tiles ad Pod Area Compared to Figure Number 1 3 4 Figure umber d) A patter rule for the area of the pod is, which has a curve for its graph. A patter rule for the border tiles is 4 + 4, which has a straight lie for its graph. e) A expressio for the patter rule is 4 + 4, so solve 4 + 4 = 6. 4 + 4 = 6 4 = = 13 Figure 13 will have 6 tiles. f) A expressio for the patter rule is. To solve = 11, try differet values for. For example, if = 11, the = 11. Figure 11 will have a pod with a area of 11 square uits. 1 3 4 6 8-4 Chapter 8: Equatios ad Relatioships
8.3 Creatig ad Evaluatig Algebraic Expressios, pp. 9 61 4. a) Let the variable m be the umber of muffis sold ad variable j be the umber of juice boxes sold at the school sack bar. The multiply each variable by the price of that type of sack, to get a expressio for the total icome from the sale of muffis ad juice boxes: 1.m + 1.j. b) To determie the total icome, let m = 3 ad j = 4. 1.(3) + 1.(4) = 9 The total icome is $9.. a) 7c 4 = 7() 4 = 31 b) (h + ). = ( + ). = 17. c) r + 3s = (1.) + 3(.3) = 9.3 d) a + b = + (7) = + 3 = 8 e) (m ) p = (() 9) 4 = 4 6. a) 8b + 6 = 8() + 6 = 46 b) p + 1. 3 =. + 1. 3 = 7 c) 1 m = 1(3.) 3.4 = 31.6 d) x 6y + 8 = ( 1) 6( ) + 8 = 1 e) 3p + 4q 8 = 3( 3) + 4() 8 = 9 7. a) Use d to represet the umber of dimes, to represet the umber of ickels, ad q to represet the umber of quarters i the stack. Sice a dime is 1 cets, a ickel is cets, ad a quarter is cets, a algebraic expressio for the total value of ay stack of dimes, ickels, ad quarters, i cets, is 1d + + q. The value i dollars is.1d +. +.q. b) For 1 dimes, 3 ickels, ad quarters, 1d + + q = 1(1) + (3) + () = 1 + 11 + 6 = 9 or $9. The total value of the stack is $9.. 8. a) Sice water is pumped through the first hose at a rate of L/mi, a algebraic expressio to represet the amout of water pumped through the first hose after t miutes is t. b) Sice water is pumped through the secod hose at a rate of 18 L/mi, a algebraic expressio to represet the amout of water pumped through the first hose after t miutes is 18t. c) The total amout of water pumped ito the pool from both hoses after t miutes is obtaied by addig the expressios i parts a) ad b): t + 18t = 4t. d) Sice there are 6 mi i oe hour,. h = 1 mi. Usig the formula i part c), with t = 1. 4t = 4(1) = 6 6 L of water have bee pumped ito the pool after. h. 9. a) Sice Kurt jumps a average of 4.6 m ad each boy s score is the sum of the legths of his jumps, if Kurt jumps k times, a algebraic expressio to represet his score is 4.6k. b) Sice Jared jumps a average of 4.4 m ad each boy s score is the sum of the legths of his jumps, if Jared jumps j times, a algebraic expressio to represet his score is 4.4j. c) Sice the term score is the sum of the scores of both boys, a algebraic expressio for the total team score is 4.6k + 4.4j. d) Set k = ad j = 4. 4.6k + 4.4j = 4.6() + 4.4(4) = 4.6 The total team score is 4.6 if Kurt jumps times ad Jared jumps 4 times. 1. a) If she gets t 3-mark questios correct, a algebraic expressio for the umber of marks that she ears for these questios is 3t. b) If she gets w -mark questios correct, a algebraic expressio for the umber of marks that she ears for these questios is w. c) To determie the total umber of marks that Amy ears o this test, add the formulas i parts a) ad b). 3t + w. d) Set t = ad w = 13. 3t + w = 3() + (13) = 41 She will get 41 out of o the test if she aswers three-mark questios ad 13 two-mark questios correctly. 11. a) Sice a box of apples has a mass of 17. kg ad a box of pears has a mass of 1. kg, a algebraic expressio for the total mass of a boxes of apples ad p boxes of pears is 17.a + 1.p. b) For a = 3 ad p = 41, 17.a + 1.p = 17.(3) + 1.(41) = 119. The total mass of a shipmet of 3 boxes of apples ad 41 boxes of pears is 119. kg. Nelso Mathematics 8 Solutios 8-
1. a) Let w represet the width of the rectagle ad l represet the legth of the rectagle. The algebraic expressio for the perimeter of the rectagle is l + w. b) The area of the rectagle is calculated by multiplyig legth by width. Therefore, the algebraic expressio for the area of the rectagle is lw. c) For the perimeter, perimeter = l + w = (11. m) + (8.4 m) = 39. m The perimeter of the rectagle is 39. m. area = lw = 11. m 8.4 m = 94.1 m The area of the rectagle is 94.8 m. 13. a) Let h represet the rate of pay per hour ad b represet the rate of pay per budle. i) weekly earigs = 4 h ii) weekly earigs = 4h + 3(4)b = 4h + 1b iii) weekly earigs = 1b b) i) 4($8.7) = $3 ii) 4($4.7) + 1($1.3) = $19 + $16 = $3 iii) 1($.7) = $33 Nick should take the secod offer sice he ears the most with that rate of pay. 14. a) Use s to represet the umber of pairs of shoes ad b to represet the umber of pairs of boots she sells. Sice she ears a commissio of $.4 for each pair of shoes she sells, ad $.8 for each pair of boots, her total commissio would be.4s +.8b. b) Use s = 8 ad b = 7..4s +.8b =.4(8) +.8(7) = 39. Shao s commissio is $39.. 1. a) Algebraic expressios for the areas of the two squares are s ad t, respectively. The the sum of the area of the two squares is s + t. b) For example, if s = 1 ad t = 1, the s + t = 1 + 1 = If s = ad t = 3, the s + t = + 3 = 13 If s = 3 ad t = 4, the s + t = 3 + 4 = c) The resultig value sometimes represets the area of a sigle square with a whole-umber side legth. For example, the last example i part b) results a value of, which ca represet the area the square with a side legth of. 16. a) Let l represet the legth of the rectagle. Sice the legth of the rectagle is cm more tha its width, we ca express the width i terms of l as l. b) The area of a rectagle is the product of its legth ad width, which ca be expressed i terms of l as l(l ). c) Let w represet the width of the rectagle. If the legth of a rectagle is cm more tha its width, the you ca express its legth i terms of w as w +. The the area of the rectagle i terms of w is w(w + ). d) The legth is cm more tha the width, so if the legth is 1, the width is 8. Area from b = l(l ) Area from c = w(w + ) = (1)(1 ) = 8(8 + ) = 8 = 8 The area of the rectagle is 8 cm. 17. a) Use s to represet the umber of cotaiers sold by the school. Sice the school is sellig the cotaiers for $4. each, 4.s is paid to the school. But because each cotaier costs the school $., you eed to subtract.s from the moey made to obtai the profit. So the first algebraic expressio for the total profit is 4.s.s. Also, sice every cotaier is sold for $4. ad costs the school $. origially, the profit for each cotaier is $4. $. = $1.7. Therefore, the secod expressio for the total profit is 1.7s. b) First Expressio: 4.s.s = 4.(174).(174) = 34. Secod Expressio: 1.7s = 1.7(174) = 34. The total sales of 174 cotaiers is $34.. Mid-Chapter Review, p. 63 1. a) Day umber Number of golf carts 1 9 1 3 1 4 7 b) Each time, the value icreases by 6 golf carts, so 6 must be icluded i the algebraic expressio. Sice o day 1, the factory produced 9 carts ad 6 of them are give by 6, you eed to add 3 to the expressio. Therefore, a algebraic expressio for the umber of golf carts produced after days is 6 + 3. 8-6 Chapter 8: Equatios ad Relatioships
c) A expressio for the umber of golf carts is 6 + 3, so a equatio for what day 1 golf carts is produced is 6 + 3 = 1. Read across the graph from 1 to fid the itersectio at day 8. The solutio to the equatio is = 8. So, 1 karts were made o day 8.. a) Figure umber Number of couters 1 11 1 3 19 b) the umber of couters icreases by 4 each time the figure umber icreases, so 4 must be i the algebraic expressio, where represets the figure umber. Figure 1 has 11 couters, 4 of which are take ito accout by 4. Therefore, you eed to add 11 4 = 7 ito the expressio, which gives 4 + 7 for a algebraic expressio for the patter rule. c) A expressio for the umber of couters is 4 + 7, so a equatio for which figure has 63 couters is 4 + 7 = 63. d) 63 4 4 36 7 18 9 Number of Couters Compared to Figure Number 4 6 8 1 Figure umber As the graph shows, figure 14 will have 63 couters. Check: 4(14) + 7 = 63 3. a) A horizotal ruler crosses the graph at 41 whe = 1. Therefore, 4 7 = 41 whe = 1. b) A horizotal ruler crosses the graph at 49 whe = 14. Therefore, 4 7 = 49 whe = 14. 4. a) 3m + 4 = 3(1.) + 4 = 7.6 b) 8 6 = 8( ) 6 = c) m + = (1.) + (1.4) = 8.8 d) 9 + 6m = 9( ) + 6( 3) = 1 14 16. a) Sice Eri makes a deposit of $1. i her savigs accout each week ad she already has $18.73 i her accout, a algebraic expressio for the amout i her savigs accout after weeks is 18.73 + 1.. b) = 8. 18.73 + 1. = 18.73 + 1.(8) = 8.73 After 8 weeks, Eri will have $8.73 i her accout. 8.4 Solvig Equatios I, pp. 66 67 4. a) Number of laws mowed Amout eared ($) 1 1 1 1 3 3 4 6 4 3 1 Mowig Earigs at $1/h 1 1 3 3 Number of laws mowed b) She ears $1 for each law she mows. Let represet the umber of laws that she mows. The she ears 1 for mowig laws. To buy her guitar ad amplifier, she eeds $. A equatio is 1 =. c) Solve 1 =. Try 3, 1(3) = 4 too low Try 3, 13(3) = She eeds to mow 3 laws. d) For example, my solutio i part c) matches my solutio by readig from the graph, so yes, I thik that my solutio is correct.. a) + 7 = 13 = 6 b) + 3 =. =. c) Solve 9.3 = 3. For example, Predict. Evaluate 3. Is the aswer 9.3? 3 3(3) = 9 No, too low. 3.1 3(3.1) = 9.3 It s correct. = 3.1 d) Solve. =.. For example, Predict. Evaluate.. Is the aswer.? 3 (3). = 3. No, too low. 4 (4). =. It s correct. = 4 Nelso Mathematics 8 Solutios 8-7
6. a) Number of laws mowed Amout eared ($) 1 18 1 18 36 3 4 6 4 3 1 Mowig Earigs at $18/h 1 1 Number of laws mowed b) For each law she mows, Rowy ears $18. Let represet the umber of laws that she mows. The she ears 18 for mowig laws. To buy her guitar ad amplifier, she eeds $. A equatio is 18 =. c) Solve 18 =. 9.17 Rowy eeds to mow 3 laws to ear the moey she eeds. d) My solutio to the equatio ad my solutio from the graph match, so yes, I thik that my solutio is correct. 7. a) For example, 9 is close to 3, ad 3 3 = 1, so my estimate is x 1. b) For example, 19 is close to 13 ad 387 is close to 39. Sice 39 13 = 3, my estimate is c 3. c) For example, Sice 3 = 7, the 3 +.4 is about 7. My estimate is a 3. d) For example, Roud 1. to 1 ad 8.3 to 9. Sice 1 = 9, my estimate is h. e) For example, Roud 1.1 dow to 1, ad 6.4 to 6. Sice. + 1 = 6, my estimate is p.. f) For example, try = 6. 4 = 4(6) = 19 This is close to the desired value. My estimate is 6. 8. a) Try = 13. 3 3 = 3(13) 3 = 36 too low The give solutio is icorrect. Use systematic trial to determie the correct solutio. For example, 3 3 Predict. Evaluate 3 3. Is the aswer 4? 14 3(14) 3 = 39 No, too low. 1 3(1) 3 = 4 It s correct. The correct solutio is = 1. b) Try m = 7. 6m + 6 = 6( 7) + 6 = 36 The give solutio is correct. c) Try t = 9. t +.1 = 9 +.1 = 14.1 too low The give solutio is icorrect. Use systematic trial to determie the correct solutio. For example, Predict t. Evaluate t +.1. Is the aswer 6? () +.1 =.1 No, too low. () +.1 = 6.1 No, just too high. 4.9 (4.9) +.1 = 6 It s correct. The correct solutio is t = 4.9. d) Try y = 3. y + 3 = (3) + 3 = 18 The give solutio is correct. e) Try x = 1.4. 3x.4 = 3(1.4).4 = 1.8 The give solutio is correct. f) Try w =.6. w 4.3 = (.6) 4.3 = 8.7 too low The give solutio is icorrect. Use systematic trial to determie the correct solutio. For example, Predict w. Evaluate w 4.3. Is the aswer 1.7? 3 (3) 4.3 = 1.7 It s correct. The correct solutio is w = 3. 9. a) By ispectio, 4 + 4 = 8, so b = 4. b) By ispectio, 4 + = 6, so x = 1. c) Use systematic trial to solve. For example, Predict. Evaluate 4 + = 6.8 Is the aswer 6.8? 1 4(1) + = 6 No, too low. 1. 4(1.) + = 6.8 It s correct. The solutio is = 1.. d) Use systematic trial to solve. For example, Predict t. Evaluate t +.. Is the aswer 1.7? 4 (4) +. = 1. No, too low. 4. (4.) +. = 11. No, too high. 4. (4.) +. = 1.7 It s correct. The solutio is t = 4.. 8-8 Chapter 8: Equatios ad Relatioships
e) Use systematic trial to solve. For example, Predict t. Evaluate t.3. Is the aswer 4.6? 4.8 (4.8).3 = 4.7 No, just too low. 4.83 (4.83).3 = 4.6 It s correct. The solutio is t = 4.83. f) Use systematic trial to solve. For example, Predict. Evaluate 1 6.. Is the aswer 3.? 4 1(4) 6. = 3. It's correct The solutio is = 4. 1. a) For example, let be the ukow umber. Multiplyig it by 6 is represeted by 6. The result is 48, so the equatio is 6 = 48. 6 = 48 = 8 b) For example, let p be the ukow umber. To double the umber meas to multiply it by, so you have p. Subtractig 1 gives p 1, ad sice the result is 37 the equatio is p 1 = 37. To solve the equatio, use systematic trial. Predict p. Evaluate p 1. Is the aswer 37? 4 (4) 1 = 38 No, just too high. 3. (3.) 1 = 37 It s correct. The solutio is p = 3.. c) For example, let w be the ukow umber. Seve times a umber is 7w, ad the addig. gives 7w +.. This is equal to 8., so the equatio is 7w +. = 8.. To solve, try w = 8. 7w +. = 7(8) +. = 8. This is the desired value, so the solutio is w = 8. 11. a) For example, use u to represet the cost of oe uiform, before tax. There were uiforms purchased, so that gives u. The sales tax must be added to this cost, so that gives u + 119.8. The total cost overall was $918.8, so this must equal u + 119.8. b) Use systematic trial to solve. For example, Is the aswer Predict u. Evaluate u + 119.8. 918.8? 3 (3) + 119.8 = 819.8 No, too low. 4 (4) + 119.8 = 919.8 No, just too high. (39.9) + 119.8 39.9 It s correct. = 918.8 The cost of oe uiform before tax was $39.9. 1. a) Some coordiates for the graph would be (4, 96) ad (8, 19). Plot these poits ad coect them with a lie. Exted the lie. 9 8 7 6 4 3 1 Packig Cas i Boxes 1 1 Number of boxes b) To determie the umber of boxes eeded to hold 744 cas, draw a horizotal lie o the graph from 744 o the vertical axis util it touches the graph. The draw a vertical lie dow to about 3 o the horizotal axis. With this graph, estimate that about 3 boxes will be eeded to pack 744 cas of food. c) Use b to represet the umber of boxes eeded. Each box ca hold 4 cas, so 4b is part of the equatio. The total umber of cas is 744, so the equatio is 4b = 744. d) I ca tell by ispectio that to solve for b, I have to divide 744 by 4. 4b = 744 b = 31 Therefore, 31 boxes are eeded to pack 744 cas. e) Yes, I thik that my solutio is correct, because I used two differet methods to solve the problem (a graph ad a equatio) ad I got almost the same aswer with both methods. 13. a) ad b) Some coordiates for the graph would be x-value y-value 1.6 1 6 Plot these poits ad exted the lie. The amout of shelvig Austi eeds to fill is 1 m m = 8 m. From the graph, draw a horizotal lie from 8 m o the vertical axis util it touches the graph. The draw a vertical lie dow to the horizotal axis ad read that umber. I get a estimate of 13 cartos eeded to fill 8 m of shelvig. 9 8 7 6 4 3 1 Stockig the Shelves 3 3 4 6 8 Number of cartos 1 1 14 16 Nelso Mathematics 8 Solutios 8-9
c) Use c to represet the umber of cartos eeded. Each carto fills 6 cm =.6 m of shelvig, so.6c is i the equatio. There are 8 m of shelvig that eeds to be filled, so the equatio is.6c = 8. d) By ispectio, you must divide 8 by.6 to solve for the umber of cartos, c..6c = 8 c 13.3 13 cartos will almost fill the shelves. e) Yes, I thik my solutio is correct, because my estimate usig a graph ad my aswer from solvig a equatio are very close. 14. Let represet oe of the cosecutive umbers. The sice they are whole umbers, the other two umbers ca be represeted by + 1 ad +. The sum of these umbers is 36. The equatio is below. + ( + 1) + ( + ) = 36 + + + 1 + = 36 3 + 3 = 36 To solve this equatio, use systematic trial. For example, Predict. Evaluate 3 + 3. Is the aswer 36? 1 3(1) + 3 = 33 No, too low. 11 3(11) + 3 = 36 It s correct. Oe of the umbers is 11. The other two umbers are: + 1 = 11 + 1 = 1 + = 11 + = 13 The three cosecutive whole umbers whose sum is 36 are 11, 1, ad 13. Check: 11 + 1 + 13 = 36 1. Let d represet Daryl s weekly earigs. Madiso s weekly earigs ca be represeted by 3d. Together they ear $41.. The equatio is below. d + 3d = $41. 4d = $41. d = $1. Madiso ears 3d = 3($1.) or $3.7. Therefore, Daryl ears $1. each week ad Madiso ears $3.7 each week. 16. Let j represet the total umber of poits Julia scored. Oe-half of this ca be writte as.j, ad 3 more tha half is.j + 3, which represets the umber of poits Holly scored. Their total was 17 poits, so the equatio is as below. j +.j + 3 = 17 1.j + 3 = 17 To solve, use systematic trial. For example, Predict j. Evaluate 1.j + 3. Is the aswer 17? 1.() + 3 = 11 No, just too high. 48 1.(48) + 3 = 17 It s correct. So Julia s score was 48. Holly s score was:.j + 3 =.(48) + 3 = 9 So Julia scored 48 poits ad Holly scored 9 poits. Holly scored 11 more poits tha Julia. 17. a) Solve 3m + 4 = 14. It looks like m must be egative. Use systematic trial. For example, Predict m. Evaluate 3m + 4. Is the aswer 14? 3( ) + 4 = 11 No, too high. 6 3( 6) + 4 = 14 It s correct. m = 6. b) Solve 3m + 4 = 13. It looks like m must be egative. Try m = 3. 3m + 4 = 3( 3) + 4 = 9 + 4 = 13 This is the desired value, so the solutio is m = 3. c) Solve t. = 4.. It looks like the variable must be egative. Use systematic trial to solve, for example, Is the aswer Predict t. Evaluate t.. 4.? 1 ( 1). =. No, too low. 9 ( 9). = 4. It s correct. The solutio is t = 9. 8. Solvig Equatios II, pp. 7 71. a) To maitai balace, if you double the umber of couters o the left side, you must also double the umber of couters ad bags o the right side. b) To maitai balace, if you subtract 4 couters o the left side, you must also subtract 4 couters o the right side. 6. Let represet the umber of couters i each bag. Sice there are two bags ad 3 extra couters o the left side of the balace, the left side of the equatio is + 3. O the right side of the balace there are 17 couters, so the right side of the equatio is 17. The equatio is + 3 = 17. To solve, Subtract 3 couters from both sides. + 3 3 = 17 3 = 14 8-1 Chapter 8: Equatios ad Relatioships
Divide the remaiig couters o each side by, sice there are two bags, to determie how may couters are i each bag. = 14 = 7 There are 7 couters i each bag. Check: + 3 = (7) + 3 = 17 7. a) To solve, add 3 to both sides. t 3 = 9 t 3 + 3 = 9 + 3 t = 1 Check: t 3 = 9 1 3 = 9 Therefore, t = 1. b) To solve, subtract 4 from both sides. 7= a + 4 7 4 = a + 4 4 3 = a Check: 3 + 4 = 7 Therefore, a = 3. c) To solve the equatio, add 1 to both sides ad the divide both sides by 3. 3b 1 = 8 3b 1 + 1 = 8 + 1 3b = 9 3b 3 = 9 3 b = 3 Check: 3b 1 = 8 3(3) 1 = 8 Therefore, b = 3. d) To solve, multiply both sides by. x = 8 x = 8 x = 16 Check: 16 = 8 Therefore, x = 16. 8. Let represet the umber of couters i each bag. O the left side there are 6 bags of couters ad 4 extra couters, givig 6 + 4. O the right side there are 8 couters. So the equatio is 6 + 4 = 8. To solve the equatio, subtract 4 from both sides ad the divide both sides by 6. 6 + 4 = 8 6 + 4 4 = 8 4 6 = 4 6 6 = 4 6 = 4 Check: 6 + 4 = 6(4) + 4 = 8 Therefore, there are 4 couters i each bag. 9. a) To solve, add 3 to both sides ad the divide both sides by. m 3 = 31 m 3 + 3 = 31 + 3 m = 34 m = 34 m = 17 Check: m 3 = (17) 3 = 31 Therefore, m = 17. b) To solve, subtract 1 from both sides ad the divide both sides by. a + 1 = 7 a + 1 1 = 7 1 a = 6 a = 6 a = 1 Check: a + 1 = (1) + 1 = 7 Therefore, a = 1. c) To solve, add 13 to both sides ad the multiply both sides by 8. ( 8) 13 = 37 ( 8) 13 + 13 = 37 + 13 ( 8) = 16 ( 8) 8 = 16 8 = 18 Check: ( 8) 13 = (18 8) 13 = 37 Therefore, = 18. d) To solve, tur the equatio aroud, subtract 19 from both sides, ad the divide both sides by 6. 14 = 6 + 19 6 + 19 = 14 6 + 19 19 = 14 19 6 = 19 6 6 = 19 6 = 3. Check: 6 + 19 = 6(3.) + 19 = 14 Therefore, = 3.. Nelso Mathematics 8 Solutios 8-11
e) To solve, add 14 to both sides ad the divide both sides by 1. 1w 14 = 9 1w 14 + 14 = 9 + 14 1w = 11 1w 1 = 11 1 w = 11 Check: 1w 14 = 1(11) 14 = 9 Therefore, w = 11. f) To solve, subtract 7 from both sides ad the multiply both sides by 4. ( 4) + 7 = 13 ( 4) + 7 7 = 13 7 ( 4) = 6 ( 4) 4 = 6 4 = 4 Check: ( 4) + 7 = (4 4) + 7 = 13 Therefore, = 4. 1. a) To solve, tur the equatio aroud ad subtract.7 from both sides. 1.8 = a +.7 a +.7 = 1.8 a +.7.7 = 1.8.7 a = 1.1 Check: a +.7 = 1.1 +.7 = 1.8 Therefore, a = 1.1. b) To solve, subtract 1.3 from both sides ad the divide both sides by. t + 1.3 = 3.9 t + 1.3 1.3 = 3.9 1.3 t =.6 t =.6 t = 1.3 Check: t + 1.3 = (1.3) + 1.3 = 3.9 Therefore, t = 1.3. c) To solve, add 4. to both sides ad the multiply both sides by 4. 4 4. = 39.8 4 4. + 4. = 39.8 + 4. 4 = 44 4 4 = 44 4 = 11 Check: 4 4. = 4 11 4. = 39.8 Therefore, = 11. d) To solve, add. to both sides. The divide both sides by 6. 6x. = 4. 6x. +. = 4. +. 6x = 47.7 6x 6 = 47.7 6 x = 7.9 Check: 6x. = 6(7.9). = 4. Therefore, x = 7.9. e) To solve, subtract 3.3 from both sides ad the divide both sides by 3. 3 + 3.3 = 86.7 3 + 3.3 3.3 = 86.7 3.3 3 = 83.4 3 3 = 83.4 3 = 7.8 Check: 3 + 3.3 = 3(7.8) + 3.3 = 86.7 Therefore, = 7.8. f) To solve, subtract 9 from both sides ad the multiply both sides by. ( ) + 9 = ( ) + 9 9 = 9 ( ) = 13 ( ) = 13 = 6 Check: ( ) + 9 = (6 ) + 9 = Therefore, = 6. 11. Let represet the umber of stoes i each bag. There are 4 bags of stoes ad 4 extra stoes, so there are 4 + 4 stoes. I all there are 8 stoes, so the equatio is 4 + 4 = 8. To solve, subtract 4 from both sides ad the divide both sides by 4. 4 + 4 = 8 4 + 4 4 = 8 4 4 = 4 4 4 = (4 4) = 6 Therefore, there are 6 stoes i each bag. 1. a) Let be the umber of stoes i each bag. There are bags of stoes ad 6 extra stoes, so there are + 6 stoes. There are a total of 66 stoes, so the equatio is + 6 = 66. To solve, subtract 6 from both sides ad the divide both sides by. + 6 = 66 + 6 6 = 66 6 = 6 = 6 = 3 Therefore, there are 3 stoes i each bag. 8-1 Chapter 8: Equatios ad Relatioships
b) Let m represet the umber of stoes i each bag. There are 3 bags, but oe bag has had 7 stoes removed from it, so there are 3 bags less 7 stoes, givig 3m 7 stoes. There are a total of 3 stoes, so the equatio is 3m 7 = 3. To solve, add 7 to both sides ad the divide both sides by 3. 3m 7 = 3 3m 7 + 7 = 3 + 7 3m = 39 3m 3 = 39 3 m = 13 Therefore, there are origially 13 stoes i each bag. 13 a) To solve, add to both sides ad the divide both sides by. x = 1 x + = 1 + x = 1 x = 1 x =.4 Check: x = (.4) = 1 Therefore, x =.4. b) To solve, subtract 4.1 from both sides ad the divide both sides by. + 4.1 = 8. + 4.1 4.1 = 8. 4.1 = 4.4 = 4.4 =. Check: + 4.1 = (.) + 4.1 = 8. Therefore, =.. c) To solve, add 7 to both sides ad the multiply both sides by 3. (x 3) 7 = (x 3) 7 + 7 = + 7 x 3 = 9 (x 3) 3 = 9 3 x = 7 Check: (x 3) 7= (7 3) 7 = Therefore, x = 7. d) To solve, subtract. from both sides ad the multiply both sides by. ( ) +. =.7 ( ) +.. =.7. =. ( ) =. = 7. Check: ( ) +. = (7. ) +. =.7 Therefore, = 7.. 14. Let b represet the cost of oe bider. Deborah bought 3 biders, ad there was o tax o them, so she paid 3b. She received $1.1 back from her $1, so the cost of the biders plus the $1.1 makes $1. So the equatio is 3b + 1.1 = 1. To solve, subtract 1.1 from both sides ad the divide both sides by 3. 3b + 1.1 = 1 3b + 1.1 1.1 = 1 1.1 3b = 8.8 3b 3 = 8.8 3 b =.9 Therefore, the cost of each bider was $.9. 1. Let t represet the cost of each ticket. Farha s gradfather bought tickets, ad the delivery charge of $3.7 was added to the cost. The total cost was $31.6, so the equatio is t + 3.7 = 31.6. To solve, subtract 3.7 from both sides ad the divide both sides by. t + 3.7 = 31.6 t + 3.7 3.7 = 31.6 3.7 t = 7.9 t = 7.9 t = 13.9 Therefore, the cost of each ticket was $13.9. 16. Let c represet the cost of 1 kg of steak. There were 4 kg of steak sold, plus a jar of barbecue sauce for $3.79, so the value of these items are added together to give 4c + 3.79. The total cost was $3.19, so the equatio is 4c + 3.79 = 3.19. To solve, subtract 3.79 from both sides ad the divide both sides by 4. 4c + 3.79 = 3.19 4c + 3.79 3.79 = 3.19 3.79 4c = 19.4 4c 4 = 19.4 4 c = 4.8 Therefore, the cost of 1 kg of steak is $4.8. 17. Let b represet the mass of each bag of cemet. The total mass of 4 bags of cemet ad a additioal 8 kg cocrete block is 98 kg, so the equatio is 4b + 8 = 98. To solve, subtract 8 from both sides, ad the divide each side by 4. 4b + 8 = 98 4b + 8 8 = 98 8 4b = 9 4b 4 = 9 4 b =. Therefore, each bag of cemet has a mass of. kg. 18. Let a represet the average mass of each apple, i kilograms. There are apples, 1 pieapple, ad 1 orage i the bag, so the total mass will be the sum of these fruits, which is a + 1. +.. The total mass of the bag is.7 kg, so the equatio is: a + 1. +. =.7 a + 1.4 =.7 Nelso Mathematics 8 Solutios 8-13
To solve, subtract 1.4 from both sides ad the divide both sides by. a + 1.4 =.7 a + 1.4 1.4 =.7 1.4 a = 1. a = 1. a =. Therefore, the average mass of each apple is. kg. 19. The blue couters represet the 4, ad the red couters represet the 6. The umber of red couters i the bag is x. To solve the equatio, remove all of the blue couters from the left side. Each blue couter has a egative value, so each time you remove a blue couter from the right side you eed to replace it with a red couter o the left side. x + ( 4) = 6 x + ( 4) ( 4) = 6 ( 4) x = 6 + 4 x = 1 Whe all 4 blue couters have bee chaged ito 4 extra red couters o the right side, there will be 1 red couters o the right side ad oe bag of couters o the left side, so there are 1 couters i the bag. Therefore, x = 1.. a) Use couters to represet x + 6 = 4. (Sice this maual is prited i black ad white, a couter with a egative sig ( ) is used to represet 1 ad a couter with a positive sig ( ) is used to represet +1.) To solve the equatio, arrage it so there are o couters o the left side, where the bag is. x There are 6 positive couters o the left side. To remove them, add 6 egative couters to the left side. To keep the sides balaced, add 6 egative couters to the right side also. x + 6 + ( 6) = 4 + ( 6) x 6 + ( 6) =, so ow there is just x o the left side. x There are 1 couters o the right side, so there must be 1 couters i the bag. So, x = 1. Check by substitutio: 1 + 6 = 4 b) Use couters to represet + ( ) = 8. There are egative couters o the left side. To remove them, add positive couters to the left side. To keep the sides balaced, add positive couters to the right side also. + ( ) + = 8 + ( ) + =, so ow there is o the left side ad 1 o the right side. = 1 There are bags o the left side, so divide the couters o the right ito equal groups. = 1 There are positive couters i each group, so =. Check by substitutio: () + ( ) = 1 + ( ) or 8. c) Model x = 3 with couters. x x There are egative couters o the left side. To remove them, add positive couters to the left side. To keep the sides balaced, add positive couters to the right side also. x + = 3 + x x + =, so ow there is x o the left side ad 18 egative couters o the right side. 8-14 Chapter 8: Equatios ad Relatioships
x = 18 e) Use couters to represet + ( ) = 6. x x There are bags o the left side, so divide the 18 couters o the right ito equal groups. x = 18 x x x = 9 There are 9 egative couters i each group, so x = 9. If x = 9 the x must be 9. Check by substitutio: (9) = 18 or 3 d) Model 3 ( ) = 14 with couters. ( ) is the same as +, so put positive couters o the left side. There are egative couters o the left side. To remove them, add positive couters to the left side. To keep the sides balaced, add positive couters to the right side also. + ( ) + = 6 + ( ) + =, so ow there is o the left side ad 4 o the right side. = 4 There are bags o the left side, so divide the couters o the right ito equal groups. = 4 Subtract the positive couters from the left side. To keep the sides balaced, subtract positive couters from the right side also. 3 = 1 There are 3 bags o the left side, so divide the 1 couters o the right ito 3 equal groups. 3 3 = 1 3 There are 4 couters i group, so = 4. Check by substitutio: 3(4) ( ) = 1 + or 14 There are egative couters i each group, so =. Check by substitutio: ( ) + ( ) = 4 + ( ) or 6. 8.6 Commuicatig about Equatios, p. 76 4. a) For example, the variable is h, so the cost of oe item could be represeted by h. The equatio cotais h, so there were items bought. The 1. ca represet aother item that was purchased for $1., ad the total before tax ca be represeted by the. i the equatio. So, my problem is this: Alec bought boxes of cereal ad a carto of milk. The milk cost $1.. If his total before tax was $., the how much did each box of cereal cost? b) Let h represet the cost of oe box of cereal. Alec bought boxes ad a carto of milk that cost $1., so add these values together. The total cost was $., so the equatio is h + 1. =.. To solve, subtract 1. from both sides ad the divide both sides by. Nelso Mathematics 8 Solutios 8-1
h + 1. =. h + 1. 1. =. 1. h = 4 h = 4 h = Check: LS = (.) + 1. =. RS =. LS = RS Each box of cereal cost $.. c) For example, I created my problem by thikig about buyig items from a store, ad lettig the cost of oe of the items be the variable ad the cost of aother item beig $1.. The total before tax of the items purchased could be $., so that was my problem. I solved my problem by balacig the equatio. I the checked my solutio to make sure that it was correct, ad wrote a cocludig statemet to my problem.. Check that the solutio is correct by solvig the equatio. First add 6 to both sides, ad the divide both sides by 1 to get the value of. 1 6 = 48 1 6 + 6 = 48 + 6 1 = 4 1 1 = 4 1 = 4. Check: 1 6 = 1(4.) 6 = 48 Therefore, the solutio is right. For example, the variable could represet the mass of a pumpki i kilograms. There could be 1 pumpkis, but these pumpkis have a mass that is 6 kg more tha average. Usually the mass of the 1 pumpkis is 48 kg. My problem is this: Braxto ad his family etered 1 equal pumpkis i a cotest. Last year the total mass was 48 kg. This year the mass is less by 6 kg. What is the mass of each pumpki? 6. For example, the improved solutio is: Let s represet the umber of sweaters the customer bought. Each sweater costs $8.7, so the customer paid 8.7s dollars for the sweaters, plus the shippig charge of $3.. The total cost was $6.7, so the equatio is 8.7s + 3. = 6.7. 8.7s + 3. = 6.7 8.7s + 3. 3. = 6.7 3. 8.7s = 7. 8.7s 8.7 = 7. 8.7 s = Check: 8.7s + 3. = 8.7() + 3. = 6.7 Therefore s =. The customer bought sweaters. 7. For example, the variable could represet a umber of people i oe group, ad the 4 would mea that there are four groups of people. The added 7 o the left side could mea that there are 7 more people, maybe parets for a field trip. The total umber of people are represeted by. So my problem is this: The day camp split all of the campers ito 4 equal groups for their field trip. There were 7 paret voluteers that also came alog. There were people i total. How may campers were i each group? To solve the problem: Let represet the umber of campers i each group. There are 4 groups ad 7 extra people, so there are 4 + 7 people. There is a total of people, so the equatio is 4 + 7 =. 4 + 7 = 4 + 7 7 = 7 4 = 48 4 4 = 48 4 = 1 Check: 4 + 7 = 4(1) + 7 = Therefore, there were 1 campers i each group. 8. For example, the umbers i the equatio are to decimal places, so the equatio ca represet the cost of buyig items ad the total amout. The total cost could be $8.89, ad the 1.39 o the left side could represet the tax. The could represet buyig items that each have a cost of. So the variable represets the cost of oe item. My problem is: Lolita bought birthday cards for her family members. The tax o the cards was $1.39. The total cost came to $8.89. How much did each card cost? 9. For example, the umbers i the equatio are decimal umbers. They could represet measurig a legth. The variable could stad for the legth of 6. items, measured i cm. There could be a extra cm to the legth of the object, say a poster, ad the whole poster is 31 cm log. My problem is: Judith has a poster with a row of equal stars o it. There are a total of 6. stars. There is a extra cm of space i total aroud the stars. If the poster is a total of 31 cm log, the what is the legth of each star o the poster? To solve the problem, let represet the legth of each star. There are 6. stars, so the legth of all stars is 6. cm. The legth of the poster is the legth of the stars plus a extra cm, ad the total legth is 31 cm. So the equatio is 6. + = 31. 8-16 Chapter 8: Equatios ad Relatioships
To solve, subtract from both sides ad the divide both sides by 6.. 6. + = 31 6. + = 31 6. = 6 6. 6. = 6 6. = 4 Check: 6. + = 6.(4) + 31 Therefore, each star is 4 cm log. 1. For example, the variable represets what is beig solved for, so the variable i this case will represet the umber of cedar trees that eed to be plated. Let c be that variable. The sides where the cedar trees eed to be plated ca be thought of as oe big lie of cedar trees. The legth of that lie is 3 times the legth of oe side of the backyard. The area of the square backyard is 144 m, so the legth of oe side is 144 = 1 m. The legth where trees eed to be plated is 3 1 m = 36 m. So the total legth is 36 m. Sice the trees will be plated every half metre, they ca be thought of as each takig up a half metre of that legth, say the half metre that is behid the tree. So c trees take up.c m of the legth. This is true for all of the trees except for the first oe, which is right agaist the house, ad does ot take up a half metre. So you have to subtract. m from the legth that the trees take up. This gives the equatio as follows:.c. = 36 To solve, add. to both sides ad the divide both sides by...c. = 36.c. +. = 36 +..c = 36..c. = 36.. c = 73 Check:.c. =.(73). = 36 Therefore, c = 73, so 73 cedars eed to be plated. Chapter Self-Test, p. 77 1. Fid o the vertical axis. Place your ruler horizotally at ad see where it touches the graph. The place your ruler vertically from the itersectio to the horizotal axis. This is at term umber 1. The solutio is = 1.. a) Figure umber Number of couters 1 1 3 9 b) The umber of couters icrease by 4 each time. So the patter rule ivolves 4, where is the figure umber. From figure 1, you ca see that the umber of couters is actually 3 less tha 4(1) = 4, ad the same with the ext two figures also. So a rule is 4 3. c) For the figure umber that has 161 couters, the rule will be equal to 161. A equatio is 4 3 = 161. d) For example, solve the equatio usig the systematic trial method with a table. For example, Predict. Evaluate 4 3. Is the aswer 161? 4 4(4) 3 = 17 No, too low. 4 4(4) 3 = 16 No, too high. 41 4(41) 3 = 161 It s correct. The solutio is = 41, so figure 41 has 161 couters. 3. a) Substitute 1. for a ad 3.7 for c. 7a + 8c = 7(1.) + 8(3.7) = 11.3 b) Substitute 3 for a ad for c. 7a + 8c = 7(3) + 8( ) = 37 4. a) Divide both sides by 6. 78 = 6x 78 6 = 6x 6 13 = x Check: 6x = 6(13) = 78 Therefore, x = 13. b) To solve, subtract from both sides ad the divide both sides by 6. 6 + = 41 6 + = 41 6 = 36 6 6 = 36 6 = 6 Check: 6 + = 6(6) + = 41 Therefore, = 6. Nelso Mathematics 8 Solutios 8-17
c) To solve, add 11 to both sides ad the divide both sides by 4. 4m 11 = 4 4m 11 + 11 = 4 + 11 4m = 6 4m 4 = 6 4 m = 14 Check: 4m 11 = 4(14) 11 = 4 Therefore, m = 14. d) To solve, divide both sides by. t = 8. t = 8. t = 14. Check: t = (14.) = 8. Therefore, t = 14... a) To solve, add to both sides. = + = + = 7 Check: = 7 = Therefore, = 7. b) To solve, add 6.1 to both sides. x 6.1 = 1. x 6.1 + 6.1 = 1. + 6.1 x = 18.3 Check: x 6.1 = 18.3 6.1 = 1. Therefore, x = 18.3. c) To solve, subtract.3 from both sides. t +.3 =.6 t +.3.3 =.6.3 t =.37 Check: t +.3 =.37 +.3 =.6 Therefore, t =.37. d) To solve, divide both sides by 1.68. 1.68h = 6.7 1.68h 1.68 = 6.7 1.68 h = 4 Check: 1.68h = 1.68(4) = 6.7 Therefore, h = 4. e) To solve, subtract 1 from both sides, ad the divide both sides by 6. 6p + 1 = 19 6p + 1 1 = 19 1 6p = 18 6p 6 = 18 6 p = 3 Check: 6p + 1 = 6(3) + 1 = 19 Therefore, p = 3. f) Add 8. to both sides, ad the divide by. 8. = 11. 8. + 8. = 11. + 8. = 11 = 11 = 4.4 Check: 8. = (4.4) 8. = 11. Therefore, = 4.4. 6. a) Let be the ukow umber. Add 7.1 to get + 7.1. The result is 1.3, so that is o the right-had side of the equatio. The equatio is + 7.1 = 1.3. b) Let be the ukow umber. Multiply it by 6, to get 6. The result is 48.6, so that is o the right-had side of the equatio. The equatio is 6 = 48.6. c) Let m be the ukow umber. Whe it is added to itself, the you get m + m. The result is 49, so that is o the right-had side of the equatio. The equatio is m + m = 49 or m = 49. d) Let p be the ukow umber. Three times a umber is 3p, ad plus. gives 3p +.. This is equal to 83., which is the right-had side of the equatio. So the equatio is 3p +. = 83. 7. a) Let s represet the amout of moey i the savigs accout. By doublig the moey you are multiplyig it by, so the amout is s. This would give $1.6, so the equatio is s = 1.6. To solve, divide both sides by. s = 1.6 s = 1.6 s = 7.31 Curretly there is $7.31 i the savigs accout. b) Let h represet the hourly wage. Three times the hourly wage is 3h. The moey eared i tips is added to the hourly wage, so the left-had side is 3h + 7.6. The total is $7.9, the right-had side. So, the equatio is 3h + 7.6 = 7.9. To solve, subtract 7.6 from both sides ad the divide both sides by 3. 3h + 7.6 = 7.9 3h + 7.6 7.6 = 7.9 7.6 3h =. 3h 3 =. 3 h = 6.7 The hourly wage is $6.7. 8. For example, use p to represet the umber of pairs of socks that Yig bought. The socks were $. per pair, so for p pairs of socks the cost was.p, or.p. She also bought a jacket, which must be added to the cost of the socks, so she spet.p + 39. The total for her items was $1., so the equatio is.p + 39 = 1.. 8-18 Chapter 8: Equatios ad Relatioships
To solve, subtract 39 from both sides ad the divide both sides by...p + 39 = 1..p + 39 39 = 1. 39.p = 1..p. = 1.. p = Therefore, p = ad so Yig bought pairs of socks. Chapter Review, p. 79 1. a) Place a ruler at horizotally from o the vertical axis to the graph. The place the ruler vertically from the itersectio dow to the horizotal axis. It meets the horizotal axis at 8, so the solutio to + 4 = is = 8. b) + 4 = 1 + 4 4 = 1 4 = 8 = 8 = 4 Check: + 4 = (4) + 4 = 1 Therefore, = 4.. a) Figure umber Number of couters 1 13 19 3 b) Let represet the figure umber. With each figure the umber of couters icreases by 6, so the expressio cotais 6. I figure 1, there are 13 couters. This is 13 6 = 7 couters more tha 6 gives. This is true i the other two figures as well. So, a algebraic expressio for the rule is 6 + 7. c) 14 1 1 7 Number of Couters Compared to Figure Number 1 1 Figure umber To determie which figure has 14 couters, locate 14 o the vertical axis ad draw a horizotal lie from there to the poit where it touches the graph. From there, draw a vertical lie dow to the horizotal axis, where it is 3. So, figure 3 has 14 couters. 3. Let represet the umber of ice cream bars that Kyra sells. She gets paid $. for each ice cream bar she sells, so if she sells bars she will receive. i commissio. She also receives $4 each day. So she ears. + 4 each day. If she sells 77 ice cream bars, replace by 77 i the expressio to determie how much she will ear that day.. + 4 =.(77) + 4 = 9. Therefore, if Kyra sells 77 ice cream bars i oe day she will ear $9.. 4. a) To solve, subtract 4 from both sides ad the divide both sides by. b + 4 = 14 b + 4 4 = 14 4 b = 1 b = 1 b = Check: b + 4 = () + 4 = 14 Therefore, b =. b) Subtract from both side ad divide by. a + = 1 a + = 1 a = 8 a = 8 a = 1.6 Check: a + = (1.6) + = 1 Therefore, a =. c) Subtract 4.3 from both sides. + 4.3 =.1 + 4.3 4.3 =.1 4.3 =.8 Check: + 4.3 =.8 + 4.3 =.1 Therefore, =.8. d) Simplify the right side ad divide by...x = 1 3.x = 18.x. = 18. x = 36 Check: LS =.x =.(36) = 18 RS = 1 3 = 18 LS = RS Therefore, x = 36. Nelso Mathematics 8 Solutios 8-19
e) To solve, subtract 1 from both sides ad the divide both sides by 4. 4t + 1 = 4t + 1 1 = 1 4t = 19 4t 4 = 19 4 t = 4.7 Check: 4t + 1 = 4(4.7) + 1 = Therefore, t = 4.7. f) To solve, subtract 7 from both sides ad the divide both sides by 4. 4m + 7 = 43.8 4m + 7 7 = 43.8 7 4m = 36.8 4m 4 = 36.8 4 m = 9. Check: 4m + 7 = 4(9.) + 7 = 43.8 Therefore, m = 9... a) Let represet the ukow umber. Addig 3.1 to this umber gives a sum of 16., so the equatio is + 3.1 = 16.. To solve, subtract 3.1 from both sides. + 3.1 = 16. + 3.1 3.1 = 16. 3.1 = 1.84 Check: + 3.1 = 1.84 + 3.1 = 16. Therefore, = 1.84. b) Let represet the ukow umber. The product of a umber ad 3.1 is represeted by 3.1, ad the result is 16.. The equatio is 3.1 = 16.. To solve, divide both sides by 3.1. 3.1 = 16. 3.1 3.1 = 16. 3.1 = Check: 3.1 = 3.1() = 16. Therefore, =. c) Let represet the ukow umber. Whe this umber is doubled, it is multiplied by, ad the subtractig 3.1 gives 3.1. The result is equal to 16., so the equatio is 3.1 = 16.. To solve, add 3.1 to both sides ad divide by. 3.1 = 16. 3.1 + 3.1 = 16. + 3.1 = 19.6 = 19.6 = 9.63 Check: 3.1 = (9.63) 3.1 = 16. Therefore, = 9.63. 6. For example, use to represet the amout of moey each team must be charged to meet the goal. There are 34 teams, so there will be 34 raised. They have already raised $31, so this is added to the amout they raise from the shoot-out challege. They eed to raise a total of $89. So the equatio is 34 + 31 = 89. 34 + 31 = 89 34 + 31 31 = 89 31 34 = 44 34 34 = 44 34 = 16 Check : 4 + 31 = 34(16) + 31 = 89 Therefore, they should charge $16 per team. 7. For example, the umerical values i this equatio look like they represet dollar values, so the variable t ca represet the umber of items bought. Sice it is multiplied by 1.99, each item costs $1.99. The righthad side of the equatio is 1.98, so the total spet is $1.98. My problem is this: Eva oticed a sale at the book store: ay book for $1.99. If her total without tax was $1.98, how may books did Eva buy? To solve: Let t represet the umber of books that Eva bought. Each book was $1.99, so she paid 1.99t, before tax. The total before tax was $1.98. So the equatio is 1.99t = 1.98. To solve, divide both sides by 1.99. 1.99t = 1.98 1.99t 1.99 = 1.98 1.99 t = Check: 1.99t = 1.99() = 1.98 Therefore, Eva bought books. 8. For example, use t to represet the cost of each ticket. Eda bought tickets, ad the delivery charge of $3.7 was added to the cost. The total cost was $4.7, so the equatio is t + 3.7 = 4.7. To solve, subtract 3.7 from both sides ad divide both sides by. t + 3.7 = 4.7 t + 3.7 3.7 = 4.7 3.7 t = 1 t == 1 t =. Therefore, the cost of each ticket was $.. 8- Chapter 8: Equatios ad Relatioships