7A 7B 7C 7D 7E 7F 7G 7H 7I Chpter Wht you will lern Equtions review (Consolidting) Equivlent equtions (Consolidting) Equtions with frtions Equtions with pronumerls on oth sides Equtions with rkets Formuls nd reltionships Applitions Inequlities (Etending) Solving inequlities (Etending) 7Equtions nd inequlities Austrlin urriulum NUMBER AND ALGEBRA Liner nd non-liner reltionships Solve liner equtions using lgeri nd grphil tehniques. Verify solutions y sustitution (ACMNA194)
Proteting se turtles Loggerhed se turtles re mgnifi ent mrine predtors tht feed on shellfi sh, rs, se urhins nd jellyfi sh. Sdly, Austrlin loggerhed turtles hve lost more thn 50% of their nesting femles in 10 yers. These diminishing numers men tht loggerhed turtles re now n endngered speies. They re under thret for numer of resons, inluding humn tivity on the ehes where the femles ly eggs nd identl deth in fi shing nets. In order to est work out how to sve the loggerhed se turtle, sientists need to work out wht effet vrious tions, suh s losing ehes, will hve on the loggerhed turtle popultion. To do this sientists use mthemtis! With omintion of dt nd equtions, they use Online resoures omputer models to predit popultion numers of future genertions of the loggerhed turtle. For emple, this simple eqution desries the numer of turtles tht there will e net yer: F = C (1 + B D) The vriles used in this eqution re: F = future popultion C = urrent popultion B = irth rte D = deth rte By mthemtilly prediting the future of the loggerhed turtle popultion, environmentl sientists n dvise governments of the est deisions to help sve the loggerhed se turtle from etintion. Chpter pre-test Videos of ll worked emples Intertive widgets Intertive wlkthroughs Downlodle HOTsheets Aess to HOTmths Austrlin Curriulum ourses
390 Chpter 7 Equtions nd inequlities Key ides 7A Equtions review CONSOLIDATING An eqution is sttement tht two things re equl, suh s: 2 + 2 = 4 7 5 = 30 + 5 4 + = 10 + y. It onsists of two epressions seprted y the equls sign (=), nd it is onsidered true if the left-hnd side nd right-hnd side re equl. True equtions inlude 7 + 10 = 20 3 nd 8 = 4 + 4; emples of flse equtions re 2 + 2 = 7 nd 10 5 = 13. If n eqution hs pronumerl in it, suh s 3 + = 7, then solution to the eqution is vlue to sustitute for the pronumerl to form true eqution. In this se solution is = 4 euse 3 + 4 = 7 is true eqution. Let s strt: Solving the equtions Find numer tht would mke the eqution 25 = (10 ) true. How n you prove tht this vlue is solution? Try to find solution to the eqution 11 = 11 +. An eqution is mthemtil sttement tht two epressions re equl, suh s 4 + = 32. It ould e true (e.g. 4 + 28 = 32) or flse (e.g. 4 + 29 = 32). A flse eqution n e mde into true sttement y using the sign. For instne, 4 + 29 32 is true sttement. An eqution hs left-hnd side (LHS) nd right-hnd side (RHS). A solution to n eqution is vlue tht mkes n eqution true. The proess of finding solution is lled solving. In n eqution with pronumerl, the pronumerl is lso lled n unknown. An eqution ould hve no solutions or it ould hve one or more solutions.
Numer nd Alger 391 Emple 1 Clssifying equtions s true or flse For eh of the following equtions, stte whether they re true or flse. 3 + 8 = 15 4 7 3 = 20 + 5 + 20 = 3, if = 10 SOLUTION EXPLANATION True Left-hnd side (LHS) is 3 + 8, whih is 11. Right-hnd side (RHS) is 15 4, whih is lso 11. Sine LHS equls RHS, the eqution is true. Flse LHS = 7 3 = 21 RHS = 20 + 5 = 25 Sine LHS nd RHS re different, the eqution is flse. True If = 10 then LHS = 10 + 20 = 30. If = 10 then RHS = 3 10 = 30. LHS equls RHS, so the eqution is true. Emple 2 Stting solution to n eqution Stte solution to eh of the following equtions. 4 + = 25 5y = 45 26 = 3z + 5 SOLUTION EXPLANATION = 21 We need to find vlue of tht mkes the eqution true. Sine 4 + 21 = 25 is true eqution, = 21 is solution. y = 9 If y = 9 then 5y = 5 9 = 45, so the eqution is true. z = 7 If z = 7 then 3z + 5 = 3 7 + 5 = 21 + 5 = 26 Note: The ft tht z is on the right-hnd side of the eqution does not hnge the proedure.
392 Chpter 7 Equtions nd inequlities Emple 1, 1 Emple 1 1 Emple 3 Writing equtions from desription Write equtions for the following senrios. The numer k is douled, then three is dded nd the result is 52. Akir works n hours, erning $12 per hour. The totl she erned ws $156. SOLUTION EXPLANATION 2k + 3 = 52 The numer k is douled, giving k 2. This is the sme s 2k. Sine 3 is dded, the left-hnd side is 2k + 3, whih must e equl to 52 ording to the desription. 12n = 156 If Akir works n hours t $12 per hour, the totl mount erned is 12 n, or 12n. Eerise 7A Clssify these equtions s true or flse. 5 3 = 15 7 + 2 = 12 + 3 5 + 3 = 16 2 d 8 6 = 6 e 4 3 = 12 1 f 2 = 8 3 3 2 If the vlue of is 3, wht is the vlue of the following? 10 + 3 5 d 6 3 Stte the vlue of the missing numer to mke the following equtions true. 5 + = 12 10 = 90 3 = 12 d 3 + 5 = 4 Consider the eqution 15 + 2 =. 5 1 4 3, 4 If = 5, find the vlue of 15 + 2. If = 5, find the vlue of. Is = 5 solution to the eqution 15 + 2 =? d Give n emple of nother eqution with = 5 s solution. 5 6(½), 8(½) If = 2, stte whether the following equtions re true or flse. 7 = 8 + 3 10 = 4 3 = 5 d + 4 = 5 e 10 = 40 f 12 + 2 = 15. 5 8(½) 6 If = 3, stte whether the following equtions re true or flse. 7 + = 10 2 + 4 = 12 8 = 5 d 4 3 = 9 e 7 + 2 = 8 f = 6 5 8(½) UNDERSTANDING FLUENCY
Numer nd Alger 393 Emple 2 Emple 3 7 Someone hs ttempted to solve the following equtions. Stte whether the solution is orret (C) or inorret (I). 8 9 5 + 2 = 4 1, proposed solution: = 3 4 + q = 3 + 2q, proposed solution: q = 10 13 2 = + 1, proposed solution: = 4 d ( + 3) = 4, proposed solution, = 4 Stte solution to eh of the following equtions. 5 + = 12 3 = 10 4v + 2 = 14 d 17 = p 2 e 10 = 20 f 16 = g 4u + 1 = 29 h 7k = 77 i 3 + = 2 Write equtions for eh of the following prolems. You do not need to solve the equtions. A numer is douled nd then 7 is dded. The result is 10. The sum of nd hlf of is 12. Aston s ge is. His fther, who is 25 yers older, is twie s old s Aston. d Fel s height is h m nd her rother Pt is 30 m tller. Pt s height is 147 m. e Coffee osts $ per up nd te osts $t. Four ups of offee nd three ups of te ost totl of $21. f Chirs ost $ eh. To purhse 8 hirs nd $2000 tle osts totl of $3600. 10 Find the vlue of the unknown numer for eh of the following. A numer is tripled to otin the result 21. Hlf of numer is 21. Si less thn numer is 7. d A numer is douled nd the result is 16. e Three-qurters of numer is 30. f Si more thn numer is 7. 11 Berkeley uys kg of ornges t $3.20 per kg. He spends totl of $9.60. Write n eqution involving to desrie this sitution. Stte solution to this eqution. 9, 10(½) 9 10(½),11, 12 12 Emily s ge in 10 yers time will e triple her urrent ge. She is urrently E yers old. Write n eqution involving E to desrie this sitution. Find solution to this eqution. How old is Emily now? d How mny yers will she hve to wit until she is four times her urrent ge? 13 Find two possile vlues of t tht mke the eqution t(10 t) = 21 true. 11 13 FLUENCY PROBLEM-SOLVING 7A
394 Chpter 7 Equtions nd inequlities 7A 14 Why is = 3 solution to 2 = 9? Why is = 3 solution to 2 = 9? Find the two solutions to 2 = 64 (Hint: one is negtive). d Eplin why 2 = 0 hs only one solution, ut 2 = 1 hs two. e Eplin why 2 = 9 hs no solutions. (Hint: onsider multiplying positive nd negtive numers.) 15 Eplin why the eqution + 3 = hs no solutions. Eplin why the eqution + 2 = 2 + is true, regrdless of the vlue of. Show tht the eqution + 3 = 10 is sometimes true nd sometimes flse. d Clssify the following equtions s lwys true (A), sometimes true (S) or never true (N). i + 2 = 10 ii 5 q = q iii 5 + y = y iv 10 + = 10 v 2 = + vi 3 = 10 vii 3 + 2z = 2z + 1 viii 10p = p i 2 + + = ( + 1) 2 e Give new emple of nother eqution tht is lwys true. 14 14, 15 16 The eqution p (p + 2) = 3 hs two solutions. Stte the two solutions. Hint: One of them is negtive. How mny solutions re there for the eqution p + (p + 2) = 3? Try to find n eqution tht hs three solutions. More thn one unknown 17 There re si equtions in the squre elow. Find the vlues of,,, d nd e to mke ll si equtions true. + 12 = 22 2 = = = = d e = 10 Find the vlue of f tht mkes the eqution e = d f true. 18 For eh of the following pirs of equtions, find vlues of nd d tht mke oth equtions true. More thn one nswer my e possile. + d = 10 nd d = 24. d = 8 nd + d = 14 d = 4 nd + d = 30 d d = 0 nd d = 7 14, 16 17, 18 REASONING ENRICHMENT
Numer nd Alger 395 7B Equivlent equtions CONSOLIDATING If we hve n eqution, we n otin n equivlent eqution y performing the sme opertion to oth sides. For emple, if we hve 2 + 4 = 20, we n dd 3 to oth sides to otin 2 + 7 = 23. The new eqution will e true for etly the sme vlues of s the old eqution. This oservtions helps us to solve equtions lgerilly. For emple, 2 + 4 = 20 is equivlent to 2 = 16 (sutrt 4 from oth sides), nd this is equivlent to = 8 (divide oth sides y 2). The ottom eqution is only true if hs the vlue 8, so this mens the solution to the eqution 2 + 4 = 20 is = 8. We write this s: Let s strt: Attempted solutions 4 2 + 4 = 20 2 = 16 2 2 = 8 Below re three ttempts t solving the eqution 4 8 = 40. Eh hs prolem. 4 8 = 40 + 8 +8 4 Attempt 1 4 = 48 = 44 4 Attempt 2 4 8 = 40 +8 8 4 = 32 4 4 = 8 4 4 Attempt 3 4 8 = 40 8 = 10 4 + 8 + 8 = 18 Cn you prove tht these results re not the orret solutions to the eqution ove? For eh one, find the mistke tht ws mde. Cn you solve 4 8 = 40 lgerilly?
396 Chpter 7 Equtions nd inequlities Key ides Two equtions re equivlent if you n get from one to the other y repetedly: dding numer to oth sides sutrting numer from oth sides multiplying oth sides y numer other thn zero dividing oth sides y numer other thn zero swpping the left-hnd side nd right-hnd sides of the eqution. To solve n eqution lgerilly, repetedly find n equivlent eqution tht is simpler. For emple: 5 + 2 = 32 2 2 5 = 30 5 5 = 6 To hek solution, sustitude the unknown s vlue in to oth sides to see if the eqution is true, e.g. LHS = 5(6) + 2 = 32 nd RHS = 32. Emple 4 Finding equivlent equtions Show the result of pplying the given opertion to oth sides of these equtions. 8y = 40 [ 8] 10 + 2 = 36 [ 10] 5 3 = 12 [+3] SOLUTION 8y = 40 8 8 y = 5 10 + 2 = 36 10 2 = 26 10 5 3 = 12 + 3 + 3 5 = 15 Emple 5 Solving equtions lgerilly EXPLANATION Write the eqution out nd then divide oth sides y 8. 40 8 is 5 nd 8y 8 is y. Write the eqution out nd then sutrt 10 from oth sides. 36 10 is 26 10 + 2 10 is 2 Write the eqution out nd then dd 3 to oth sides. 12 + 3 is 15 5 3 + 3 is 5 Solve the following equtions lgerilly nd hek the solution y sustituting. 4 = 16 2u + 7 = 17 40 3 = 22
Numer nd Alger 397 SOLUTION EXPLANATION 4 = 16 + 4 = 20 + 4 So the solution is = 20. 2u + 7 = 17 7 7 2u = 10 2 2 u = 5 So the solution is u = 5. 40 3 = 22 40 40 3 = 18 3-3 = 6 So the solution is = 6. Eerise 7B By dding 4 to oth sides of the eqution, we get n equivlent eqution. Chek: 20 4 = 16 To get rid of the +7, we sutrt 7 from oth sides. Finlly we divide y 2 to reverse the 2u. Rememer tht 2u mens 2 u Chek: 2(5) + 7 = 10 + 7 = 17 We sutrt 40 from oth sides to get rid of the 40 t the strt of the LHS. Sine 3 = 18, we divide y 3 to get the finl solution. Chek: 40 3(6) = 40 18 = 22 1 For eh of the following equtions dd 4 to oth sides to otin n equivlent eqution. 3 = 10 7 + k = 14 5 = 2 2 For eh eqution fill in the lnk to get n equivlent eqution. 5 = 10 10 2 = 20 + 2 + 2 + 5 + 5 5 + 2 = 15 2 = 3q + 4 = d 7z + 12 = 4z + 10 4 4 10 10 3q = 12 3 Consider the eqution 4 = 32. Copy nd omplete the following working. 4 = 32 4 4 = Wht is the solution to the eqution 4 = 32? 1 4 2 4 7z + 2 = UNDERSTANDING
398 Chpter 7 Equtions nd inequlities 7B Emple 4 Emple 5 Emple 5 Emple 5 4 To solve the eqution 10 + 5 = 45, whih of the following opertions would you first pply to oth sides? A Divide y 5 B Sutrt 5 C Divide y 10 D Sutrt 45 5 For eh eqution, show the result of pplying the given opertion to oth sides. 10 + 2 = 30 [ 10] 4 + q = 12 [ 2] 13 = 12 q [+5] d 4 = 8 [ 3] e 7p = 2p + 4 [+6] f 3q + 1 = 2q + 1 [ 1] 6 Copy nd omplete the following to solve the given eqution lgerilly. 10 = 30 q + 5 = 2 4 + 2 = 22 10 10 5 5 2 2 = = 4 = 4 4 = d 30 = 4p + 2 e 20 4 = 8 f p 3 + 6 = 8 2 2 20 20 6 6 = 4 = 12 p 3 = 2 4 4 = = = 7 8 9 5 6, 7 9(½) 5 10(½) Solve the following equtions lgerilly. + 5 = 8 t 2 = 14 7 = q 2 d 11 = k + 2 e 19 = + 9 f 30 = 3h g 36 = 9l h g 3 = 3 Solve the following equtions lgerilly. Chek your solutions using sustitution. 5 + 9h = 32 9u 6 = 30 13 = 5s 2 d 18 = 6 3w e 12 = 5 + 8 f 44 = 10w + 6 g 8 = 8 + 8 h 4y 8 = 40 Solve the following equtions lgerilly nd hek your solutions. 20 4d = 8 34 = 4 5j 21 7 = 7 d 6 = 12 3y e 13 8k = 45 f 44 = 23 3n g 13 = 3 + 4 h 22 = 14 9 10 The following equtions do not ll hve whole numer solutions. Solve the following equtions lgerilly, giving eh solution s frtion. 2 + 3 = 10 5 + 3q = 6 12 = 10 + 7 d 15 = 10 + 2 e 15 = 10 2 f 13 + 2p = 10 g 22 = 9 + 5y h 12 2y = 15 6 10(½) UNDERSTANDING FLUENCY
Numer nd Alger 399 11 For eh of the following, write n eqution nd solve it lgerilly. The sum of p nd 8 is 15. The produt of q nd 3 is 12. 4 is sutrted from doule the vlue of k nd the result is 18. d When r is tripled nd 4 is dded the result is 34. e When is sutrted from 10 the result is 6. f When triple y is sutrted from 10 the result is 16. 12 Solve the following equtions lgerilly. More thn two steps re involved. 14 (4 + 2) = 140 8 = (10 4) 2 12 = (3 ) 4 13 The following shpes re retngles. By solving equtions lgerilly, find the vlue of the vriles. Some of the nswers will e frtions. 10 13 5y + 3 17 2 2 4 4 Perimeter = 40 d 11 11 13 25 12 y 25 70 3p 10 + 5 4q 1 14 Sidney works for 10 hours t the norml rte of py ($ per hour) nd then the net three hours t doule tht rte. If he erns totl of $194.88, write n eqution nd solve it to find his norml hourly rte. 15 Prove tht 7 + 4 = 39 nd 2 + 13 = 3 re equivlent y filling in the missing steps. 7 + 4 = 39 4 4 7 = 35 7 7 = 2 2 = + 13 + 13 2 + 13 = 3 15 15, 16 12 14 16, 17 PROBLEM-SOLVING REASONING 7B Prove tht 10k + 4 = 24 nd 3k 1 = 5 re equivlent.
400 Chpter 7 Equtions nd inequlities 7B 16 Prove tht 4 + 3 = 11 nd 2 = 4 re equivlent. Try to use just two steps to get from one eqution to the other. Are the equtions 5 + 2 = 17 nd = 5 equivlent? Prove tht 10 2 = 13 nd 14 + 7 = 20 re not equivlent, no mtter how mny steps re used. 17 A student hs tken the eqution = 5 nd performed some opertions to oth sides. = 5 4 4 4 = 20 + 3 + 3 4 + 3 = 23 2 2 (4 + 3) 2 = 46 Solve (4 + 3) 2 = 46 lgerilly. Desrie how the steps you used in your solution ompre with the steps the student used. Give n emple of nother eqution tht hs = 5 s its solution. d Eplin why there re infinitely mny different equtions with the solution = 5. Dividing whole epressions 18 It is possile to solve 2 + 4 = 20 y first dividing oth sides y 2, s long s every term is divided y 2. So you ould solve it in either of these fshions. 2 + 4 = 20 4 4 2 = 16 2 2 = 8 2 + 4 = 20 2 2 + 2 = 10 2 2 = 8 Note tht 2 + 4 divided y 2 is + 2, not + 4. Use this method of dividing first to solve the following equtions nd then hek tht you get the sme nswer s if you sutrted first. 2 + 6 = 12 4 + 12 = 16 10 + 30 = 50 d 2 + 5 = 13 e 5 + 4 = 19 f 3 + 2 = 5 g 7 = 2 + 4 h 10 = 4 + 10 i 12 = 8 + 4 18 REASONING ENRICHMENT
Numer nd Alger 401 7C Equtions with frtions Rell from lger tht frtion suh s 3 represents 3. This mens tht to solve n eqution with 3 on one side, we should first multiply oth sides y 3. For emple: 3 = 10 3 3 = 30 Let s strt: Prtising with frtions If is numer greter thn 1, evlute these epressions nd put them into sending order (smllest to lrgest). ( ) ( 2 + 1 2 2 2 + 1 2 2 + 2 2 + 1 ) + 1 2 2 Investigte how the order might hnge if ould e numer less thn 1. mens. To solve n eqution with frtion on one side, multiply oth sides y the denomintor. For emple: q = 12 4 4 4 q = 48 Emple 6 Solving equtions with frtions Solve the following equtions lgerilly. 4 3 = 8 4y + 15 = 3 9 4 + 5 2 = 29 SOLUTION 4 = 8 3 3 3 4 = 24 4 4 = 6 d 7 2 3 = 5 EXPLANATION Multiplying oth sides y 3 removes the denomintor of 3. Both sides re divided y 4 to solve the eqution. Key ides
402 Chpter 7 Equtions nd inequlities 4y + 15 = 3 9 9 9 d 4y + 15 = 27 15 15 4y = 12 4 4 y = 3 4 + 5 = 29 2 4 4 5 = 25 2 2 2 5 = 50 5 5 = 10 2 7 = 5 3 7 7 2 = 2 3 1 1 2 = 2 3 3 3 2 = 6 2 2 = 3 Eerise 7C 1 If = 4 find the vlue of 2 + 6. If = 4 find the vlue of + 6 2. Are 2 + 6 nd + 6 equivlent epressions? 2 2 Fill in the missing steps to solve these equtions. = 10 3 3 3 = q d 11 = 2 = q Multiplying oth sides y 9 removes the denomintor of 9. The eqution 4y + 15 = 27 is solved in the usul fshion (sutrt 15, divide y 4). We must sutrt 4 first euse we do not hve frtion y itself on the left-hnd side. One there is frtion y itself, multiply y the denomintor (2). Sutrt 7 first to get frtion. When oth sides re negtive, multiplying (or dividing) y 1 mkes them oth positive. 1 3 2, 3 m = 2 5 5 5 m = p = 7 10 p = UNDERSTANDING
Numer nd Alger 403 Emple 6 Emple 6, 6, 6d 3 Mth eh of these equtions with the orret first step to solve it. 4 5 4 = 7 4 2 = 5 A Multiply oth sides y 2. B Add 4 to oth sides. C Multiply oth sides y 4. D Sutrt 4 from oth sides. Solve the following equtions lgerilly. e i 5 = 4 2l 5 = 8 3m 7 = 6 f j g 10 = 2 7w 10 = 7 2n 7 = 4 g k 2 4 = 7 5 = 3 3s 2 = 9 7j 5 = 7 Solve the following equtions lgerilly. Chek your solutions y sustituting. t 8 2 = 10 h + 10 = 4 3 e 1 = s 2 8 f 5j + 6 = 2 8 7q + 12 i = 6 j 4 = f 15 5 3 m q 6 = 5 8 7 7m 12 = 5 12 g k 4 5(½) + 12 = 2 5 3 = 7v 12 + 10 15 = 3 12l 5 n 5u 7 4 = 2 o 5k + 4 8 = 3 r 4 + 7y 8 = 3 s 4 = p 15 3 4 5(½) 6 6(½), 7 d d h l d h l p t 4 + 4 = 3 k 6 = 3 5v 4 = 15 6f 5 = 24 7 2 = 5 4n 9 6 = 2 9 4r 7 = 5 3 + 13 20 = 7 g 3 5 = 1 6 For the following puzzles, write n eqution nd solve it to find the unknown numer. A numer is divided y 5 nd the result is 7. Hlf of y is 12. A numer p is douled nd then divided y 7. The result is 4. d Four is dded to. This is hlved to get result of 10. e is hlved nd then 4 is dded to get result of 10. f A numer k is douled nd then 6 is dded. This result is hlved to otin 10. 7 The verge of two numers n e found y dding them nd then dividing the result y 2. If the verge of nd 5 is 12, wht is? Solve the eqution + 5 = 12 to find out. 2 The verge of 7 nd p is 3. Find p y writing nd solving n eqution. The verge of numer nd doule tht numer is 18. Wht is tht numer? d The verge of 4 nd 6 is 19. Wht is the verge of 6 nd 4? (Hint: Find first.) 4 5(½) 6(½) 7, 8 UNDERSTANDING FLUENCY PROBLEM-SOLVING 7C
404 Chpter 7 Equtions nd inequlities 7C 8 A resturnt ill of $100 is to e pid. Blke puts in one-third of the mount in his wllet, leving $60 to e pid y the other people t the tle. Write n eqution to desrie this sitution, if represents the mount in Blke s wllet efore he pys. Solve the eqution lgerilly, nd hene stte how muh money Blke hs in his wllet. 9 In solving 2 = 10 we hve first een multiplying y the denomintor, ut we ould hve 3 ( written 2 = 10 nd divided oth sides y 2. 3) ( Solve 2 = 10. 3) d e Is the solution the sme s the solution for 2 = 10 if oth sides re first multiplied y 3? 3 Solve 147q = 1470 y first: 13 i multiplying oth sides y 13 ii dividing oth sides y 147 Wht is one dvntge in dividing first rther thn multiplying? Solve the following equtions. i 20p 14 = 40 ii 13q 27 = 39 iii 4p 77 = 4 iv 123r 17 = 246 10 To solve n eqution with pronumerl on the denomintor we n first multiply oth sides y tht pronumerl. 10 30 = 10 30 = 10 3 = 10 Use this method to solve the follows equtions. 12 = 2 15 = 5 9 9, 10 1 + 3 = 4 d 4 + 20 = 14 e 16 + 1 = 3 f 5 = 10 + 3 9, 10(½) PROBLEM-SOLVING REASONING
Numer nd Alger 11, 12 11 Solve the following equtions. Note tht the solutions should e given s frtions. 4 + 3 = 12 5 8 + 3 =6 5 7C U N SA C O M R PL R E EC PA T E G D ES ENRICHMENT Frtionl solutions 405 1 7= + 4 3 d 2= 10 3 4 12 Rell from Setion 5E (Adding nd sutrting lgeri frtions) tht lgeri frtions n e omined y finding ommon denomintor, for emple: 2 5 8 15 + = + 3 4 12 12 = 23 12 Use this simplifition to solve the following equtions. 2 5 + = 46 3 4 d 4= 2 3 + = 22 5 6 10 = e 6 2 + = 28 5 3 f 4= + 2 3 3 7 3
406 Chpter 7 Equtions nd inequlities 7D Equtions with pronumerls on oth sides So fr ll the equtions we hve onsidered involved pronumerl either on the left-hnd side, for emple, 2 + 3 = 11 or on the right side, for emple, 15 = 10 2. But how n you solve n eqution with pronumerls on oth sides, for emple, 12+5 = 16+3? The ide is to look for n equivlent eqution with pronumerls on just one side. The eqution 12 + 5 = 16 + 3 n e thought of s lning sles. 12 Then 3 n e removed from oth sides of this eqution to get 12 The eqution 12 + 2 = 16 is strightforwrd to solve. 16 Let s strt: Moving pronumerls 16 You re given the eqution 11 + 5 = 7 + 3. Cn you find n equivlent eqution with just on the left-hnd side? Cn you find n equivlent eqution with just on the right-hnd side? Try to find n equivlent eqution with 9 on the left-hnd side. Do ll of these equtions hve the sme solution? Try to find it.
Numer nd Alger 407 If oth sides of n eqution hve term dded or sutrted, the new eqution will e equivlent to the originl eqution. If pronumerls re on oth sides of n eqution, dd or sutrt terms so tht the pronumerl ppers only one side. For emple: 2 10 + 5 = 13 + 2 10 + 3 = 13 2 4 + 12 = 89 3 + 3 + 3 7 + 12 = 89 Emple 7 Solving equtions with pronumerls on oth sides Solve the following equtions nd hek your solutions using sustitution. 7t + 4 = 5t + 10 6 + 4 = 22 3 2u = 7u 20 SOLUTION 7t + 4 = 5t + 10 5t + 3 4 4 2 9 2t + 4 = 10 2t = 6 t = 3 2 6 + 4 = 22 3 9 + 4 = 22 9 = 18 = 2 2u = 7u 20 2u + 20 5 0 = 5u 20 20 = 5u 4 = u u = 4 5 4 9 5t 4 2u + 20 + 3 EXPLANATION Pronumerls re on oth sides of the eqution, so sutrt 5t from oth sides. One 5t is sutrted, the usul proedure is pplied for solving equtions. LHS = 7(3) + 4 = 25 RHS = 5(3) + 10 = 25 Pronumerls re on oth sides. To get rid of 3, we dd 3 to oth sides of the eqution. Alterntively, 6 ould hve een sutrted from oth sides of the eqution to get 4 = 22 9. LHS = 6(2) + 4 = 16 RHS = 22 3(2) = 16 Choose to get rid of 2u y sutrting it. Note tht 2u 2u is equl to 0, so the LHS of the new eqution is 0. LHS = 2(4) = 8 RHS = 7(4) 20 = 8 Key ides
408 Chpter 7 Equtions nd inequlities Emple 7 Emple 7 Emple 7 Eerise 7D 1 If = 3 re the following equtions true or flse? 5 + 2 = 4 1 7 = 6 + 5 2 + 8 = 12 d 9 7 = 3 + 11 2 Fill in the lnks for these equivlent equtions. 5 + 3 = 2 + 8 9q + 5 = 12q + 21 2 2 9q + 2p = 8 3p + 9 = 5 2p = + 2p d 1 3 2, 3 = 3p + 21 9q 15k + 12 = 13 7k + 7k + 7k = 3 To solve the eqution 12 + 2 = 8 + 16, whih one of the following first steps will ensure tht is only on one side of the eqution? A Sutrt 2 B Sutrt 8 C Add 12 D Sutrt 16 E Add 20 4 5 6 Solve the following equtions lgerilly. Chek your solution using sustitution. 10f + 3 = 23 + 6f 10y + 5 = 26 + 3y 7s + 7 = 19 + 3s d 9j + 4 = 4j + 14 e 2t + 8 = 8t + 20 f 4 + 3n = 10n + 39 g 4 + 8y = 10y + 14 h 5 + 3t = 6t + 17 i 7+5q = 19 + 9q Solve the following equtions lgerilly. 9 + 4t = 7t + 15 2 2 = 4 6 6t 3 = 7t 8 d 7z 1 = 8z 4 e 8t 24 = 2t 6 f 2q 5 = 3q 3 g 5 + 8 = 6 1 h 8w 15 = 6w + 3 i 6j + 4 = 5j 1 Solve the following equtions lgerilly. 1 4 = 7 6 6 7g = 2 5g d 2 + 8u = 37 + 3u e 21 3h = 6 6h g 13 7 = 8 2 h 10 + 4n = 4 2n j 10v + 14 = 8v k 18 + 8 = 2 m 6n 47 = 9 8n n 3n = 15 + 8n f i l o 12 8n = 8 10n 37 4j = 7 10j 10 + 32 = 2 2t + 7 = 22 3t 38 10l = 10 + 4l 4 5(½) 7 Solve the following eqution, giving your solutions s improper frtions where neessry. 3 + 5 = + 6 5k 2 = 2k 3 + m = 6 + 3m d 9j + 4 = 5j + 14 e 3 j = 4 + j f 2z + 3 = 4z 8 4 6(½) 4 7(½) UNDERSTANDING FLUENCY
Numer nd Alger 409 8 Write n eqution nd solve it lgerilly to find the unknown numer in these prolems. Douling nd dding 3 is the sme s tripling nd dding 1. If z is inresed y 9, this is the sme s douling the vlue of z. The produt of 7 nd y is the sme s the sum of y nd 12. d When numer is inresed y 10 this hs the sme effet s tripling the numer nd sutrting 6. 9 Find the vlue of nd y in the following retngles. 3 + 2 y + 3 3y 17 5 10 10 Find the re nd the perimeter of this retngle. 8 21 2y 4 y + 8 4 + 7 3y 18 2 + 4 11 At newsgeny, Preet ought 4 pens nd $1.50 newspper, while her husnd Levy ought 2 pens nd $4.90 mgzine. To their surprise the ost ws the sme. 8 8 10 y 4 Write n eqution to desrie this, using p for the ost of single pen in dollrs. Solve the eqution to find the ost of pens. If Fred hs $20 note, wht is the mimum numer of pens tht he n purhse? 9 11 PROBLEM-SOLVING 7D
410 Chpter 7 Equtions nd inequlities 7D 12 12, 13 13, 14 12 To solve the eqution 12 + 3 = 5 + 2 you n first sutrt 3 or sutrt 5. Solve the eqution ove y first sutrting 3. Solve the eqution ove y first sutrting 5. Wht is the differene etween the two methods? 13 Prove tht the retngulr shpe shown to the right must e squre. (Hint: First find the vlues of nd y.) 3 + 5 6y 7 5y 1 2 + 13 14 Try to solve the eqution 4 + 3 = 10 + 4. This tells you tht the eqution you re trying to solve hs no solutions (euse 10 = 3 is never true). Prove tht 2 + 3 = 7 + 2 hs no solutions. Give n emple of nother eqution tht hs no solutions. Geometri equtions 15 Find the vlues of the unknown vriles in the following geometri digrms. (6 5) (2z + 7) e (4 + 35) (6k 32) (3k + 4) (2) (3) d f (9) ( + 10) (3y) (6 + 5) (7 5) (100 2) (26 + 2) 15 REASONING ENRICHMENT
Numer nd Alger 411 7E Equtions with rkets In Chpter 5 it ws noted tht epressions with rkets ould e epnded y onsidering retngle res. So 4( + 2) nd 4 + 8 re equivlent. This eomes quite helpful when solving n eqution like 4( + 2) = 5 + 1. We just solve 4 + 8 = 5 + 1 using the tehniques from the previous setion. Let s strt: Arhitet s dilemm In the following house plns, the kithen nd dining room re seprted y dividing door. If the width of the divider is, wht is the re of the kithen? Wht is the re of the dining room? 3 Try to find the width of the divider if the res of the two rooms re equl. Is it esier to solve 3(7 + ) = 4( + 4) or 21 + 3 = 4 + 16? Whih of these did you solve when trying to find the width of the divider? To epnd rkets, use the distriutive lw, whih sttes tht ( + ) = +. For emple, 3( + 4) = 3 + 12. 4 2 4 = 4 4 2 = 8 Are = 4( + 2) Are = 4 + 8 7 Dining room Kithen ( ) =. For emple, 4( 2) = 4 8. Like terms re terms tht ontin etly the sme vriles nd n e olleted to simplify epressions. For emple, 5 + 10 + 7 n e simplified to 12 + 10. Equtions involving rkets n e solved y first epnding rkets nd olleting like terms. Emple 8 Solving equtions with rkets Solve the following equtions y first epnding ny rkets. 3(p + 4) = 18 12(3q + 5) = 132 4(2 5) + 3 = 57 d 2(3k + 1) = 5(2k 6) SOLUTION 3(p + 4) = 18 3p + 12 = 18 12 12 3p = 6 3 3 p = 2 EXPLANATION 4 4 Divider Use the distriutive lw to epnd the rkets. Solve the eqution y performing the sme opertions to oth sides. Key ides
412 Chpter 7 Equtions nd inequlities 12(3q + 5) = 132 36q + ( 60) = 132 36q 60 = 132 + 60 + 60 36q = 72 36 36 q = 2 d 4(2 5) + 3 = 57 8 20 + 3 = 57 11 20 = 57 + 20 + 20 11 = 77 11 11 = 7 2(3k + 1) = 5(2k 6) 6k + 2 = 10k 30 6k 6k 2 = 4k 30 + 30 + 30 32 = 4k 4 4 8 = k k = 8 Eerise 7E Use the distriutive lw to epnd the rkets. Simplify 36q + ( 60) to 36q 60. Solve the eqution y performing the sme opertions to oth sides. Use the distriutive lw to epnd the rkets. Comine the like terms: 8 + 3 = 11. Solve the eqution y performing the sme opertions to oth sides. Use the distriutive lw on oth sides to epnd the rkets. Solve the eqution y performing the sme opertions to oth sides. 1 Fill in the missing numers. 4(y + 3) = 4y + 7(2p 5) = p 35 2(4 + 5) = + d 10(5 + 3q) = + q 2 Mth eh epression d with its epnded form A D. 2( + 4) A 4 + 8 4( + 2) B 2 + 4 2(2 + 1) C 2 + 8 d 2( + 2) D 4 + 2 3 Rolf is unsure whether 4( + 3) is equivlent to 4 + 12 or 4 + 3. Fill out the tle elow. 0 1 2 4( + 3) 4 + 12 4 + 3 Wht is the orret epnsion of 4( + 3)? 1 4 1, 4 4 Simplify the following epressions y olleting like terms. 4 + 3 7p + 2p + 3 8 2 + 4 d 2k + 4 + 5k e 3 + 6 2 f 7k + 21 2k UNDERSTANDING
Numer nd Alger 413 Emple 8 Emple 8 Emple 8 Emple 8d 5 6 7 8 5, 6 7(½) 5, 6 8(½) Solve the following equtions y first epnding the rkets. 2(4u + 2) = 52 3(3j 4) = 15 5(2p 4) = 40 d 15 = 5(2m 5) e 2(5n + 5) = 60 f 26 = 2(3 + 4) Solve the following equtions involving negtive numers. 6(4p + 4) = 24 2(4u 5) = 34 2(3v 4) = 38 d 28 = 4(3r + 5) e 3(2 2) = 48 f 6 = 3(2d 4) Solve the following equtions y epnding nd omining like terms. 4(3y + 2) + 2y = 50 5(2l 5) + 3l = 1 4(5 + 3w) + 5 = 49 d 49 = 5(3 + 5) 3 e 28 = 4(3d + 3) 4d f 58 = 4(2w 5) + 5w g 23 = 4(2p 3) + 3 h 44 = 5(3k + 2) + 2k i 49 = 3(2 5) + 4 Solve the following equtions y epnding rkets on oth sides. 5(4 4) = 5(3 + 3) 6(4 + 2r) = 3(5r + 3) 5(5f 2) = 5(3f + 4) d 4(4p 3) = 2(4 + 3p) e 2(5h + 4) = 3(4 + 3h) f 4(4r 5) = 2(5 + 5r) g 4(3r 2) = 4(2r + 3) h 2(2p + 4) = 2(3p 2) i 3(2 + 1) = 11( 2) 9 Solve the following equtions lgerilly. 2(3 + 5r) + 6 = 4(2r + 5) + 6 3(2l + 2) + 18 = 4(4l + 3) 8 2(3 5) + 16 = 3 + 5(2 5) d 3(4s + 3) 3 = 3(3s + 5) + 15 e 4(4y + 5) 4 = 6(3y 3) + 20 f 3(4h + 5) + 2 = 14 + 3(5h 2) 10 Desmond notes tht in 4 yers time his ge when douled will give the numer 50. Desmond s urrent ge is d yers old. Write n epression for Desmond s ge in 4 yers time. Write n epression for doule his ge in 4 yers time. Write n eqution to desrie the sitution desried ove. d Solve the eqution to find his urrent ge. 11 Rhd s usul hourly wge is $w. She works for 5 hours t this wge nd then 3 more hours t n inresed wge of $(w + 4). 10, 11 11, 12 Write n epression for the totl mount Rhd erns for the 8 hours. Rhd erns $104 for the 8 hours. Write nd solve n eqution to find her usul hourly wge. 5 9(½) 11 13 FLUENCY PROBLEM-SOLVING 7E
414 Chpter 7 Equtions nd inequlities 7E 12 Kte s ge 5 yers go, when douled, is equl to triple her ge 10 yers go. Write n eqution to desrie this, using k for Kte s urrent ge. Solve the eqution to find Kte s urrent ge. 13 Retngles A nd B hve the sme re. Wht is the vlue of? Stte the perimeter of the shpe shown t right. 6 5 10 A 14 Arhm is sked how mny people re in the room net door. He nswers tht if three more people wlked in nd then the room s popultion ws douled, this would hve the sme effet s qudrupling the popultion nd then 11 people leving. Prove tht wht Arhm sid nnot e true. 15 Ajith lims tht three times his ge 5 yers go is the sme s nine times how old he will e net yer. Prove tht wht Ajith is sying nnot e true. 16 A ommon mistke when epnding is to write 2(n + 3) s 2n + 3. These re not equivlent, sine, for emple, 2(5 + 3) = 16 nd 2 5 + 3 = 13. Prove tht they re never equl y trying to solve 2(n + 3) = 2n + 3. Prove tht 4(2 + 3) is never equl to 8 + 3 ut it is sometimes equl to 4 + 12. Chllenging epnsions B 15.5 14 14,15 17 Solve the following equtions. Note tht, in generl, your nswers will not e integers. 2(3 + 4) + 5(6 + 7) = 64 + 1 5(3p + 2) + 5(2p + 3) = 31 10(n + 1) + 20(2n + 13) = 7 d 4(2q + 1) 5(3q + 1) = 11q 1 e + 2( + 1) + 3( + 2) = 11 f m 2(m + 1) 3(m 1) = 2(1 4m) 15,16 17 PROBLEM-SOLVING REASONING ENRICHMENT
Numer nd Alger 415 7F Formuls nd reltionships Some equtions involve two or more vriles tht re relted. For emple, you know from mesurement tht the re of retngle is relted to its length nd width, given y the formul A = l w nd its perimeter is given y P = 2l + 2w. Although these re often used s definition for the re nd definition of perimeter, they re lso just equtions two epressions written on either side of n equls sign. Let s strt: Retngulr dimensions You know tht the re nd perimeter of retngle re given y A = l w nd P = 2l + 2w. w A = l w P = 2l + 2w If l = 10 nd w = 7, find the perimeter nd the re. If l = 2 nd w = 8, find the perimeter nd the re. Notie tht sometimes the re is igger thn the perimeter nd sometimes the re is less thn the perimeter. If l = 10, is it possile to mke the re nd the perimeter equl? If l = 2, n you mke the re nd the perimeter equl? Disuss. The sujet of n eqution is pronumerl (or vrile) tht ours y itself on the left-hnd side. For emple, V is the sujet of V = 3 + 2y. A formul or rule is n eqution ontining two or more vriles, one of whih is the sujet of the eqution. To use formul, sustitute ll known vlues nd then solve the eqution to find the unknown vlue. Emple 9 Applying formul Apply the formul for retngle s perimeter P = 2l + 2w to find: P when l = 4 nd w = 7 l when P = 40 nd w = 3 SOLUTION P = 2l + 2w P = 2(4) + 2(7) P = 22 P = 2l + 2w 40 = 2l + 2(3) EXPLANATION 40 = 2l + 6 6 6 34 = 2l 2 2 17 = l l = 17 Write the formul. Sustitute in the vlues for l nd w. Simplify the result. Write the formul. Sustitute in the vlues for P nd w to otin n eqution. Solve the eqution to otin the vlue of l. l Key ides
416 Chpter 7 Equtions nd inequlities Eerise 7F 1 Sustitute = 4 into the epression + 7. Sustitute = 2 into the epression 3. Sustitute p = 5 into the epression 2p 3. d Sustitute r = 4 into the epression 7r. 1 3 1, 3 2 If you sustitute P = 10 nd = 2 into the formul P = 3m +, whih of the following equtions would you get? A 10 = 6 + B 10 = 3m + 2 C 2 = 3m + 10 D P = 30 + 2 3 If you sustitute k = 10 nd L = 12 into the formul L = 4k + Q, whih of the following equtions would you get? A 12 = 40 + Q B L = 40 + 12 C 12 = 410 + Q D 10 = 48 + Q Emple 9 4 Consider the rule A = 4p + 7. Find A if p = 3. Find A if p = 11. Find A if p = 2. d Find A if p = 13 2. Emple 9 5 Consider the rule U = 8 + 4. Find if U = 44. Set up nd solve n eqution. Find if U = 92. Set up nd solve n eqution. If U = 12, find the vlue of. 6 Consider the reltionship y = 2 + 4. Find y if = 3. By solving n pproprite eqution, find the vlue of tht mkes y = 16. Find the vlue of if y = 0. 7 Use the formul P = mv to find the vlue of m when P = 22 nd v = 4. 8 Assume tht nd y re relted y the eqution 4 + 3y = 24. If = 3, find y y solving n eqution. If = 0, find the vlue of y. If y = 2, find y solving n eqution. d If y = 0, find the vlue of. 9 Consider the formul G = k(2 + p) +. If k = 3, = 7 nd p = 2, find the vlue of G. If G = 78, k = 3 nd p = 5, find the vlue of. 4 7 4 9 5 9 UNDERSTANDING FLUENCY
Numer nd Alger 417 10 The ost $C to hire ti for trip of length d km is C = 3 + 2d. Find the ost of 10 km trip (i.e. for d = 10). A trip hs totl ost of $161. i Set up n eqution y sustituting C = 161. ii Solve the eqution lgerilly. iii How fr did the ti trvel? (Give your nswer in km.) 11 To onvert temperture etween Celsius nd Fhrenheit, the rule is F = 1.8C + 32. Find F if C = 10. Find C if F = 95. Vinod s ody temperture is 100 Fhrenheit. Wht temperture is this in degrees Celsius? Answer orret to one deiml ple. 12 The formul for the re of trpezium is A = 1 h ( + ). 2 Find the re of the trpezium shown t right. Find the vlue of h if A = 20, = 3 nd = 7. Find the missing vlue in the trpezium t right. 10 10, 11 8 7 Are = 72 13 Kty is sientist who tries to work out the reltionship etween the volume of gs, V ml, nd its temperture T C. She mkes few mesurements. Wht is possile rule etween V nd T? Use your rule to find the volume t temperture of 27 C. (V 10)2 Prove tht the rule T = + 10 would lso work for Kty s results. 20 14 Consider the rule G = 120 4p. If p is etween 7 nd 11, wht is the lrgest vlue of G? If p nd G re equl, wht vlue do they hve? 13 13, 14 12? 11, 12 V 10 20 T 10 15 15, 16 h 6 8 PROBLEM-SOLVING REASONING 7F
418 Chpter 7 Equtions nd inequlities 7F 15 Mrie is sientist who is trying to disover the reltionship etween the volume of gs V, its temperture T nd its trnspreny A. She mkes few mesurements. Test 1 Test 2 V 10 20 A 2 5 T 15 12 Whih one or more of the following rules re onsistent with the eperiment s results? A T = 3V A B T = V + 2A C T = 17 A 16 Tempertures in degrees Fhrenheit nd Celsius re relted y the rule F = 1.8C + 32. By sustituting F = nd C =, find vlue suh tht the temperture in Fhrenheit nd the temperture in Celsius re equl. By sustituting F = 2 nd C =, find temperture in Celsius tht doules to give the temperture in Fhrenheit. Prove tht there re no Celsius tempertures tht n e multiplied y 1.8 to give the temperture in Fhrenheit. Moile phone plns 17 Two ompnies hve moile phone plns tht ftor in the numer of minutes spent tlking eh month (t) nd the totl numer of lls mde (). Compny A s ost in ents: A = 20t + 15 + 300 Compny B s ost in ents: B = 30t + 10 d e In one month 12 lls were mde, totlling 50 minutes on the phone. Find the ost in dollrs tht ompny A nd ompny B would hve hrged. In nother month, ompny A user ws hrged $15 (1500 ents) for mking 20 lls. How long were these lls in totl? In nother month, ompny B user tlked for 60 minutes in totl nd ws hrged $21. Wht ws the verge length of these lls? Briony noties one month tht for her vlues of t nd, the two ompnies ost etly the sme. Find possile vlue of t nd tht would mke this hppen. Briony revels tht she mde etly 20 lls for the month in whih the two ompnies hrges would e the sme. How muh time did she spend tlking? 17 REASONING ENRICHMENT
Numer nd Alger 419 38pt 7A 38pt 7A 38pt 7A 38pt 7B 38pt 7C 38pt 7D 38pt 7E Progress quiz 1 2 3 4 5 6 7 For eh of the following equtions, stte whether they re true (T) or flse (F). 5 + 11 = 8 2 + 9 = 18, if = 4 ( 4) = 2, if = 6 Stte solution to eh of the following equtions (no working required). k + 5 = 21 7 = 6 4m = 32 Write equtions for the following senrios. You do not need to solve the equtions. A numer n is douled nd then 5 is dded. The result is 17. Arhie s ge is. Arhie s mother, who is 26 yers older thn Arhie, is triple Arhie s ge. Solve the following equtions lgerilly. + 8 = 15 12 = 9 k 42 = 6h d 5 + 3y = 29 e 4u 8 = 40 f 52 = 8j 4 g 68 12d = 8 h 59 = 13 9m Solve the following equtions lgerilly. 5u 2 = 100 3h 7 = 6 Solve the following equtions lgerilly. 4n + 3 = 2n + 17 9w 7 = 4w 17 d 10 + t = 3t 3 + 4 3 = 15 2w + 7 d = 5 3 e + 8 = 28 3e Solve the following equtions y first epnding ny rkets. 6( + 2) = 42 3(4w 6) = 114 5(2q 1) 3q = 30 d 8(2 p) = 3(2p 8) 38pt 7F 8 Apply the formul for retngle s perimeter P = 2l + 2w to find the length l, when P = 32 m nd w = 6 m. The rule for the re of tringle is A = h. By solving n pproprite eqution, find the se 2 length, for tringle of height h = 3 m nd re A = 24 m 2. 38pt 7B/C 38pt 7E 38pt 7F 9 10 11 For eh of the following, write n eqution nd solve it lgerilly to find the unknown numer. The produt of q nd 6 is 30. Two thirds of numer m gives result of 12. A numer k is tripled nd then 4 is dded. This result is hlved to otin 13. d The verge of 3 nd 10 is 14. Mddie s ge 8 yers go when multiplied y 5, is the sme s triple Mddie s ge in 2 yers time. Write nd solve n equtions to find Mddie s urrent ge. The formul for the re of trpezium is A = h ( + ). 2 Find the vlue of the height h, in trpezium with A = 162 m 2, = 12 m, = 15 m. h
420 Chpter 7 Equtions nd inequlities 7G Key ides Applitions Although knowing how to solve equtions is useful, it is importnt to e le to reognise when rel-world situtions n e thought of s equtions. This is the se whenever it is known tht two vlues re equl. In this se, n eqution n e onstruted nd solved. It is importnt to trnslte this solution into meningful nswer within the rel-world ontet. Let s strt: Siling sum John nd his elder sister re 4 yers prt in their ges. If the sum of their ges is 26, desrie how you ould work out how old they re. Could you write n eqution to desrie the sitution ove, if is used for John s ge? How would the eqution hnge if is used for John s sister s ge insted? The differene in two people s ges n e epressed s n eqution. An eqution n e used to desrie ny sitution in whih two vlues re equl. To solve prolem follow these steps. 1 Define pronumerls to stnd for unknown numers (e.g. let j = John s urrent ge). 2 Write n eqution to desrie the prolem. 3 Solve the eqution lgerilly if possile, or y inspetion. 4 Ensure you nswer the originl question, inluding the orret units (e.g. dollrs, yers, m). Your finl solution should e heked to see if it solves the prolem orretly. Emple 10 Solving prolem using equtions The weight of 6 identil ooks is 1.2 kg. Wht is the weight of one ook? SOLUTION Let = weight of one ook in kg. 6 = 1.2 6 = 1.2 6 6 = 0.2 The ooks weigh 0.2 kg eh, or 200 g eh. EXPLANATION Define pronumerl to stnd for the unknown numer. Write n eqution to desrie the sitution. Solve the eqution. Answer the originl question. It is not enough to give finl nswer s 0.2; this is not the weight of ook, it is just numer.
Numer nd Alger 421 Emple 11 Solving hrder prolem using equtions Purhsing 5 pples nd $2.40 mngo osts the sme s purhsing 7 pples nd mndrin tht osts 60 ents. Wht is the ost of eh pple? SOLUTION Let = ost of one pple in dollrs. 5 + 2.4 = 7 + 0.6 5 0.6 5 + 2.4 = 7 + 0.6 2 2.4 = 2 + 0.6 1.8 = 2 0.9 = 2 Apples ost 90 ents eh. 5 0.6 EXPLANATION Emple 12 Solving prolems with two relted unknowns Define pronumerl to stnd for the unknown numer. Write n eqution to desrie the sitution. Note tht 60 ents must e onverted to $0.6 to keep the units the sme throughout the eqution. Solve the eqution. Answer the originl question. It is not enough to give finl nswer s 0.9; this is not the ost of n pple, it is just numer. Jne nd Luke hve omined ge of 60. Given tht Jne is twie s old s Luke, find the ges of Luke nd Jne. SOLUTION Let l = Luke s ge l + 2l = 60 3 3l = 60 l = 20 3 Luke is 20 yers old nd Jne is 40 yers old. EXPLANATION Define pronumerl for the unknown vlue. One Luke s ge is found, we n doule it to find Jne s ge. Write n eqution to desrie the sitution. Note tht Jne s ge is 2l euse she is twie s old s Luke. Solve the eqution y first omining like terms. Answer the originl question.
422 Chpter 7 Equtions nd inequlities Eerise 7G 1 Mth eh of these worded desriptions with n pproprite epression. The sum of nd 3 A 2 The ost of 2 pples if they ost $ eh B + 1 The ost of ornges if they ost $1.50 eh C 3 d Triple the vlue of D + 3 e One more thn E 1.5 2 For the following prolems hoose the eqution to desrie them. The sum of nd 5 is 11. A 5 = 11 B + 5 = 11 C 5 = 11 D 11 5 1 3 2 The ost of 4 pens is $12. Eh pen osts $p. A 4 = p B 12p C 4p = 12 D 12p = 4 Josh s ge net yer is 10. His urrent ge is j. A j + 1 = 10 B j = 10 C 9 D j 1 = 10 d The ost of n penils is $10. Eh penil osts $2. A n 10 = 2 B 5 C 10n = 2 D 2n = 10. 3 Solve the following equtions. 5p = 30 5 + 2 = 23 12k 7 = 41 d 10 = 3 + 1 Emple 10 4 Jerry uys 4 ups of offee for $13.20. d Choose pronumerl to stnd for the ost of one up of offee. Write n eqution to desrie the prolem. Solve the eqution lgerilly. Hene stte the ost of one up of offee. 5 A omintion of 6 hirs nd tle osts $3000. The tle lone osts $1740. d Define pronumerl for the ost of one hir. Write n eqution to desrie the prolem. Solve the eqution lgerilly. Hene stte the ost of one hir. 6 The perimeter of this retngle is 72 m. Write n eqution to desrie the prolem, using w for the width. Solve the eqution lgerilly. Hene stte the width of the retngle. 4 6 4 7 4 m w m 5 7 UNDERSTANDING FLUENCY
Numer nd Alger 423 Emple 11 Emple 12 7 A plumer hrges $70 ll-out fee nd $52 per hour. The totl ost of prtiulr visit ws $252. 8 d Define vrile to stnd for the length of the visit in hours. Write n eqution to desrie the prolem. Solve the eqution lgerilly. Stte the length of the plumer s visit, giving your nswer in minutes. A numer is tripled, then 2 is dded. This gives the sme result s if the numer were qudrupled. Set up nd solve n eqution to find the originl numer. 9 A squre hs perimeter of 26 m. 10 Solve n eqution to find its width. Hene stte the re of the squre. w Perimeter = 26 m Alison nd Flynn s omined ge is 40. Given tht Flynn is 4 yers older thn Alison, write n eqution nd use it to find Alison s ge. 11 Rell tht in qudrilterl the sum of ll ngles is 360. Find the vlues of nd y in the digrm shown. 12 The sum of three onseutive numers is 357. Use n eqution to find the smllest of the three numers. Wht is the verge of these three numers? If the sum of three onseutive numers is 38 064, wht is their verge? 13 The width of retngulr pool is 5 metres longer thn the length. The perimeter of the pool is 58 metres. Drw digrm of this sitution. Use n eqution to find the pool s length. Hene stte the re of the pool. 9 8 11 100 2 11 13 y FLUENCY PROBLEM-SOLVING 7G
424 Chpter 7 Equtions nd inequlities 7G 14 14 14, 15 14 The verge of two numers n e found y dding them nd dividing y 2. If the verge of nd 10 is 30, wht is the vlue of? If the verge of nd 10 is 2, wht is the vlue of? If the verge of nd 10 is some numer R, rete formul for the vlue of. 15 Sometimes you re given n eqution to solve puzzle, ut the solution of the eqution is not tully possile for the sitution. Consider these five equtions. A 10 = 50 B 8 + = 10 C 10 + = 8 D 10 = 8 E 3 + 5 = + 5 d You re told tht the numer of people in room n e determined y solving n eqution. Whih of these equtions ould e used to give resonle nswer? If the length of n inset is given y the vrile m, whih of the equtions ould e solved to give resonle vlue of? Eplin why eqution D ould not e used to find the numer of people in room ut ould e used to find the length of n inset. Give n emple of puzzle tht would mke eqution C resonle. Unknown numers 16 Find the unknown numer using equtions. The nswers might not e whole numers. The verge of numer nd doule the numer is 25.5. Adding 3 to twie numer is the sme s sutrting 9 from hlf the numer. The verge of numer nd doule the numer gives the sme result s dding one to the originl numer nd then multiplying y one-third. d The produt of 5 nd numer is the sme s the sum of four nd twie the originl numer. e The verge of 5 numers is 7. When one more numer is dded the verge eomes 10. Wht numer ws dded? 16 REASONING ENRICHMENT
Numer nd Alger 425 7H Inequlities EXTENDING An inequlity is like n eqution ut, insted of inditing tht two epressions re equl, it indites whih of the two hs greter vlue. For emple, 2 + 4 < 7, 3 5 15 nd 10 re ll inequlities. The first two re true, nd the lst one ould e true or flse depending on the vlue of. For instne, the numers 9.8, 8.45, 7 nd 120 ll mke this inequlity true. We ould represent ll the vlues of tht mke 10 true sttement. Let s strt: Smll sums Two positive whole numers re hosen: nd y. You re told tht + y 5. How mny possile pirs of numers mke this true? For emple, = 2 nd y = 1 is one pir nd it is different from = 1 nd y = 2. If + y 10, how mny pirs re possile? Try to find pttern rther thn listing them ll. If ll you know out nd y is tht + y > 10, how mny pirs of numers ould there e? An inequlity is sttement of the form: LHS > RHS (greter thn). For emple, 5 > 2 LHS RHS (greter thn or equl). For emple, 7 7 or 10 7 LHS < RHS (less thn). For emple, 2 < 10 LHS RHS (less thn or equl). For emple, 5 5 or 2 5 Inequlities n e reversed: 3 < nd > 3 re equivlent. Inequlities n e represented on numer line, using losed irles t the end points if the vlue is inluded, or open irles if it is eluded. Closed irle indites 5 is inluded 5 < 6 4 5 6 7 4 5 6 7 Open irle indites 6 is eluded A rnge n e represented s segment on the numer line using pproprite losed nd open end points. Emple 13 Representing inequlities on numer line Represent the following inequlities on numer line. 4 < 6 1 < 5 8 9 10 11 Key ides
426 Chpter 7 Equtions nd inequlities SOLUTION 3 3 0 4 4 1 5 5 2 6 6 3 7 4 5 6 EXPLANATION Emple 14 Using inequlities to desrie rel-life situtions A irle is pled t 4 nd then the rrow points to the right, towrds ll numers greter thn 4. The irle is filled (losed) euse 4 is inluded in the set. A irle is pled t 6 nd then the rrow points to the left, towrds ll numers less thn 6. The irle is hollow (open) euse 6 is not inluded in the set. Cirles re pled t 1 nd 5, nd line goes etween them to indite tht ll numers in etween re inluded. The irle t 1 is open euse the inequlity is < not. Desrie the following situtions s n inequlity, using to stnd for the unknown quntity. Fred is shorter thn 160 m. John is t lest s old s Mri, who is 10. Rose s test sore is etween 40 nd 50 inlusive. SOLUTION EXPLANATION < 160 Using to stnd for Fred s height in m, must e less thn 160. 10 John is t lest 10, so his ge is greter thn or equl to 10. 40 50 is etween 40 nd 50. The word inlusive Eerise 7H tells us tht 40 nd 50 re oth inluded, so is used (rther thn < if the word elusive is used). 1 4 4 1 Clssify the following sttements s true or flse. 5 > 3 7 < 5 2 < 12 d 13 < 13 e 13 13 f 4 2 2 Mth eh of these inequlities with the pproprite desription. > 5 < 5 3 d 3 A The numer is less thn 5. B The numer is greter thn or equl to 3. C The numer is less thn or equl to 3. D The numer is greter thn 5. UNDERSTANDING
Numer nd Alger 427 Emple 13, 13 Emple 13 Emple 14 3 For eh of the following, stte whether they mke the inequlity > 4 true or flse. = 5 = 2 = 4 d = 27 4 If = 12, lssify the following inequlities s true or flse. > 2 < 11 13 d 12 5 6 Represent the following inequlities on seprte numer lines. > 3 < 10 2 d < 1 e 1 f 4 g < 5 h < 9 i 2 j < 6 k 3 l 5 m 10 > n 2 < o 5 p 3 5(½), 6 5(½), 6, 7(½) List whih of the following numers mke the inequlity 2 < 7 true. 8, 1, 3, 4, 6, 4.5, 5, 2.1, 7, 6.8, 2 Represent the inequlity 2 < 7 on numer line. 7 Represent the following inequlities on seprte numer lines. 1 6 4 < 11 2 < 6 d e 2 < 5 f 8 < < 1 g 7 < 8 h 0 < < 1 8 8 3 For eh of the following desriptions, hoose n pproprite inequlity from A H elow. John is more thn 12 yers old. Mrik is shorter thn 150 m. Mtthew is t lest 5 yers old ut he is younger thn 10. d The temperture outside is etween 12 C nd 10 C inlusive. 5(½), 6, 7(½) A < 150 B < 12 C > 12 D 150 E 10 12 F 12 10 G 5 < 10 H 5 < 10 9 It is known tht Tim s ge is etween 20 nd 25 inlusive, nd Nik s ge is etween 23 nd 27 inlusive. 8 8, 10 If t = Tim s ge nd n = Nik s ge, write two inequlities to represent these fts. Represent oth inequlities on the sme numer line. Nik nd Tim re twins. Wht is the possile rnge of their ges? Represent this on numer line. 9, 10 UNDERSTANDING FLUENCY PROBLEM-SOLVING 7H