RESULTS, HINTS, ETC., FOR THE EXERCISES ON EQUATIONS OF THE FIRST DEGREE. 1+ - {1 } or VI. 1. x=-;. 2. x=4. 3. x=0 4. x=11. 5. x=7. 6. The equation may be reduced to the form 1 1 =l+ X-3 2 1 1 1 1 -1 -1 and X - 3 2-4 (0 - 3)(x - 1)(x-6)(x – 7)* 7. U= =20. 8. x=771. 9. 23 10. x=-111. a-2 11. x=1} 12. x=2. 13. x=51. 14. x=2. 4 15. The equation is 2+1+. 2- -=X+3+ -X+4- X + 4 19 16. The equation is -4+; 19 +2 9+ +73x+1 2x +7 2+2 19 6 Whence 1 and 2+1 114 VII. a- 1-6 2. 2 d 3. 2= 6+c-a a-1° a-26 a+0 ab ad ac (a - b)c 13. X = 14. 2= 15. 2 bc a+b ab ab 16. X= a? -6 17. The equation is an identity if ab=bc; if not, it is impossible. ac + bd - 62 18. 33. 19. 2= - (a? + 2abc – 62) 20. x= (a-c). -1 a-6 =X+ ab a VIII. 1. x=1. 2. (a+b)x=(c+d)x, an identity if a+b==c+d, if not, what explanation can be given ? 3. 2? +(a+b).x + ab 2? +(c+d)x+cd 2 + a + b a+c+d (a+b)cd-(c+d)ab ab-cd 5. 2 6. i= Wa+c-b). (a-c). {u(6-c)+(a - b)} b+c - (92b + ab? +a?c + b2c + abc) 355a + 3486 7. x= 8. = ab + ac + bc 37 - (ab2 + bc? + ca?) ab? + bc2 +ca? + abc 9. 2= 10. 2= a'b+b2c + ca a2b +6*c+c?a 11. X = (62 – ca)bc +(c? – a?)ac +(a? – 62)ab =a a+b+c. 13. 2= a+b+c+d 14. 3 2(a +36) 2a-C m+n cd - ab 15. X = (m? + n)b - (m.- n)a 16. [=1. 17. 2= m? - m - 2n a+b-c-à 18. The equation when reduced and simplified becomes a: +68+63 – 3abc= -3{a+62 +62 – ab – ac— be}", and as + b3 +28 - 3abc a+b+c 3(a? + b3 + c2 - ab - ac- bc) 3 bd - ab 19. x= 3. 20. (a -- d)bc' . 1. Let x denote one number, then x+10 is the other, and x+x+10=100... 2x=90, and x=45 the less, and x+10 - 55 the greater. Otherwise, if x denote the larger number, then 2 – 10 denotes the smaller, and x+2- 10 -100,... 2x=110, and x=55 the greater, and a – 10 = 45 the less. 2. x= 105. 3. x-3. 4. x= 108. 5. 2=60. 6. The parts are 80 and 160. 7. Here 1 of the first part= } of the second = 4 of the third. Let 2x denote th first part, then 3x is the second part, and 4x the third part . . 2x+ 3x + 4x=144. 8. X=90. 9. The original number is 24. 10. The fraction is *** 14. Let x denote the number when the first digit is removed, then 100000+ x expresses the number ; but 10x+1=3 times the number =300000+ 3x. .. 7x=299999, and x= =42857 ... the number is 142857. 15. The parts are 1+ m +mn 1+ m + mn 1+ m + mn та na n-1 XI. 1. If A be supposed older than B, from the birth of B they advance equally in age, and it is obvious that A may become x times as old as B some years after the given epoch, or some years before that epoch. Suppose A becomes m times as old as B in 3 years after the epoch, Then a+x=n(6+x), and (n-1)x=a – nb, a - nb ; and x will be positive or negative according as a - nb is positive or negative; the former refers to time after, the latter to time before the epoch. 2. x+44=3(2x + 4), x=16, 3+40=56. 3. The father is 75 years of age. 4. A's age 62, B's 22. 5. x=-20, indicating time past, A was therefore twice as old as B in 1830 - 20 or in 1810. 6. One fired 14 shots, the other 18. XII. 1. They meet when A has travelled 298miles from Cambridge, and when B has travelled 2634 miles from London. 2. 500 feet. 3. 90 miles. 7. After 10 days, 20 days' provisions are left. If be the reinforcement, then 20 days' provisions gives 5(x + 1000) rations; but the 20 days' provisions would have given to 1000 men 20 ~ 1000 rations... 5(x+1000)=20 x 1000. + 3.0 5(1000—2)=40. + XIII. 1. A invested £24,000 in the 3} per cents., and B 25,000 in the 4 per cents. 4550x5420(x+1) 2. Let x the rate on £4,550, then = 453. 100 100 3. Let x be the sum invested in the 3 per cents., then 1063 4. Consols were at 81 and at 90. 3x - 1000 5 5. Let 3x, 2x denote their incomes, then and the incomes were 2.2 -- 10003 £6,000 and £4,000. xp 2a + b 3 3 8. 240 sovereigns, 720 shillings, 960 pence. 9. £8 and 10 half-crowns. 10. Cost price 6s. 8d. 11. =£2,200, the original sum. 12. £540 original income. m na na na ах n =a. + m n + m n XIV. 1. In the time ezbile. The problem may be extended to 3, 4, &c., agents. estateat, 2. Herc a men or b boys dig m acres in n days. .. 1 man digs and i boy m(x – P)(1 – p) digs acres in 1 day ... (a-p) men dig acres in (n-p) days. nb mn- ploc Suppose x boys are wanted, then x boys dig acres in man-p) days, and nb the whole is (m +p) acres. m(a – p)(n-P), mem--p)x=m+p, from which x can be found. nb 3. C will finish the portion left in 76 days. 4. The hare takes 240 leaps. 7. Let the cistern contain a gallons, and suppose x hours the time required when pipe A is opened. Then in x hours pipe A pours in gallons, pipe B pours in a(x-1) a(x-2) P ax a(x - 1) a(x - 2) P P in the former case, in the latter case, (m+n)P- mn under the coulition that in both cases (m+p) is greater than mn; and in the second also that p is greater than 2n. abcd 8. 1975 men. 9. If x be the number of hours, x= abc + abd + acd + bcd 10. The chain rule is applicable to questions of equivalents, as the comparison of the standards of moneys, weights, and measures of different nations, and in almost all transactions of international commerce where equivalents are considered. Questions of this nature may always be most conveniently solved by aid of this a, units of the first kind is equivalent to b, of the second kind, ca third a, first then a,b,c par=a,b,capar, and therefore a,b,c,=a,b,cz. Hence, if any five of these six quantities be given, the sixth may be found by a simple equation. The number of equivalents may be extended to 4, 5, or to any number whatever. The following example will illustrate the rule : If 3 pounds of tea be worth 4 pounds of coffee, and 15 pounds of coffee be equal to 37 pounds of sugar, how many pounds of sugar are equal in value to 180 pounds of tea. IIere 3lb. tea=4lb. coffee, Then 3 x 15 xx=4x37 x 180. 151b. coffee=371b. sugar, 4 x 37 x 180 ..2= Let oll). sugar=180lb. tea. =5921b. sugar. 1. See page 3. XV. 2. See page 1. 4. Here x - a=}(C — ), -a=1(2-c), 3-a=(x-d). By substituting the value of 2, 6+c+d=f(a +c+d), 6+c+d=f(a+b+d), 6+c+d=f(a+b+c), from these equations the relations may be found. 5. (5--c)bep – (c-a)acq_(c-a)9+(6–c)p. 3abc 6. = 10. a= b(c-1)-c 2(ab +ac+bc) 1+b+c . XVI. 1. x=7, y=6. 2. x=371, y=511. 3. x=2, y=1. 4. x=10, y=11. 5. The equations are when reduced, x – 2y=1, 2 – - 2y=-6. Explain. 6. x= =5, y=7. 7. 23 - 3, y=1. 8. x=1, y=1. 9. x=741%, y=321. 10. x=}, y=1. 11. 2=4, y=16. 12. x= 1, y=tó: XVII. 1. x=12, y=3. 2. x=39.812.. ..., y=58.542... 3. x='02, y=2.9. 4. The reduced equations are •125y=-3x, 3x= -25y. Explain them. 5. x=4, y=5. 6. x=1703, y=$11. 1. x=25, y=12. 4. x=9, y=12. 7. x= =-251, y=2413. 10. x=6, y=. XVIII. 3. 3 = = 415, y=56 6. x=12, y=8. 9. x=21/1, y=2775. 12. x=18, y=40. XIX. 1. Incompatible equations. From (1) 10y=2, from (2) 3x=74y. 2. x=411, y=-*1. 3. x=-4, y=-2. 4. x=21, y=20. 5. x=1215, y=-2195. 6. The equations when reduced are 6x + 55Y=328, 49x – 3y=111. 7. x=811, y=131. 8. The reduced equations are 9113 — 478y=1695, 9x – 10y=3. 3. x=ff, y=47. 10. x=414, y=5814. 11. x=105, y=745. 12. x=-*t, y=-24%. C ab Y= =ag + bp ac-bd , y= ac XX. 6 + ma 1. 2= a - mb Y= a2 + ab +62 2. = a + 6 a+b 3. 2= -pc+d2 md + nb na-mc y= 4. x= bd ad + bc ad + bc 6+d-a-c = 26 7. X= (6-c)d-(+c) (6-c)d-(a+b)e y= (6—c)(a -c) (6-c)(c-a) 6 8. The equations may be reduced to Y & y C 62 +a? y= b-a bta y=-(a+b+c-d). +=1, and b a =1, |