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MATHEMATICS. ALGEBr. A. [SOLUTIONS TO THE EXERCISES. 



each copper coin U worth as many peuoo ai there are 

 ilver coina, and each ailvcr coin in worth as many 

 pence M then are copper coin* ; and the whole u worth 

 18. ; how many are there of each I 



.i.poee there were x of one sort, and therefore 

 21 jc of the other : then the worth of tho whole in 



- x> = 210, or 2(24 - x)x - 216 



feting the square, x 24* + 144 - 36, . . x 12 

 _ 2 6, .; x - 18, or 6, .'. 24 x - 6 : hence there 

 were 18 silver coins and 6 copper coins, or C silver and 



(Menriw. Let tho number of coins be x and y : then 



X + y ~* -I. :m.l -Vy = 216 



... ( X + ,,)- 670, and4xy = 432. 

 Hence by subtraction, (x y)* - 144, . . x y - + 12. 

 by adding to and subtracting from the first equation, 

 wohave2x = 36,2y-12; .-.x = 18,y = 6. 



Tho negatin values are suppressed, because they are 

 inadmissible, from the nature of the question. 



6. Two messengers, A and B, were dispatched to the 

 same place, 90 miles distant. A, by riding one mile an 

 hour more than B, arrives at his destination an hour 

 before him. How many miles did each travel ? 



Suppose A travels x miles an hour, then he reaches 



90 

 his destination in hours ; and B, by travelling 1 mile 



an hour more that is, x + 1 miles an hour reaches it 



And, by the question, 

 90 90 , ]_ or 90 _ 91_+ x 



90 



111 , , hours. 



Clearing fractions, 90 x + 90 = 91x + x 1 



Transposing, x s + x = 90 

 . . Rule II., 2x + 1 = V(300 + 1) = ^361 = + 19, 



... x= , = 1 + 19 == 9. .-.x + l=lft 



Hence A travels 9 miles an hour, and B 10 miles an 

 hour. The negative value of x is suppressed, as it is ex- 

 cluded by the conditions of the question. You will 

 observe that algebra gives all the values of the unknown 

 quantity that can satisfy the equation : if there are any 

 restrictions in the question embodied in that equation, 

 the final results must, of course, conform to those re- 

 strictions. 



0. A grocer sold 801bs. of mace and lOOlbs. of cloves 

 for 65 ; but he sold 601bs. more of cloves for 20, 

 than he did of mace for 10. What was the price of 

 1 Ib. of each ? 



Suppose tho price of the mace per Ib. was x pounds, 

 and that of tho cloves y pounds ; then by the question, 

 80x + lOOy = 65. Also, the number of Ibs. of cloves 



for 20 is , and of mace for 10 is - ; and by the 



question, J- CO :" hence, tho tw.o equations are 



''+ lOOy - 65, .'. 10t + 20y = 13 . . . (1) 



y x ' y x 



Clearing tho last of fractions, .'. 2x y = C.ry ... (3) 



Substituting in (3) tho expression for * given by (1> 

 viz., * ~^, it becomes 



13 20y 39y COy* 



s ~v~- g 



Multiplying by 8, /. 13 20y 8y = 39y GOy' 

 Trannponing, COy 1 C7y - 13 



.-. llulo II., 120i/ 07 - V( 13 X 240 + 67 1 ) 



67 + 37 1 13 

 -V 13C9 - + 37.'. V= I20~~~4 >0r i5 



13 20i/ 13 5 1 

 .'.- - - - - "* o ! I" other value of x is 



negative: hence tho price of lib. of maoo was i" 

 10 , and of 1 Ib. of cloves J, or 5*. 



Othcrwite. Multiply (1) and (2) together; we t!icn 

 have 



32 Z 20^ 10 + 40-78,.-. 32y 20^-54, 



Multiplying (2) by 8x, 16 - 8 - 48x 

 Subtracting, 10 - 8 = 48x 27 



Dividing (1) by 2x, 8 + 10 J - . -. 10 J -g - 8- 

 Hence, by substitution in tho preceding equation, 



Clearing the fraction, 



96x s 22x = 13, . . Rule II., 192* 22 

 V (90x52 + 22*)= +74 



'. x=- 



22 + 74 1 



13 



192 



= 2' or "~48 



and, rejecting the negative value of x, we have from (1) 

 13 10.e = 13 8 = 1 

 20 20 = 4 



Therefore, the price of lib. of mace was 1, or 10s. ; and 

 of lib. of cloves }, or 5s., as before. 



7. The product of two numbers is 240, and they are 

 such that if one of them be increased by 4, and the other 

 diminished by 3, the product of tho results is still 240. 

 Find the numbers. 



Let x and y represent the numbers ; then by the 

 question, 



xy = 240, and (* + 4)(y 3) = 240, 

 That is, xy 3* + 4y 12 = 210 

 Subtracting this from the first equation, 3x 4y + 12 



,5_ 



3 !/ 



3-^ = 240, .-.4> 12y= 720 

 180, .-. Rule II., 2y 3= ^(720+9) 



.y = -^_=15,or-12 



240 

 .-.x = -- =16, or 20. 



Hence the numbers are either 16 and 15, or 20 and 



12. 



8. A and B set out at the same time for a place 150 

 miles distant. A travels 3 miles an hour faster than B, 

 and arrives at the place 8J hours before lum. How 

 many miles did each travel per hour ? 



Suppose A travels x miles per hour, then B travels x 



3 miles an hour ; and the number of hours occupied 



by A is the number occupied by B is -i?L ; but by 



y ' y. o 



this latter number exceeds tho former 

 150 . 150 



tho question, 

 by 8*, 



Clearing fractions, 450x 1350 + 25x 75* = 450x 

 Transposing, 25x 2 75x = 1350, . . x a 3* = 54, 



. . Rule II., 2* 3 - v/ (216 + 9) - ^225 = + 16 



.. I = _ 3 tl 5 . 9)0r _ c , 



.-.x 3 = 6. 



Hence, A travels at the rate of 9 miles an hour, and B 

 at the rate of 6 miles an hour. 



!i \Vliiit nuniliiT is tliat tlio sum of whoso dibits is 15, 

 and if III be added to their product, tho digits will lo 



