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MATHEMATICS. ALGEBRA. 



[SOLUTIONS TO THE EXEBCHES. 



Utter 15(50 z) pence ; hence, as they received together 

 480 peace, we must have the equation 



15(50 r) 4- 9x - 480 ; 

 That u, 750 15* + 9x - 480 ; 

 Tran*poing, 270 = Ox, .'. x 4ft, and 60 * 5 ; 

 Consequently, 45 received 9./. each, and 5 received 1 .' '. 

 6. A penou ordered a quantity of mm and brandy, 

 for which ho raid 19 4*. ; the brandy was 9*. a quart, 

 and the rum <k He has, however, forgotten the exact 

 quantity of each which he has to receive ; but he 

 I (-members that if his brandy hod been rum, and hi; 

 rum brandy, his outlay would have been 1 13*. less. 

 llow many quarts of each did he buy} 



SimpoM lie bought x quarts of rum and y quarts of 

 brandy ; then, by the question, the number of shillings 

 for the two was 6x + 9y ; but if the brandy had 

 DM C*. a quart and the rum 9*., the number of shil- 

 lings paid would liave been 9x 4- 6i/. The actual price 

 paid was 384*.; the sum that would have been paid in 

 tlu> latter case, we are told, is 33*. less than this, that is, 

 361*. ; hence we have these two equations viz., 

 Cz + 9y=384 . 2z + 3y = 128 

 9x 4- Ct/ = 351 ' ' 3x 4- 2y = 117. 

 By adding these two equations, we have 



5x 4- by = 245 . . y + z = 49 

 By subtracting, y x = 11 



Therefore adding and subtracting, 2y = 60, 2r = 38, 

 . . z = 19, and y = 30 ; so that he had 19 quarts of 

 rum and 30 quarts of brandy. 



OM-trtrtse. Suppose he had x quarts of rum ; for this 

 ho paid 6x shillings by the question ; therefore 384 Gs 

 is the number of shillings paid for the brandy ; and as 1 

 quart of brandy cost 9*. the number of quarts must 



have been 



384 6* 



128 2x 

 or o Now, by the ques- 



9 ' "' 3 

 lion, if these quarts had cost Cs. each, and the x quarts 

 9*. each, the number of shillings paid would have been 

 only 351 : hence we have the equation 



128 -*" 351, 



or, 



and 



3 X 6 + 9x = 



256 4x4- 9x = 351, 

 . . 5z = 95, . . x = 19, quarts of rum, 



128 2* 128 38 



-~ = o = 30, quarts of brandy. 



It is plain that the former is the easier mode of solu- 

 tion ; and it will usually be found that both thought 

 and work are diminished when as many unknown 

 symbols are employed in the solution of a question as 

 there are distinct conditions embodied in that question. 

 It is a mistake to suppose that it is always easier to 

 solve a question with only one unknown quantity than 

 with two, when two are implied in the conditions ; the 

 contrary is more frequently true. Some of the present 

 examples will sufficiently snow this. 



0. A person has spirits at 12*. a gallon, and at 1 a 

 gallon ; how much of each must ho take to make a 

 gallon worth 14*.? Let the fractional part of a gallon 

 at 12*. be represented by x, and that of a gallon at 20*. 

 by y ; then 



and the price of the mixture = 12x 4- 20y 14, by the 

 question. 



Multiplying the first of these equations by 12, and 

 subtracting 



8y 2 . . y ; and z 1 y = j 



Consequently the mixture must consist of j gillon at 

 20*., and J gallon at 12*. 



Otherwise. Suppose ho take z gaL at 12*.; then ho 



Blunt take 1 z at 20*. ; the price of the former is 12x 



hilling*, and that of the latter 20 (1 x) shillings ; 



therefore the price of the whole gallon of the mixture is 



12x -f- 20 20z = 14, by the question. 



O -, 



Transposing, 6 - Sr, .-. ^=z ; .-. y - 1 z = j 

 o that there mast be J gaL at 12*., and } gal. at 20*. 



7. A merchant lias spirits at a shillings a gallon, and 

 at 6 shillings a gallon ; how much of each must ho take 

 to make a mixture of d gallons worth c shillings a gallon I 

 Suppose ho take z gallons at a shillings, and y at b shil- 

 lings ; then the wortli of the mixture of z 4- y gallon 

 that is, of d gallons is 



oz 4- fry = cd, by the question ; 

 also z 4- y = d. 

 Mult, by a, ax + ay = ad 



Subtract, (a % - (o e)d, .: y=^ ~ ^ '. 



In like manner, multiplying by 6, and subtracting, 



OUierwlse. Suppose he take x gallons at o shillings, 

 then, by the question, ho must take d x gallons at 6 

 shillings : the price of the mixture is, 



ax + b(d x) cd, by the question ; that is, 

 (a 6)x = (c V)d, .-. z = g _ fr , gals, at o shillings, 



and d - x = d - fc^J* 1 - ( ^g, gals, at 6 shillings. 



8. In a composition of a certain quantity of gun- 

 powder, two-thirds of the whole 4- lOlbs. was nitre ; 

 one-sixth of the whole 4J Ibs. was sulphur ; and the 

 charcoal was one-seventh of the nitre, all but 2 Ibs. 

 How many Ibs. of gunpowder were there ? 



Ibs. 

 Suppose there were x of gunpowder ; 



2x 

 then there were -- 4- 10 of nitre ; 



3 



.. 4J of sulphur; 



+ 10) 2 of charcoal ; 

 consequently, adding the ingredients, we have 



or, multiplying by 6, 



39 = 5z. 



Multiplying by 7, 



32x4- 480 4- 273 = 0.-..-, 

 .'. 207 = 3*, .-. x = 69. 

 Consequently there were 69 Ibs. of gunpowder. 



9. A farmer wishes to mix 28 bushels of barley, at 

 2s. 4<l. a bushel, with rye at 3*. a bushel, and wheat at 

 4s. a bushel, so that the whole may make 100 bushels 

 worth 3*. 4d. a bushel : how much rye and wheat must 

 he use ? 



Suppose x bushels of rye, and y of wheat ; then x + 

 y 4- 28 must moke 100. 



Now the price of x bushels of rye is 36x ponco 

 V wheat is 48i/ ,, 



28 ,, barley is 28 X 28 pence, 



. . the price of the x -f y 4- 28 bushels is SO* -f 48y 

 4- 28 X 28 pence ; which, by the question, is equal 

 to 100 X 40 : hence wo have 



x 4- y 4- 28 = 100 .... (1) 

 36z 4- 48y 4- 28 x 28 = 100 x 40 ; 

 or, dividing the latter equation by 4, 



9x 4- 12y 4- 28 X 7 = 100 X 10 

 that is, 9x 4- 12y 4- 196 = 1000 .... (2). 



Multiplying (1) by 9, 9x 4- 9y 4- 252 =900 



Subtracting, 



3y 66 -100 

 = 52 





