11. * Given x-+y+z=13, and x+y+u=17, 2+z+u=18, and ytutu=21, to find x, y, z, and u. Then, assume S=x+y+z+u, and the above equations will be transformed into the following ones: S-u: 13, S-2 - 17, S-y=18, and S-x=21. Add all these equations together, and we have 4S-1-y2-u=69, that is, 45-(x+y +z+u)=69. But x+y+z+u=S, therefore 4S-S69, or 3S=69, and S=23, and, consequently, by substituting for S its value in the four transformed equations, we shall have 23-U=13, 23—2=17,23—y=18, and 23—1=21. And, consequently, x=2, y=5, z=6, and u=10. MISCELLANEOUS QUESTIONS, PRODUCING SIMPLE EQUATIONS. The usual method of resolving algebraic questions, is first to denote the quantities that are to be found by : and y, or some of the other final letters of the alphabet ; then, having properly examined the state of the question, perform with these letters, and the known quantities, by means of the common signs, the same operations and reasonings that it would be necessary to make if the quantities were known, and it was required to verify them, and the conclusion will give the result sought. Or, it is generally best, when it can be done, to denote only one of the unknown quantities by x, y, or z, and then to determine the expression for the others from the nature of the question; after which the same method of reasoning may be followed, as above. And, in some cases, other methods of proceeding may be used ; which practice and observation alone can suggest. 1. What number is that whose third part exceeds its fourth part by 16 ? Let x= the number required. x=48, or 4x43x=192. x * This can be resolved by proceeding after the same manner as equations involving three unknown quantities; but the resolution of it may be greatly facilitated, by introducing into the calculation, besides the principal unknown quantities, a new unknown quantity arbitrarily assumed ; such as, for example, the sum of all the rest : and, when a little practised in such calculations, they become easy Let x = 2. It is required to find two numbers such that their sum shall be 40, and their difference 16. Let x denote the least of the two numbers required, Then will x+16= the greater number, And x++16=40, by the question. That is, 2x=40—16, or x==12= least number. 3. Divide $1000 between A, B, and C, so that A shall have $72 more than B, and C $100 more than A. B's share of the given sum ; then will x+72= A's share, and x+172 = C's share. Hence their sum is x+x+72 , +*+172, or 3x+244=1000, by the question; that is, 3=1000 -2445756 : or, x=26=$252= B's share. Hence x+ 72=$324= A's share. And x+172=$4243 C's share. Also, as above, $252= B's share. Sum of all $1000, the proof. 4. It is required to divide $1000 between two persons, so that their shares of it shall be in the proportion of 7 to 9. Let x = person's share. That is, 9x=(1000—2)x1=7000-7x, Or 9x +7=7000, or x = zile $4371 ist share, and 1000-2 $4371 = $5621 = 2d share. 5. The paving of a square court with stones, at 2s. a yard, will cost as much as the enclosing it with pallisades at 5s. a yard; required the side of the square. Let x = length of the side of the square sought; Then 4x = number of yards of enclosure, And 22 number of yards of pavement. Hence 4xX5= 20x = price of enclosing it, , And w X2 2x = the price of the paving. Therefore 2x=20x, by the question, Or 2x=20, and x= 10, the length of the side required. 6. Out of a cask of wine, which had leaked away a third part, 21 gallons were afterwards drawn, and the cask being then gauged, appeared to be half full; how much did it hold? Let x = the number of gallons the cask is supposed to have held. Then would it have leaked away fx gallons; whence there had been taken out of it, altogether, 21+ x gallons, and therefore 21+ha=4x, by the question ; that is, 63+=iz, or 126 +22=3x : consequently, 3x-2x==126, or x=126, the number of 2x, = gallons required. =1000 = 2 X =}, and yti=, by the question. Therefore 3 {21}+3=y, or 3y+3+13=Ay; And } a 7. What fraction is that, to the numerator of which if 1 be added, its value will be ļ, but if one be added to the denominator, its value will be { ? Let the fraction required be represented y Then 2+1 = y y+1 Hence 3x+3=y, and 4x=y+1, or = 4 4 That is, y=15, and x= y+1_15+1 =1604; 4 Whence the fraction that was to be found is is. 8. A market woman bought in a certain number of eggs at 2 a penny, and as many others at 3 a penny, and having sold them out again, altogether, at the rate of 5 for 2d., found she had lost 4d.; how many eggs had she? Let x = the number of eggs of each sort; Then will x = the price of the first sort, = the price of the second sort. But 5: 2:: 2x (the whole number of eggs) : 82; Whence x = the price of both sorts, when mixed together, at the rate of 5 for 2d. And consequently 4x+12—x=4, by the question ; That is, 15ż+10-——24.x 120, or <= 120, the number of eggs of each sort, as required. 9. If A can perform a piece of work in 10 days, and B in 13, in what time will they finish it, if they are both set about it together ? Let the time sought be denoted by x. Then = the part done by A in one day, 10 And in = the part done by B in one day; 13 Consequently =1, (the whole work ;) 13 That is, 13x+10x 130, or 23x = 130 : Whence x = = 523 days, the time required. 10. If one agent, A, alone, can produce an effect, e, in the time e, a, and another agent, B, alone in the time b; in what time will both of them together produce the same effect ? io ti ex ex ac a Let x = Let the time sought be denoted by I. ex Then a: e:: 2 : = part of the effect produced by A, a And b:e:: 2 : =part of the effect produced by B. b ex Hence + =e, (the whole effect,) by the question. a 6 2 Or, -+ = 1, by dividing each side by e. Therefore at a 6 6 ab = 2, or bxtax=ab; consequently, x= =time required. ato 11. How much rye, at 4s. 6d. a bushel, must be mixed with 50 bushels of wheat, at 6s. a bushel, so that the mixture may be worth 58. a bushel ? the number of bushels required. 2 = 12. Á laborer engaged to serve for 40 days, on condition that for every day he worked he should receive 20d., but for every day he was absent he should forfeit 8d. ; now, at the end of the time, he had to receive 11. 11s. 8d. ; how many days did he work, and how many was he idle ? Let the number of days that he worked be denoted by x; then will 40—« be the number of days he was idle; also 20x the sum earned, and (40—2) X 8, or 32048x, the sum forfeited; whence 20x—(32048x)=380d. (=11. 11s. 8d.), by the question : that is, 20x—320+8x = 380, or 28x=380+320=700 ; consequently, x=200525, the number of days he worked, and 40—=40_25 =15, the number of days he was idle. 13. It is required to divide a line, of 15 inches in length, into two such parts, that one may be three-fourths of the other. Let one of the parts Then the other will be And by the question, 15m=x, or 60~4x=3x, or 7x=60; whence x=6=84, the one part; and 15—X=64, the other. 14. My purse and money together are worth 20s., and the money is worth 7 times as much as the purse. How much is there in it? Let the value of the purse be z; then the money 72; and by the question 7x+=20, or 8x=20, or x=0, 2s.6d., the value of the purse, and consequently 17s. 6d., the money contained 28 X. 15 — X, a 3x; 15. A shepherd being asked how many sheep he had in his flock, said, 'If I had as many more, half as many more, and 7 sheep and a half, I should have just 500. How many had he ? Let the number of sheep = x; then, by the question, x + + 12+7=500; or 24a=4921; mult. by 2, 5x=985; Whence x=995–197, the number sought. 16. A post is one-fourth of its length in the mud, one-third in the water, and 10 feet above the water. What is its whole length ? Let the length of the post = x; then by the question 4x + 42 fx +10=r; mult. by 12, 3x+4x+120=12x ; Transposing, 5.x=120, or x=ize=24 feet, the answer required. = 17. After paying away 4 of my money, and then f of the remainder, I had 72 guineas left. What had I at first? Let the number of guineas = x; then x-**- =72; or mult. by 20, 20x52—3x=1440 ; whence 12x 1440, or x= 1440=120 guineas. 18. It is required to divide $300 between A, B and C, so that A may have twice as much as B, and C as much as A and B together. Let B's share = X; then, by the question, A's share = 2x, C's share Consequently, 2+2x+3x==300, or 6x=300; whence x= = 280 =$50, B's share ; 2x=$100, A's share, and 3x=$150, C's share. 19. A person, at the time he was married, was 3 times as old as his wise ; but, after they had lived together 15 years, he was only twice as old. What were their What were their ages on the wedding day? Let the age of the wife at the time of the marriage =X, and that of the husband 3x ; Then after 15 years their ages will be x+15, and 3x+15; and by the question, 3x+15=2(x+15), or 3x+15=2x+30; = whence, by transposing, x=15, the age of the wife; and 32—45, the age of the husband. 20. What number is that from which if 5 be subtracted, twothirds of the remainder will be 40 ? Let the number ght =X; then by the question 2–5 will 2(x-5) be the remainder, and -40; whence 2x—10=120, or 3 2x=130 ; and consequently ize=65, the number sought. 21. At a certain election, 1296 persons voted, and the successful candidate had a majority of 120. How many voted for each? Let the less number of voters equal x; then the greater number =x+120, and by the question. a |