ܪ ber of independent simple equations ; but, in cases of this kind, there are frequently shorter and more commodious methods of operation, which can only be learnt from practice. EXAMPLES. x+y +z =29) 1x+y+z=10 x 2 1 z Whence 29-4-2=62-2y—3z, 2 1 And, also, 29-y-z=20 – 2y-*. 34m From the first of which y=33-22, 3 3 . x=29-y-2=8. 6x+7y-2 =63 Here, multiplying the first equation by 6, the second by 3, and the third by 2, we shall have 12x +244 - 18z=132, 12x+147- 2z=126. And, subtracting the second of these equations succes sively from the first and third, there will arise 30y-33z=78, K and 2. Or, by dividing the first of these two equations by 3, and then multiplying the result by 2, 20y - 22z=52, 20y-172=72. Whence, by subtracting the former of these from the latter, we have 52=20, or z=4. And, consequently, by substitution and reduction, y=7 and x=3. 3. Given x+y+z=53, x+2y+32=105, and x+3y +4z=134, to find the values of x, y, Ans. x=24, y=6, and z=23 1 1 1 1 1 + x 4 1 1 + 5y + 2=12, to find the values of X, Y, 6 Aps. x=12, y=20, z=30. 5. Given 7x + 5y + 2z = 79, 8x+7y + 9z 122, and x+4y+52=55, to find the values of x, y, and z. Ans. x=4, y=9, z=3 6. Given x+y=a, x+z=b, and y+z=c, to find the values of x, y, and z. 4. Given x+y+z=32, 3x + x + === z=15, and co and z. MISCELLANEOUS QUESTIONS, PRODUCING SIMPLE EQUATIONS. The usual method of resolving algebraical questions, is first to denote the quantities, that are to be found, by x, 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 or y, and then 32, 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, the substituting for the sums and differences of quantities ; or availing ourselves of any other mode, that a proper consideration of the question may suggest, will greatly facilitate the solution. 1. What number is that whose third part exceeds its fourth part by 16 ? Let x= the number required, 1 1 1 Then its part will be and its - part 2x. 4 4 4 3 That is x X548, or 4x - 3x=192, 4 = 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= to the greater number, 24 That is 2x=40-16, or 1> =12= least number. 2 And x+16=12+16=28= the greater number required. 3. Divide 10001. between A, B, and c, so that a shall have 721. more than b, and c 1001. more than A. Let x=b's share of the given sum, And x +172= c's share. 756 3 317305 Hence x+72=3241.=a.'s share, 5. Sum of all =10001. the proof. Let x= the first person's share, 7000 16 and 1000x=1000-4371. 10s. = 5621. 105. = 2d share. i ne paving or a square court with stoñės; at 2s. a Then 4x= number of yards of enclosure, Therefore 2x2=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 aivay a third part, 21 gallons were afterwards drawn, and the cask being then guaged, appeared to be half full ; how much did it hold ? Let x = the number of gallons the cask is supposed to have held, 1 Whence there had been taken out of it, altogether, 1 x 3 1 1 And therefore 21 to x=-*, by the question, 21+ gallons, 3 That is 63+x=*, or 126+21=3%, Consequently 3x - 2x=126, or r=126, the number of gallons required. 7. What fraction is that, to the numerator of which if 1 1 be added, its value will be but if 1 be added to the 3 1 denominator, its value will be 4 Let the fraction required be represented by y' 3+1 1 1 Then and 3 yti - I Hence 3x+3=y, and 4x=y+1, or x= by the question. 15+1 16 Therefore 3(+1) +3=y, or 33+3+12=9y, y #1 = Whence the fraction that was to be found, is 16 That is y=15, and x= 4 4 15 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 ort, |