1. By making the two first terms equal : Mult. the 1st equ. by 2, gives 10.7 by = 18; 9 + 3y Mult. the 1st equat. by 5, gives 25.x - 15y = 45; 31.1. = 93; 3. 5.7 9 Hence, from the 1st equ. y = 2. 5 MISCELLANEOUS EXAMPLES. * +8 1. Given = 5.1 23 4 3 find .x and y. Ans. x = 4, and y = 3. 3х ^x 2. Given + 10 = 13, and = 4 2 to find x and yo Ans. * = 5, and y = 3. 6.7 3. Given + = 10, and 5 3 6 to find x and y. Ans. * = 8, and y 4. Given 3x + 4y = 38, and 4.x - 3y = 9; to find x and y. Ans. x = 6, and 5. 2y + b=14 14; PROBLEM II. To Exterminate Three or More Unknown Quantities; Or, to Reduce the Simple Equations, containing them, to a Single one. RULE. This may be done by any of the three methods in the last problem: viz. 1. AFTER the manner of the first rule in the last problem, find the value of one of the unknown letters in each of the given equations : next put two of these values equal to each other, and then one of these and a third value equal, and so on for all the values of it ; which gives a new set of equations, with which the same process is to be repeated, and so on till there is only one equation, to be reduced by the rules for a single equation. 2. Or, as in the 2d rule of the same problem, find the value of one of the unknown quantities in one of the equations only; then substitute this value instead of it in the other equations; which gives a new set of equations to be resolved as before, by repeating the operation. 3. Or, as in the 3d rule, reduce the equations, by multiplying or dividing them, so as to make some of the terms to agree : then, by adding or subtracting them, as the signs may require, one of the letters may be exterminated, &c, as before. EXAMPLES 9 2y 3%, 16 - 2y 4%; x + y + z= 1. Given * + 2y + 32 =16 ; to find x, y, and e. * + 3y + 4z = 21 ) 1. By the 1st method : Transp. the terms containing y and zin each equa. gives 9 y 3y - 42; Then putting the 1st and 2d values equal, and the 2d and 3d values equal, give 9 % = 16 2y 3z, 32 21 3y Trans. 5 and 2z, gives z = 2. From the 1st equa. x = 9-y-z; 9 + y + 2z = 16, 9 + 2y + 32 = 21; 2 = 2. 3dly. By 2z; 38 3dly. By the 3d method : subtracting the 1st equ. from the 2d, and the 2d from the 3d, gives y + 2% = 7, y + z = 5; 2 + y + z = 18 2. Given x + 3y + 2z s to find x, y, and z. x + ty + z = 10 Ans. x = 4, y 6,= 8. x + y + z 3. Given x + 3y + z = 20°; to find x, y, and z. x + 4y + šz Ans. x = 1, y = 20, z = 60. 4. Given x - y = 2, x - x 3, and y - x =l; to find x, y, and z. Ans. r = 7; y= 5; z = 4. $2,5 2.x + 3y + 4z = 34 5. Given 3.x + 4y + 5z = 46 ; to find x, y, and z. 4x + 5y + 6z = 58 = 272 = 16 A COLLECTION OF QUESTIONS PRODUCING SIMPLE EQUATIONS. X Quest. 1. To find two numbers, such, that their sum shall be 10, and their difference 6. Let x denote the greater number, and y the less *. Then, by the 1st condition x + y = 10, y = 6, X = 10 6 + y; y = 2. X = 6 +y = 8. Y, * In all these solutions, as many unknown letters are always used as there are unknown numbers to be found, purposely the better to exercise the modes of reducing the equations : avoiding the short ways of notation, which, though giving a shorter solution, are for that reason less useful to the pupil, as affording less exercise in practising the several rules in reducing equations. QUEST. 2 QUEST. 2. Divide 100). among A, B, C, so that A may have 201. more than B, and B 101. niore than c. Let x = A's share, y = B's, and z = c's. 2 Iy + 20, y=% + 10. In the 1st substit. y + 20 for x, gives 2y + x + 20 = 100; In this substituting % + 10 for y, gives 3z + 40 = 100; By transposing 40, gives 37 = 60; And dividing by 3, gives z = 20. Hence y = x + 10 = 30, and x = y + 20 = 50. QUEST. 3. A prize of 5001. is to be divided between two persons, so as their shares may be in proportion as 7 to 8; required the share of each. Put x and y for the two shares; then by the question, 7:8::8:y, or mult. the extremes and the means, 7y and * 4- 5 = 500; And hence x = 500- y = 233;. QUEST. 4. What number is that whose 4th part exceeds its 5th part by 10? Let x denote the number sought. By mult. by 5, it gives x = 200, the number sought. QUEST. 5. What fraction is that, to the numerator of which if i be added, the value will be ; but if 1 be added to the denominator, its value will be ? 8x, Let 3 denote the fraction. y X + 1 Then by the quest. =, and y y + 1 The 1st mult. by 2 and y, gives 2x + 2 = y; The 2d mult. by 3 and y + 1, is 3x = y +l; The upper taken from the under leaves x - 2 =l; By transpos. 2, it gives .r = 3. And hence y = 2.x + 2 = 8; and the fraction is QUEST. 6. Quest. 6. A labourer engaged to serve for 30 days on these conditions: that for every day he worked, he was to receive 20d. but for every day he played, or was absent, he was to forfeit 10d. Now at the end of the time he had to receive just 20 shillings, or 240 pence. It is required to find how many days he worked, and how many he was idle ? Let x be the days worked, and y the days idled. and 20.x Toy = 240; 30.r = 540; This div. by 30, gives = 18, the days worked; Hence y = 30– X = 12, the days idled. Quest. 7. Out of a cask of wine, which had leaked away, 30 gallons were drawn; and then, being gaged, it appeared to be half full; how much did it hold? Let it be supposed to have held x gallons, And transposing *, gives x = 120 the contents. Quest. 8. To divide 20 into two such parts, that 3 times the one part added to 5 times the other may make 76. Let x and y denote the two parts. x + y = 20, and 3x + 5y = 76. Mult, the 1st by 3, gives 3x + 3y = 60; Subtr. the latter from the former, gives 24 = 16; And dividing by 2, gives 8. Hence, from the 1st, x = 20 - y = 12., Y = QUEST. 9. A market woman bought in a certain number of eggs at 2 a penny, and as many more at 3 a penny, and sold them all out again at the rate of 5 for two-pence, and by so doing, contrary to expectation, found she lost 3d.; what number of eggs had she? Let x = number of eggs of each sort. But |