When there are more equations and unknown quantities than one, a single equation, involving only one of the unknown quantities, may sometimes be obtained by the rules before laid down for the solution of simple equations; and, in this case, one of the unknown quantities being determined, the others may be found by substituting its value in the remaining equations. 1. Given x2 + y2 65 EXAMPLES. to find the values of x and y. 28 Here, from the second equation, we have y = ; and 784 by substituting this in the first, a2+ 784. 65, or x1 65x2 Whence, by the common rule before given, we have (1225-784 Or, by reducing the parts under the last radical, and ex Or the solution, in cases of this kind, may often be more readily obtained by some of the artifices frequently made use of upon these occasions; which can only be learned from experience thus, taking, as before, (1.) x2 + y2 -- 65, (2.) xy 28, we shall have, as in the former method, by multiplying by 2, 2xy = 56, and, by adding this equation to the first, and subtracting it from the same, x2 + 2xy + y2. 121, and x2 2xy + y2 9. Whence, by extracting the square roots of each of these last equations, there will arise x+y=±11, and xy=3, and consequently, by adding and subtracting these, we shall have 2x: 14, or x=7, or - 7, and y = 4, or - 4. It will also sometimes facilitate the operations, by substituting for one of the unknown quantities the product of the other, and a third unknown quantity; which method may be applied with advantage whenever the sum of the dimensions of the unknown quantities is the same in every term of the equation. 2. Given{23+xy = 56 xy+2y=60} to find the values of x and y. Here, agreeably to the above observation, let xvy, then v2y2+vy2=56, and vy2+2y260, whence, from the of these two expressions, we shall have 60v2 + 60v = 56v + 112. And, by transposing 56v, and may also be sometimes shortened, by substituting for the unknown quantities, the sum and difference of two other quantities; which method may be used when the unknown quanticies, in each equation, are similarly involved. Here, according to the above observation, let there be assumed x zv, and y = z — v. Then, by adding these two equations together, we shall have x+y 2z 12, or ≈≈ 6, also, since x=6+v, y 6 - v, and by the first equation a3+y3 18xy, we shall obtain, by substitution, (6+v)3 +(6v)3 18 (6 + v) (6 — v), or, by involving the two parts of the first member, and multiplying those of the second, 43236v 6481802, whence, by transposition, 54v2 = 216; and by division, v2 —: 216 54 4; or v = ± 2. And therefore, by the first assumption, and the first part of the process, we have x z +v=6±2=8, or 4, and y = z อ 6+2 4, or 8. QUESTIONS PRODUCING QUADRATIC The methods of expressing the conditions of questions of this kind, and the consequent reduction of them, till they are brought to a quadratic equation, involving only one unknown quantity and its square, are the same as those already given for simple equations. 1. To find two numbers such that their difference shall be 8, and their product 240. Let x equal the least number. Then will x + 8 = And x(x+8) Whence x = x2 + 8x = the greater, 240, by the question, 4+√256 by the 4 + √(16 +240) Therefore x = 16 4 12, the less number, 2. It is required to divide the number 60 into two such parts, that their product shall be 864. x Let x the greater part, Then will 60 х the less, And ∞ (60-x) = 60x —x2 864, by the question, Or, by changing the signs on both sides of the equation, And consequently, x = 30 + 6 = 36, or 30 — 6 = 24, the two parts sought. 3. It is required to find two numbers such, that their sum shall be 10 (a), and the sum of their squares 58 (b). Let the greater of the two numbers, Then will a =2x2 the less, 2ax+a2=b, by the question, b — a2, by transposition, by division. And x2 + (α — x)3 a a2 And x2 2 by the rule, And if 10 be put for a, and 58 for b, we shall have 10 1 + ✓ (116100)7, the greater number, 4 Having sold a piece of cloth for 347., I gained as much what was the price of the cloth? pounds the cloth cost, per cent. as it cost me; Let x= Whence x Then will 24 But 100: :: x: 24 x= the whole gain, x, by the question, Or, x2 = 100 (24 50 + √ (2500 + 2400) by the rule, 100x, 507020, And consequently, 201. price of the cloth. 5. A person bought a number of sheep for 801., and if he had bought four more for the same money, he would have paid 11. less for each; how many did he buy? Let x represent the number of sheep, 80 Then will be the price of each, And 80x + 320 80x + x2 + 4x, by the same, Or, by leaving out 80x on each side, x2+4x 320, Whence x= 2 + √(4 +320) = 2 + 18, by the rule, And consequently, x 16, the number of sheep. 6. It is required to find two numbers, such that their sum, product, and difference of their squares, shall be all equal to each other. y2 Hence 1 x + y x2 - y2 y -the less. by the question. =x-y, or x = y + 1, by 2d equation. And (y + 1) + y = y (y + 1), by 1st equation, Where ... denotes that the decimal does not end. + √5=2.6180... 2 2 7. It is required to find four numbers in arithmetical progression, such that the product of the two extremes shall be 45, and the product of the means 77. Let x= least extreme, and y: =common difference, Then x, x + y, x+2y, and x + 3y, will be the four numbers, § x (x + 3y) = x2+3xy = 45 Hence {x+y) { (x + y) (x+2y) = x2 + 3xy+2y=77}, by the question, And 2y2 = 77-4532, by subtraction, Whence the numbers are 3, 7, 11, and 15. 8. It is required to find three numbers in geometrical progression, such that their sum shall be 14, and the sum of their squares 84. Let x, y, and z, be the three numbers, Then, xz = y2, by the nature of proportion, And { ++ y2+22 x y + z 14 by the question, 84 Hence, x+= 14-y, by the second equation, z And x2+2zx + z2 196 -- 28y + y2, by squaring both sides, That is, x2+ y2+22=196-28y, by subtraction, Hence y = 4, by transposition and division. 16. Again, xz y2 = 16, or x = by the 1st equation, Whence z=5± √ (25 — 16) — 5±3 = 8, or 2 by the rule, Therefore, the three numbers are 2, 4, and 8. 9. It is required to find two numbers, such that their sum shall be 13 (a), and the sum of their fourth powers 4721 (b). the difference of the two numbers sought, Let x= 1 1 Then will-a + = x, or willa+, a + x the greater number, |