Thus, if ax = ab 4a; then dividing by a, gives a = b 4. In like manner, if ax + 3ab = 4c2; then by dividing by a, it is a + 3b = III. When the unknown' term is divided by any quantity, we must then multiply all the terms of the equation by that divisor; which takes it away. [Note. If there be several terms in which fractions appear, it is often best to multiply the numerator of every term of the equation by the least common multiple of all the denominators, and divide by the corresponding denominator. All the terms are thus cleared of fractions at once, whether known or unknown.] Thus, if = 3 + 2: then mult. by 4, gives x = 12 + 8 = 20. And if a = 3b + 2c — d: then mult. by a, it gives x = 3ab + 2ac ad. IV. When the unknown quantity is included under any root or surd expression transpose the rest of the terms, if there be any, by rule 1; then raise each side to such a power as is denoted by the index of the surd; viz. square each side when it is the square root; cube each side when it is the cube root; and so on, which removes that radical from the equation. Thus, if √x 34: then transposing 3, gives x = 7; and squaring both sides gives x = 49. Also, if 3/3x + 4 + 3 = 6 : then by transposing 3, it is 3/3x + 4 = 3 and by cubing, it is 3x + 4 = 27; and by rules I. II. x = 73. V. When that side or member of the equation which contains the unknown quantity is a complete power, or can easily be reduced to one, by rule I. II. or 3; then extract the root of the said power on both sides of the equation; that is, extract the square root when it is a square power, or the cube root when it is a cube, and so on. Thus, if x2+8x + 16 = 36, or (x + 4)2 = 36; then by extracting the root, it is x + 4 = 6; whence = 2. = 30; and multiply Also, if 2x2 624 then transposing 6, gives x2 ing by 4, gives 3x2 = 120; then dividing by 3, gives x2 extracting the root, gives a = √ 40 6·324555. VI. When there is any analogy or proportion, it is to be changed into an equation, by multiplying the two extreme terms together, and the two means together, and making the one product equal to the other. Thus, if 2x : 9 :: 3 : 5, gives 10x = 27; and by rule II. x = 20. And if xa: 56: 2c; then ca = = 5ab: hence by rules II. III. = 10ab 3c VII. When the same quantity is found on both sides of an equation, with the same sign, either plus or minus, it may be cancelled or left out of both; and when every term in an equation is either multiplied or divided by the same quantity, that quantity may be struck out of them all. The following example furnishes specimens of all the rules just laid down : 5a2 Let 2x + 2 √ a2 + x2 = be given; to find x. √a2 + x2 Multiplying by √ a2 + x2, gives 2x √ a2 + x2 + 2a2 + 2x2 = 5a2; transposing 2a2 + 2x2, gives 2x √a2 + x2 = 3a2 2x2; squaring both sides, 4x2 x (a2 + x2) = (3a2 - 2x2); that is, 4a2x2 + 4x4 = 9a1- 12a2x2 + 4x4; can celling 4x4 from both sides, we have 4a2x2 = 9a1 — 12a2x2; transposing 12a2x2, gives 16a2x2 = 9a1; dividing by a2 gives 16x2 = 9a2; dividing by 16, gives x2 = a2; and lastly, extracting the square root, gives x = a. 13. Given a + x = √ a2 + x √4b2 + x2; to find x. ; to find x. √2a + x Ans. x 2a2 2a. 4a4 Ans. xa. b α =2b; to find x2. Ans. x2 = 4b b2-a2 Ans. x = a Ans. 2125. a2c b b' ... 29. If the recurring decimal 082082 be multiplied by x2, and the square of the result divided by 3, gives the same value, '082 ...; what is the value SIMULTANEOUS OR COEXISTING EQUATIONS. WHEN an equation contains two or more unknown quantities, it is obviously insufficient for the determination of the value of any one of them. The methods hitherto laid down enable us to obtain the value of any one symbol which is involved in the equation in terms of the remaining ones, whether they be numeral or literal, known or unknown; but if there be more than one unknown quantity, the expression of the value of any one of them that we may select will involve the remaining unknowns, and such value will therefore be indeterminable till such other conditions are added as shall enable us to assign the values of all these last-named unknowns. If now a second equation, different in its composition from the former, but involving the same unknowns, be given; then also the value of the selected one can be obtained as in the preceding case; and if this second equation also express a second condition to which the relation of the unknowns is subjected, and which, therefore, must exist simultaneously with the former, the two values of the unknown must be identical. There may hence be formed a third equation, which will also be true simultaneously with the two former. This equation will involve one unknown less than either of the others. If there remain more than one unknown in this equation, it is still incapable of furnishing the value of either of them, as before; and there must hence be still other relations given to render the problem determinate. Suppose then a third equation to express a third condition, which is simultaneous with the other two. Then from this also we can obtain a value of the same unknown that we at first selected, in terms of the remaining ones, and this value equated to either of the other values, will furnish a second equation, containing only the remaining unknowns. Having thus two equations containing the remaining unknowns, we can find two values of a second unknown, and equate them; which will give us one equation which is freed from both the forementioned unknowns. If this yet contain more than one unknown, we shall still want other conditions, and must proceed in the same manner to eliminate them one by one from each pair of equations that is either given, or which results from the previous elimina tions; till, at last, we arrive at an equation which contains only one unknown, the value of which must be determined as has been already explained and practised. It will be quite apparent from this reasoning, that there must be as many equations simultaneously given as there are unknown quantities involved in all of them together; and that though some of the equations should not contain all the unknowns, yet they may be conceived to do so by considering 0 as the coefficient of any one that is absent. Other processes besides that explained above can sometimes be used more advantageously; and as facility in the practice of elimination is best attained by exercise upon the simple cases, the following rules have been adapted to the case of two unknowns. The extension of the same kind of processes to three or more simultaneous equations will then become easy and obvious. It is, however, to be understood, that any involutions, transpositions, multiplications, or divisions by which the equations can be reduced to simpler forms than the given ones, must be executed previous to the application of any of the special rules here laid down. TWO SIMULTANEOUS EQUATIONS. To exterminate or eliminate one of the two unknown quantities from two simultaneous equations; that is, to reduce the two simple equations containing them, to a single one. I. FIND the value of one of the unknown letters, in terms of the other quantities, in each of the equations, by the methods already explained. Then put those two values equal to each other for the new equation, involving only one unknown. The value of this is to be found as before. It is evidently best to begin by determining the values of that letter which is easiest to be found from the two proposed equations. 2. Given x + 2y = a, and 1⁄2 x-2y=b; to find x and y. Ans. xab, and ya-4b. 9. Given ax + b,y=c,2, and ax + b„y=c„‚2; to find x and y. II. Find the value of one of the unknown letters, in one of the equations, as in the former rule, and substitute this value instead of that unknown quantity in the other equation: then there will arise a new equation, with only one unknown quantity, whose value is to be found as before. It is evidently best to begin with that letter whose value is most easily found in either of the given equations. Here x = Or, finding x from the second equation. 14 +2y, which, substituted in (1) gives y = 3; and hence x = 4. 28+ 4y + 3y = 17, or 5 In a similar way we may begin to operate by finding y from either of the equations and substituting its value in the other. 2. In 2x + 3y = 29, and 3x - 2y = 11, we have x = 3. In + y = 14, and x 4. In xy:: 3 : 2, and 5. In +3y=21, and = 7, and y = 5. y = 2, we have x = 8, and y = 6. y2 = 20, we have a = 6, and +3x=29; x = 9, and y = 6. 6. Given 10 = 4, and "=" + 2 − 2 = and y. + y=4. y3 = 37; to find the product of a2 and y3, Ans. x2 y3 = 432, x 8. From x + y = a(x − y), and x2 + y2 = b2; find æ2 and y2. y=1. |