(and taking the positive sign in the root of the right mem Hence, by taking the half sum and half difference of the 5. Given, a axb, to find a. + = Dividing by x, and transposing, we readily get x and = x -a; hence (as in the last question), we have + 018118218 are the roots of the given equation. CASE VI. 1. If the unknown letter enters one or more of four arithmetical proportionals, we may put the sum of the first and fourth (or the extremes) equal to that of the second and third (or the means); and if the unknown letter enters one or more of three arithmetical proportionals, we may put the sum of the first and third terms (or the extremes) equal to twice the second (or the mean). 2. If the unknown letter enters one or more of four geometrical proportionals, put the product of the extremes equal to that of the means, and the proportion will be reduced to an equation; similarly, if the unknown letter enters one or more of three geometrical proportionals, we reduce the proportion to an equation, by making the product of the extremes equal to the square of the mean. 3. If the unknown letter enters one or more of three expressions in harmonical proportion, then since the first must be to the third as the difference between the first and second is to the difference between the second and third, we easily (by 2) reduce the harmonicals to an equation. Similarly, if the unknown letter enters one or more of four harmonicals, because the first must be to the fourth as the difference between the first and second is to the difference between the third and fourth, we easily (by 2) reduce the harmonicals to an equation. EXAMPLES. 1. Given, 3, 8, x, and 2x + 1, in arithmetical proportion, to find x. Since the sum of the extremes is 2x + 4, and that of the means x + 8, we get the equation 2x + 4 = x + 8; consequently, a 4, and the arithmeticals become 3, 8, 4, and 9. = 2. Given, 5, 7 + x, and 3, in arithmetical proportion or progression, to find x. Because the sum of the extremes is 5 + 3x, and that twice the mean is 14 + 2x, we have the equation 5 + 3x = 14 + 2a; consequently, x = 9, and the progressionals are 5, 16, and 27. 3. Given, the geometrical proportion 3: 4 :: +1:x+6, to find x. Because the product of the extremes is 3x + 18, and that of the means is 4x + 4, we get the equation 3x + 18 = 4x + 4; consequently, x = 14, and the proportion becomes 3:4 :: 15:20. x. 4. Given, the geometrical proportion 4: :: : 9, to find Ans. x6. 5. Given, 4, 3, and x, in harmonical proportion or progres sion, to find x. From the proportion 4: :: 4-3: 3-x we get the equation x = 12 - 4x, or x = 2 . 4. 6. Given, 2, 3, x, and x + 4, in harmonical proportion, to find x. The proportion 2: +4 :: 3-2x+4x gives the equation x + 4 = 8, or = 4; consequently, the harmonicals are 2, 3, 4, and 8. From the preceding cases we deduce the following rule for the solution of any (simple) determinate equation or proportion, which contains only one unknown letter (which is usually called the unknown quantity). RULE. 1. When the unknown letter is contained in an equation. 1st. Free the equation from fractions by Case III., then unite like terms into one, by transposing according to Case I., when necessary. 2d. If the unknown letter enters the monomial terms of the equation with one or more negative indices; then change the signs of such indices, and write the letter with its index changed (in any term) for a divisor of the corresponding coefficient, and free the equation from the resulting fractional terms by Case III. 3d. Free the unknown letter from any surds which may affect it by Case IV., or transform the equation into a new equation containing a new unknown letter free from surds. 4th. Free the equation thus obtained from terms which contain the unknown letter with negative indices, as in 2d, and unite like terms into single terms as in 1st; noticing if the terms can be so arranged as to make the terms which contain the unknown letter a complete power, the root must be extracted as in Case V. Hence, if the terms are all transposed to the first member, the equation will be reduced to the form A+ A-1 + An-2 + + An-1x + An= 0; in which a is the unknown letter, n a positive whole number, A, A1, A2, etc., known coefficients, and A, the absolute term which is known; noticing that, by Case III., the preceding equation may be changed to another in which the coefficient of the highest power of the unknown letter equals 1; see the Remark. 2. When the unknown letter enters one or more of the terms of a proportion. Reduce the proportion to an equation, by Case VI., then reduce the equation to its proper form, according to the preceding directions. 3. If the final equation is a simple equation, it will clearly be of the form Ax + A1 = 0; which evidently results from the equation given above, by putting 1 for n. Transposing A, in the preceding equation, and dividing by A, the coefficient of x, we shall, by Case II., get x = A1 Hence, if an equation after the required reductions is a simple equation, transpose its known terms to the right member, and its unknown terms (or those which contain the unknown letter) to the left member; then the unknown letter equals the quotient arising from the division of the right member by the coefficient (or multiplier) of the unknown letter; observing the rule of signs in Division. A1 " A Wo 4. Remarks.-1st. Resuming the equation x = observe if A1 = 0, while A is finite, that we shall clearly have x = : 0. 1 is 0 or an in 2d. If A = 0, while A, is finite, it is clear that a must be A1 unlimitedly great; because the divisor in finitesimal. 3d. If A = 0 and A1 = 0, we have x = 0 ; which is of 0 an undetermined form. If A and A, are reduced to 0, by assigning a particular value to some letter which is common to them, it is clear that they must have a common divisor. Hence, to get the true value of x= we must reduce A A Α' the fraction to its lowest terms (by dividing its numerator and denominator by their greatest common divisor), before the particular value is assigned to the common letter. If, after freeing the terms of the fraction of their greatest common divisor, we still have x = it is clear that x is indeterminate or arbitrary; but if a is not of the preceding form, it must clearly be either 0, finite, or unlimitedly great. 5. To illustrate the preceding rule take the following EXAMPLES. 1. Given, 3x+7= 15+ x, to find x. x 2. Given, x+ + + 25, to find x. 2 3 4 3. Given, ax-b2cx+d, to find x. Ans. x = 4. Ans. x 12. 4. Given, ax + bx c + de-fx, to find x. Ans. x = a + b + f 8. Given, (3 + x) × (5 + x) − 28 = x2 + 27 and (a — 2y) × (b − y) = 2y2 - 7cd, to find x and y. = ab + 7cd a + 26 a2+363y, to find 9. Given, a + x = x and y. a2 + 3b2x and a + y = 2a, and y = 0 or — 363. a-y 11. Given, √3 + x + √3 −x = 3 and a2 + y2 — a2 = y, 3√3 to find x and y. Ans. x = ± and y = ± √ a2 + 1. , 2 |