Again, if still greater accuracy be required, let 4.264 and 4.265 be the assumed numbers, then x = 4.265 3 77.581310 as 99.972448 100.036535 results 100.036535 4.265 100 true result, 99.972448 4.264 99.972448 .064087 .001 :: .027552 .0004299; consequently, x=4.264+.0004299-4.2644299, very nearly the truth. EXAMPLES FOR PRACTICE. 1. Given x3-15x2+63x=50, to find the value of x. Ans. x=1.028039. 2. Given x3-12x=84, to find the value of x. Ans. x 5.2822. 3. Given x3-17x2+54x-350-0, to find the value of x. Ans. 14.954. 4. Given x-3x*—2x2=8, to find the value of x. Ans. x=2. 5. Given 2x1- 16x3 +40x2-30x+1=0, to find the value of x. Ans. 1.284724. 6. Given x+2x1+3x3—4x2+5x-1=54320, to find the value of x. Ans. x 8.4144. 7. Find two numbers, whose product multiplied by the greater will make 405, and their difference multiplied by the less will make 20. Ans. 9 and 5. INDETERMINATE ANALYSIS, OR UNLIMITED PROBLEMS. It has already been determined how the value of one unknown quantity may be found from one equation, and also how the values of two unknown quantities may be found from two equations, &c. having as many independent equations as there are unknown quantities to determine, which must be the case when the given equations or questions require some certain or definite answer. When a question does not furnish as many equations as there are unknown quantities to be determined, some of these must remain undetermined, and depend on the will; for which reason, such questions are said to be indeterminate, and frequently admit of many answers, forming the subject of a particular and useful branch of Algebra, called Indeterminate Analysis. CASE I. When the given equation contains two unknown quantities. RULE. 1. Find the value of one of the unknown quantities in terms of the rest, as in step first, in the first example. 2. Divide the numerator by the denominator, if divisible, as in step second, and put the remainder equal to a whole number, w, if the coefficient of the unknown quantity in this remainder be 1, as in the fourth step. 3. Clear this equation of the fraction, and find the value of the unknown quantity, in terms of the rest, as in the fifth step. 4. Then w may be taken any number that will answer the conditions of the question, as in step sixth. Or the value of this unknown quantity may be substituted in the first step of the equation, as in the seventh step, where w may be taken any number that will answer the conditions of the question, as in the eighth step. 5. But if the coefficient of the unknown quantity be more than one after the division, or that the numerator cannot be divided by the denominator, as mentioned in the second part of this rule (as in the fourth step of the second and third examples), take any multiple of this quantity that may be convenient, and add it to the unknown quantity, if the unknown quantity in the fraotional part be negative, but subtract it if the unknown quantity be positive, in order to exterminate the coefficient of the unknown quantity in the fractional part, then put this sum or remainder equal to w, and proceed by the third part of the rule. EXAMPLES. х 1. Given 4x-5y=10, to find the values of x and y in whole numbers. 1. By transposition, r= 10+5y which must be a 2. Dividing the numerator by the denominator, 10+5y 4 2+y =2+y+ which must also be a whole number. 3. But 2+y being a whole number, it may be rejected, and the remainder 2+ (the unknown quantity y having 1 for a coefficient,) must be put equal to a whole number w. 5. Clearing this equation of the fraction, &c. 6. Let w be taken any number that will answer the conditions of the question; .. if w=1, then y=2 and x=5, two numbers which will answer the conditions of the question. 7. Or by substituting the value of y in the equation of the first step of this solution, X= 10+20w-10 ; 8. Then, as above, if w=1, x=5, and y=2, two numbers that will answer the conditions of the question. N. B. Several sets of numbers may be found that will answer the conditions of the question, by assum. ing w any other value, as follows: If w=2, then x= =10, and y=6. w=3, then x=15, and y=10. w=4, then x=20, and y=14, &c. all of which numbers will answer the conditions of the question. 2. Given 3x=8y-16, to find the values of x and y in whole numbers. 4. Then 2y-5 being a whole number, it remains to make the remainder, , a whole number: 3y but the unknown quantity in this remainder has 2 for a coefficient, therefore, if this remainder be subtracted from y, or its equal, as 3y_21_+1, which must be a whole follows, 3 3 = 3 3' number; here the coefficient of the unknown quantity y being 1, the expression may be put equal to a whole number w. 7. Let w be taken equal to 2, then y=5, and x=8, two numbers that will answer the conditions of the question. 8. Or, by substituting the value of y in the first step of this solution, as in the first example, 24w-8-16 If w be taken 2, then y= 5, and x= 8, numbers that will answer the conditions of the question. 3. Given 17x+15y=130, to find x and y in whole positive numbers. 130-17x 1. Here y= which must be a whole 2 15 |