164.* In case C involves the unknown quantities, the equation and, besides, all the solutions of the equation C = 0; because, for any system of values assigned to the unknown quantities, the expressions C and A-B take finite values. If one of these factors is zero the product is zero. C(A—B) Equation (2) is more general than equation (1). The solutions of equation C 0 are called foreign solutions which are introduced by multiplication. Thus, let the equation be = 4x7 53 — 2x. By multiplying both its members by x-5, the equation (4x-7) (x-5)= (532x) (x-5) is formed, which is more general than the given equation. x = 10 is a root of the first equation and also a root of the second, as is seen by substituting. x = 5 is a root of the second equation, but not of the first equation. Hence x = 5 is called a foreign root which belongs only to the second equation. 165.* If, however, the expression A-B is not integral in the unknown quantities and the multiplier Cis integral in these unknown quantities, then equation (2), possesses all of the solutions which equation (1) has, namely, because the factor has a finite value for any system of finite values of the unknown quantities which make A . B zero. Thus, consider * A thorough grasp of the ideas discussed in §164 and 2165 need not be insisted upon in the first reading, though they are essential to a full understanding of the theorem. in 163. satisfies equation (2) and also equation (1), but C, or 432 16 +2(43) +15, which is finite and different It can not further be affirmed that equation (2) possesses also all the solutions of the equation, C = 0, because, in this case, for certain values of the unknown quantities the factor might be zero and the factor A - B infinite. Then the value of the product C (AB) is not determinate and can not be said to be zero. Return to the specific example, C=x2 — 8x + 15 = (x − 3) (x — 5). which is indeterminate (273, 2), and therefore, C(A — B) = [(x − 3) (x — 5)] ( x 7 3 x+4), for x = 5, is 3 (5 — 3) (5 5) ( which is also indeterminate and can not be said to be zero. 166. Application.-Removal of denominators.-Theorem 2, 163, makes it possible to replace an equation containing terms which are fractions by an equation which is integral; this process is called the clearing of fractions, or the removal of denominators. If the denominators of the given equation do not involve the unknown quantities, the new equation is equivalent to the first. Consider the equation, All the terms can be reduced to the same denominator, 20, and the equation written as follows: Hence, by multiplying both members of the equation by 20 the equation will be transformed into which contains only integral terms, and is equivalent to the given equation (163). Therefore, to remove the denominators of an equation, reduce all the terms to a common denominator and omit this common denominator. Or, multiply both members of the equation by the L. C. M. of the denominators. Thus, In practice the process is abbreviated. Write immediately the equation obtained by removing the common denominator. consider the equation, 2. 5 3x 5 4x 3 4 12 3 12 is a common denominator. Proceed as though to reduce all the terms to fractions having a common denominator 12, but instead of writing these fractions and then finally omitting the common denominator, write only the numerators, and obtain at once the equation, 24x-209x = 5 - 16x. 167.* * In case the denominators involve unknown quantities, reduce all the terms to the same common denominator, the simplest possible; e. g., the least common denominator (140). Then multiply the two members by this common denominator, thus suppressing it. A new equation is thus formed which contains all the roots of the given equation, but which can contain, besides the roots of the given equation, foreign roots also (164). These foreign roots can be introduced only through the multiplier and are obtained by equating to zero the common denominator by which the two members of the given equation were multiplied. EXAMPLE 1. Consider the equation, and put the equation in an integral form by multiplying both mem bers by x(x-3)(x+3). Equation (2) follows: *See Note to §§ 164, 165. (2) x2+x+3= (x − 3) (5 x + 3). This new equation contains necessarily all the roots of equation (1), (3163); but it might have as roots, values of x which reduce the multiplier, to zero; e. g., the values, x (x-3) (x+3), x = 0, x = 3, x = −3. But equation (2) is satisfied by x = 4, or 3, and not by x = 0, x = 3, or x = 3, hence the multiplication does not introduce any foreign roots, and the two equations (1) and (2) are equivalent. EXAMPLE 2.-Let the equation be and the equation to an integral form by multiplying both members by this common denominator. Thus equation (2) is obtained: (2) 6x-3(x+2) = x2-4. This new equation has necessarily all the roots which equation (1) has, and, besides, may have as roots the values of x which reduce the multiplier, to zero; that is to say the values (2165) x = 0, x = 2, x = 2. 2; Equation (2) is not satisfied either for x = 0 or for x = but it is satisfied for x = 2. The number 2 may, therefore, be a foreign root introduced by multiplication. Put 2 in the first member of equation (1). Then the value of the first member is not determinate, because it is the difference between two indeterminate expressions (273, 2), namely, The first member of equation (1) has a determinate value for a value of x which is different from 2, but which may be as nearly equal to 2 as one would like. It is desired to find the value which which for all values of x different from 2 has the same value as the fraction, 3 8 obtained by dividing both terms by x-2. For x 2 the last fraction becomes 3. Therefore when x tends toward 2, the value of the first member of equation (1) tends toward. But the second member of equation (1), namely 2; and therefore 2 is not a root of equation (1). 8 x 2 as approaches Hence equation (2) has as roots, not only the roots of equation (1), but also a foreign root equal to 2. SOLUTION OF AN EQUATION OF THE FIRST DEGREE IN ONE UNKNOWN QUANTITY 168. It is a simple matter now to solve by the aid of these theorems an equation of the first degree in one unknown quantity. To solve the equation, reduce all the terms of both members of the equation to equivalent fractions with the least common denominator 12. this denominator, or multiply both members by 12. Then suppress Equation (2) is the result, and is equivalent to equation (1), (2) 48 x 84 x 60-3x. Transpose all the terms involving a to the first member (161) |