First, reduce all the fractions to the same denominator, by the known rule; the equation then becomes If now, both members of this equation be multiplied by 72, the equality of the members will be preserved, and the common denominator will disappear; and we shall have or dividing by 6, 48x 54x + 12x = 792; 89. The last equation could have been found in another manner by employing the least common multiple of the denominators. The common multiple of two or more numbers is any number which each will divide without a remainder; and the least common multiple, is the least number which can be so divided. The least common multiple can generally be found by inspection. Thus, 24 is the least common multiple of 4, 6, and 8; and 12 is the least common multiple of 3, 4, and 6. We see that 12 is the least common multiple of the denominators, and if we multiply each term of the equation by 12, divi ding at the same time by the denominators, we obtain the same equation as before found. 90. Hence, to transform an equation involving fractional terms to one involving only entire terms, we have the following RULE. Form the least common multiple of all the denominators, and then multiply every term of the equation by it, reducing at the same time the fractional to entire terms. EXAMPLES. 1. Reduce + 3 = 20, to an equation involving entire terms. 5 4 We see, at once, that the least common multiple is 20, by which each term of the equation is to be multiplied. that is, we reduce the fractional to entire terms, by multiplying the numerator by the quotient of the common multiple divided by the denominator, and omitting the denominators. Hence, the transformed equation is 91. When the two members of an equation are entire polynomials to transpose certain terms from one member to the other. 5x 6= 8 + 2x. 5x 6 2x = 8+2x-2x; 5x -6-2x = 8. Whence we see that the term 2x, which was additive in the second member, becomes subtractive in the first. In the second place, if we add 6 to both members, the equality will 5x 6 -2x+6=8+6; still exist, and we have or, since 6 and +6 destroy each other 5x - 2x = 8 + 6. Hence, the term which was subtractive in the first member, passes into the second member with the sign plus. For a second example, take the equation Hence, we have the following principle: Any term of an equation may be transposed from one member to the other by changing its sign. 92. We will now apply the preceding principles to the resolution of equations. 1. Take the equation 4x -3 2x + 5. By transposing the terms 3 and 2x, it becomes Now, if 4 be substituted in the place of x in the given equation, it becomes Hence, 4 is the true value of x; for, being substituted for x in the given equation, that equation is verified. 2. For a second example, take the equation 5x 4x 12 3 8 6 By making the denominators disappear, we have 10х32х - 312 21-52x by transposing 10x -32x + 52x = 21 + 312 a result which, being substituted for x, will verify the given equation. 3. For a third example let us take the equation (За - x) (a - b) + 2ax = 46 (x + a). It is first necessary to perform the multiplications indicated, in order to reduce the two members to polynomials, and thus be able to disengage the unknown quantity & from the known quantities. Having performed the multiplications, the equation be comes, by transposing by reducing or, (Art. 48), 3a2 ах 3ab + bx + 2ax = 4bx + 4ab; - ax + bx + 2ax - 4bx = 4ab + 3ab За2, ах 3bx = 7ab - За2; (a-3b)x=7ab - 3a2. Dividing both members by a - 3b, we find 93. Hence, in order to resolve any equation of the first degree, we have the following general RULE. I. If the equation contains fractional terms, reduce it to one in which all the terms shall be entire, and then transpose all the terms affected with the unknown quantity into the first member, and all the known terms into the second. II. Reduce to a single term all the terms involving the unknown quantity: this term will be composed of two factors, one of which will be the unknown quantity, and the other all its co-efficients connected by their respective signs. III. Then divide both members of the equation by the multiplier of the unknown quantity. |