The L. C. M. of 3, 4, 6 and 8 is 24. Multiplying each term of the equation by 24 we have: Multiplying through by 20, the L. C. M. of 4, 5 and 10, we NOTE. When a fraction, whose numerator is a polynomial, is preceded by a minus sign, the sign of each term of the numerator should be carefully changed when the denominator is removed. In such a case, the numerator may be enclosed in a parenthesis, as shown in the above example. ILLUSTRATIVE EXAMPLES. 1. Solve the equation denominators, we have: Multiplying each term by x- 1, the L. C. M. of the NOTE. When the denominators are partly monomial and partly polynomial, it often simplifies matters to clear partially of the fraction at first, multiplying by a quantity which will remove the monomial denominators. *NOTE: To obtain the L. C. M. of the denominators multiply the denominators together. 7. SOLUTION OF LITERAL EQUATIONS. ILLUSTRATIVE EXAMPLES. 1. Solve the equation: 2ax 36 Transposing and uniting, 5axx x + c3ax. 3b + c = Factoring the first member, x(5a − 1) = 3b + c Dividing by 5a — 1 2. Solve the equation (b-cx) — (a — cx)2 = b(b — a). Performing the operations indicated, (a2 2acx + c2x2) = b2 — ab. x = Ans. 2c Solve the equation .2x .01 .03x.113x + .161. Or, we may solve the equation as follows: Transposing, .2. - .03x -.113x= .01 + .161 Uniting terms, Dividing by .057, .057x.171 x 3. Ans. 9. The following leads to simple equations which contain but one unknown quantity. The solution of a problem in algebra must depend in large measure upon the natural ingenuity of the student. No general rule, applying to all cases, can be given. The following suggestions however may be found useful: (a.) Represent the unknown quantity, or one of the unknown quantities, by one of the final letters in the alphabet. (b.) Try to discover, from the given conditions, expressions for the other unknown quantities, if such exist, in the problem. (c.) Equations should be formed in accordance with the conditions given in the problem. (d.) Solve the equation thus formed. ILLUSTRATIVE EXAMPLES. 1. What is the number which if four-sevenths of itself be added, the sum will equal twice the number diminished by 27? Let x = the number. NOTE.- Do not be confused by the expression, "Let x = B's age." x can of course represent only an abstract number, -the number of years in B's age. How much had each 3. A had twice as much money as B; but after giving B $35, he had only one-third as much as B. at first? By the conditions therefore x + 35 = 3 (2x x + 356 x 140 1. What number is that whose double exceeds its half by 45? 2. A is 62 years of age, and B is 36. it since A was three times as old as B? Ans. 30. How many years is 3. A gentleman left an estate of $1,872 between his wife, three sons, and two daughters. to receive three times as much as either of the Ans. 23. to be divided The wife was daughters, and each son one-half as much as each of the daughters. How much did each receive? Ans. wife, $864, a daughter, $288, a son, $144. ILLUSTRATIVE EXAMPLES. 1. Divide a into two parts such that m times the first part shall be equal to n times the second. |