The L.C.M. of 3, 4, 6, and 8 is 24. Multiplying each term of the equation by 24, we have 16x-30 =20x-27 We have then the following rule for clearing an equation of fractions: Multiply each term by the lowest common multiple of the denominators. 12. Solve the equation 3x-1 4x-5 =4+ 7x + 5. 10 Multiplying through by 20, the L.C.M. of 4, 5, and 10, 15 х - 5 - (16 x - 20) = 80+ 14x+10 Note. If a fraction whose numerator is a polynomial is preceded by a sign, care must be taken to change the sign of each term of the numerator when the denominator is removed. It is convenient, in such a case, to enclose the numerator in a parenthesis, as shown in the above example. 21. 5x-2_3x+4_7x+2 -10. 3 4 6 2 7 22. (x + 1) - 2x - 5 - 11 x + 5 - 13. 1 х - 2 3 5 10 24. 4+ (3x-2) - 11+ 2(2-9x). 7 25. = 2 14 1 3 Multiplying each term by x2 - 1, the L.C.M. of the denominators, 2 (x + 1) - 3 (x - 1) − 1 = 0 2x+2-3x + 3 - 1 = 0 2x - 3x = -2-3+1 Multiplying by 7x-16, 28x - 64 = 30x - 60 -2x=4 x= -2, Ans. Note. If the denominators are partly monomial and partly polynomial, it is often advantageous to clear of fractions at first partially; multiplying by a quantity which will remove the monomial denomi 45. (x + 1)2 - 4. x-2 176. 1. Solve the equation 2ax-36=x+с-Зах. Transposing and uniting terms, 5 ax - x=3b+c. Factoring the first member, x(5-1)=3b+c. Dividing by 5 а - 1, x= 3b+c. Ans. 5α-1' 2. Solve the equation (b - cx)2 - (а - сx)2 = b(b − a). Performing the operations indicated, b2 - 2bcx + c2x2 - (a2 - 2 асх + c2x2) = b2 - ab - c2x2 = b2 - ab 2 асх - 2 bcx = a2 - ab b2 - 2bcx + c2x2 - а2 + 2 асх Factoring both members, 2 cx(a−b) = a(a - b) |