43. + + 44. + 45. X 46. b-C 47. + + 1 y XY-42 3(2-2) + y(x-7)+z(y-2) - a C-a-6 b+c – 9z2 " (2y + 3x)2 — 2:20 + (3y + 4z)2 - 4.x2 (42 + 2.2) -9y2* 1 1 12+ a2 X+ + a2 1 a+c 1 a2 + x2 a2 + x2 (.t+a)(v+b)–(y+a)(y+b)_(72–a) (y – b) – (623–6) (17 –a) a-6 a+b 48. + 49. 50. + CHAPTER XVII. FRACTIONAL AND LITERAL EQUATIONS. 149. In this chapter we propose to give a miscellaneous collection of equations. Some of these will serve as a useful exercise for revision of the methods already explained in previous chapters; but we also add others presenting more difficulty, the solution of which will often be facilitated by some special artifice. The following examples worked in full will sufficiently illustrate the most useful methods. X-9 X-9 1 Example 1. Solve 4 8 22 2 Multiply by 88, which is the least common multiple of the denominators, and we get 352–11( 29)=4x — 44 ; removing brackets, 352 – 11x +99=4x — 44 ; transposing, – 11x— 4x= -44-352-99; collecting terms and changing signs, 15x=495 ; .. X=33. NOTE. In this equation is regarded as a single term with 8 1 the minus sign before it. In fact it is equivalent to (2—9), the line between the numerator and denominator having the same effect as a bracket. 8x+23 5x+2 23+3 Example 2. Solve 1. 20 3x+4 5 Multiply by 20, and we have 20(5x+2) 8x+23 -=8x+12-20. 3x+4 20(5x+2) By transposition, 31= 3x+4 Multiplying across, 93x+124=20(5x+2), 84=73; .. X=12. X - 9 X-8 X-7 + -6 C - -9 This equation might be solved by clearing of fractions, but the work would be very laborious. The solution will be much simplified by proceeding as follows: Transposing, X-5 X-7 X-4 -10 7 -6 Simplifying each side separately, we have (x-8)(x-7) – (2–5)(x–10) - (x-7)(x-6) – (x-4) (0-9); (2x – 10) (x — 7) (x — 9)(x-6) 22 — 15x +56 — (x2 – 15x+50)_32–13x+42- (x2 – 13x+36) (3 — 10)(-7) (3-9) (3-6) 6 6 (3C-10) (2-7)(3-9) (2—6) Hence, since the numerators are equal, the denominators must (3-10) (C-7)=(3-9) (3-6), .:. 16=2x; be equal ; that is, X - 13 X-6 X-14 X 5x – 64 2x – 11 4x - 55 --6 Example 4. Solve X - 13 X-6 X - 14 XC - -7 1 We have 5+ 2+ 1+ 2- 1 13 -6 14 -7 Simplifying each side separately, we have 7 7 2x = 20; 150. To solve equations whose coefficients are decimals, we may express the decimals as common fractions, and proceed as before; but it is often found more simple to work entirely in decimals. 1 1 Example 1. Solve •6x+-25–5x=1.8--75x 3 Expressing the decimals as common fractions, we have 3x+1-313-13-3; clearing of fractions, 24x+9–4x=68—27.–12; transposing, 24x - 4x +27x=68—12-9, 473=47; .. X=1. Example 2. Solve •375x-1.875=:12x+1.185. Transposing, •375x — •12x=1:185+1.875; collecting terms, (-375--12)=3.06, that is, .255x=3:06; 3:06 .255 =12. 5. 5.C+5) 2(2-3) 7 :52%: + 3 76. 7. 6. + 2-3)=957 8. 3+=(-1)-6+(1-3) 6 (x+10)-(2-3)= 17 |