10. x2 — 3x — 3x(3 — x) — x { (x − 3) (x — 2) — 3(3 — x)} ̧ x- 2 = x { (x − 3) (x −2) + 3(x — 3) } __ x(x — X X- 2 the opera ADDITION AND SUBTRACTION OF FRACTIONS. EXAMPLES FOR EXERCISE. PAGE 199. In this seventh example, the terms of the first fraction are multiplied by, and those of the second by x; the denominators thus become each equal to xyz. Multiplying the terms of the first fraction by 2"-1, and will serve t a + b a b +2= = ̧ a2+b2+a2 — b2— (a2 — b2) + 2(a2 —b2) MULTIPLICATION OF FRACTIONS. EXAMPLES FOR EXERCISE. PAGE 200. In this tenth example it is seen at once that the numerator of the second fraction is divisible by the denominator of the first, because if a be put for x in x3 — a3, the result is 0. (See p. 19, Key). b2x X a2 7. 9. 2y2 2x 1 3x+1 X x+2 a xm X = a = x 3 bx xm + ክ + n Gy DIVISION OF FRACTIONS. EXAMPLES FOR EXERCISE. PAGE 201, a4 - bi = ab a-b a = b (x3 — y3) (x2 + y3) a2 b2 a2 + b2 -- (x − y) (x2 + y2) Otherwise, 1 = SIMPLE EQUATIONS IN GENERAL. EXAMPLES IN EQUATIONS WITH ONLY ONE UNKNOWN QUANTITY. PAGE 203. +22+ Multiplying the terms of the second fraction by 3, 6. This might also have been readily solved by multiplying each side of the proposed equation by 30, the least number divisible by all the denominators; the equation thus cleared of fractions would have been 14x+x2,.. 14x 42, .. x = 3. 7. √(x − a) — b = 0, .. √ (x — a) = b, .. x — a = b2, .. x = a + b2. x = 100. Vx 2 2 Applying the principle explained at page 203, we have = √x + 38 = we have Vx+ 6 Applying the principle referred to in the last example, 10. = Vx + b 2a b = 4a 36 2a + b b ; that is, 2a b 4a 36 4a 36 2a Multiplying by (2ab) (4a36), - b Sayx-6b√x— 4α √x + 2b √x = (2a — b) (4a + 3b) — (2a + b) (4a — 36) .. (4a — 4b)√x= Sab+12ab= 4ab, Vax + a 3√ ax+5a Applying the foregoing principle, Clearing, 7ax=61'ax + 3a, :. Vax = .. ax = 9a2, .. x=9a. 12. 1(x-24)=√x-2. Squaring each side, x- 24=x- 4 √x +4. Transposing, 4/ x = 28, .. 1' x = За = 7, .. x 49. 13. (4x+21) = 2√x + 1. Squaring each side, 4x+214x + 4√x + 1. Transposing, 20 = 4√x, .. √x = 5, .. x = 25. 3 14. † (2x+3)=3. Cubing each side, 2x+3=27, .'. 2x = 24, .. x = 12. of the squares of the two quantities in the denominator; therefore, the numerator is actually divisible by the denominator, so that the equation is the same as x= SIMPLE EQUATIONS WITH TWO UNKNOWN QUANTITIES. 1st. Proceeding by Rule I., we have from the first equation and from the second a " 2 pressions for x, 2nd. Proceeding by Rule II., we have from the second equation x= Substituting this in the first we get 10 + 24 = 2 + 2/3. 3rd. Proceeding by the third method, multiplying the first equation by 2, the second by 3, and then adding the results, we have 19x: 76, . x4. Again, multiplying the first equation by 5, the second by 2, and subtracting the results, we get 19y95, .. y = 5. |