RULE III. 197. If in the equation there be any irreducible fractions, in which the unknown quantity is concerned, multiply every term of the equation by the denominators of the fractions in succession, or by their least common multiple; and then proceed according to Rules I. and II. 2x Ex. 1. Given +1-x-9, to find the value of x. 4 Multiplying by 4, 2x+4=4x-36, by transposition, 2x-4x-36-4, by collecting the terms, -2x=—40, by changing the signs, 2x=40, by 4, 12x-8x+72=120-6x, by transposing, and collecting, 10x=48 Or, it is more concise and simple to multiply the equation by the least common multiple of the denominators; because, then the equation is reduced to its lowest terms; thus, Multiplying by 12, the least common multiple of 2, 3, and 4, we have, 6x-4x+36-60—3x, by transposition, 5x=24, 5x 24 by division, Ex. 3. Given x- ·1= x Here 30 is the least common multiple of 3, 5, and 6 ; to find the value of x. Ex. 4. Given -α= -3, to find the value of x. x X 5 Here 20, the product of 4 and 5, being their least common by collecting the coefficients, (a—b)x=2a, Here 2ac, the product of 2, a, and c, being the least common multiple, Multiplying by 2ac, 4a2x+3abcx=10cx+6ac, by transposition, and collecting the coefficients, we shall have (4a+3abc-10c)x=6ac, ..by division, x= 4a2+3abc-10c of x. Multiplying by 12, the least common multiple, by transposition, 36x-3x-20x-56-1+48-12, or 13x=91, 13x 91 by division, = ; .. x=7. 13 13 198. If the unknown quantity be involved in a proportion, the proportion must be converted into an equation (Art. 190); and then proceed to resolve this equation according to the foregoing Rules. Ex. 1. Given 3x of x. 2:4: Ex-9: 2, to find the value Multiplying extremes and means, we have 2(3x-2)=4(5x-9), or 6x-4=20x-36, by transposition, 6x-20x=-36+4. Ex. 2. Given 3a :x ::b+5 : x-9, to find the value of x. Multiplying extremes and means, we have 3a.(x-9)=x.(b+5), or 3ax-27a=bx+5x, by transposition, 3ɑx-bx-5x=27α, collecting the coeff's., (3a-b-5)x=27a, 27a .. by division, x= 3a-b-5 by clearing of fractions, 9x-45=32x-160, by transposition, 9x-32x=45-160, collecting and changing signs, 23x=115, 23x 115 by division, 23 23 x=5 Ex. 4. Given 2x-3: x- -1 :: 4x: 2x+2, to find the value of x. Multiplying extremes and means, we shall have (2x-3).(2x+2)=4x(x-1), by tranposition. &c., 2x=6, .. by division, x=3. Ex. 5. Given a+x: b::c-x: d, to find the value of x in terms of a, b, c, and d. Multiplying extremes and means, ad+dx-bc-bx, by transposition, bx+dx-bc-ad, or (b+d)x-bc-ad, bc-ad, .. by division, x= b+d' 1 Ex. 8. Given x+3: a :: b: -, to find the value of x. 199. It is necessary to observe that an equation expressing but a relation between abstract numbers or quantities, may agree with many questions whose enunciations would differ from that of the one proposed: but the principles of the resolution of equations being independent of any hypothesis upon the nature and magnitude of quantities; it follows, therefore, that the value of the unknown quantity substituted in the equation, will always reduce it to u=0, although it may not agree with the particular question. This is what will happen, when the value of the unknown quantity shall be negative; for it is evident that when a concrete question is the subject of inquiry, it is not a negative quantity which bught to be the value of the unknown, or which could satisfy the question in the direct sense of the enunciation. The negative root can only verify the primitive equation of a problem, by changing in it the sign of the unknown; this equation will therefore agree then with a question in which the relation of the unknown to the known quantities shall be different from that which we had supposed in the first enunciation. We see therefore that the negative roots indicate not an absolute impossibility, but only relative to the actual enunciation of the question. The rules of Algebra, therefore, make not only known certain contradictions, which may be found in enunciations of problems |