.*. 5x—20a=4x-60, by transposition, 5x-4x=20α-60, by collecting the coefficients, (a—b)x=2a, 2α ... by division, x=- α -b° 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 (4a2+3abc-10c)x=6ac, Multiplying by 12, the least common multiple, we have 36x-3x+12-48-20x+56-1, by transposition, 36x-3x-20x-56-1+48-12, 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-2:4:: 5x-9: 2, to find the value of x. Multiplying extremes and means, we have 2(3x-2)=4(5x-9), or 6x-4=20x-36, by transposition, 3x-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, 3ax-bx-5x=27a, collecting the coeff's., (3a-b-5)x=27a, by clearing of fractions, 9x-45=32x-160, by transposition, 9x-32x=45-160, collecting and changing signs, 23x=115, 23x 115 = ·; •*• x=5. 23 23 3: x 1:: 4x: 2x+2, to find Multiplying extremes and means, we shall have (2x-3).(2x+2)=4x(x-1), or 4x2 -2x-6=4x2 —4x, by transposition, &c., 2x=6, 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' clearing of fractions, 4x-4=9x+18, ... by division, x=— 22 =-43. § III. EXAMPLES IN SIMPLE EQUATIONS, Involving only one unknown Quantity. 198. It is necessary to observe that an equation expressing but a relation between abstract numbers or quantities, may agree with many questions whose enunciation 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 0=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 ought 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. |