10 .. by subtraction, we have 2y=10, or y= =5. 2 It has already been shewn that x=16-4y= (since y=2, and. 4y=8) 16-8=8. Clear equs. (A) of fract'.x+2+24y= 93, or x+24y= 91(C) (B)....y +5+40x=768, or y+40x=763(D) From From equation (C), x=91-24 y; by RULE III, substitute this value of x in equation (D); then we have, y+40(91—24 y)=763 or y + 3640–960y=763 .. 959y=3640-763=2877 By referring to equation (C), we have x=91-24 y= (since y=3, and .:. 24y=72)91-72=19. Ex. 4. Let 3x+4y=29 (A). 17x-3y=36 (B). In this example, the Rule mentioned in Art. 74. must be applied. Multiply equation (A) by 3, then 9x+12y= 87 (C) (B) by 4 68x-12y=144 (D) ... Add equation (D) to (C), then 77x=231 or x= From equation (A) we have 4y=29-3x=(since x=3, and 20 .. 3x=9) 29-9-20; hence y==5. 76. When three unknown quantities are concerned, the most general form under which simple equations can be expressed, is ax+by+cz=d (E) ax+by+cz=ď (F) a'x+by+ c′′x=d' (G), and the mode of solution may be conducted in the following manner. 1. Multiply eq". (E) by a', then a a'x+aby+a'cz=a'd (H) (F) by a... aax+aby+acx=ad (K) Subtract (K) from (H), then (ab-ab')y+(acac)x=a'd-aď (L). By multiplying (F) by a", and (G) by a', and, subtracting the latter result from the former, we obtain in the same manner, \(a"baby+(a"c'-a'c")za"d'-a'd'"(M). 11. Next, let the coefficients in equation (L) be represented by a, ß, respectively; and those in equation (M) by a', ', ' respectively; then those equations may be reduced to the following form; viz. ay+ ay+Bz=y. ay+szy. From which, by making the proper substitutions in RULE I, and in Art. 73, we have, substituting the values of y and z just now found, we obtain This mode of operation might be easily extended to equations containing any number of unknown quantities. 1. Multiply (E) by 3, then 6x+9y+12x=87 (H) (F) by 2 ...6x+4y+10x=64 (K). ... Subtract (K) from (H). . 5y + 2x=23 (L). 5y+ Multiply(F) by 4, then 12x+8y +20z=128 (G) by 2,...12x+9y+ 6x=75 II. Hence the given equations are reduced to, Multiply (M) by 5, then-5y+70%=265 The Solution of Questions producing Simple Equations. In the reduction and management of equations, we have proceeded by fixed and stated rules; but in the solution of questions we have no such rules to guide us. Every particular question requires a distinct process of reasoning, to bring it into an algebraic form; and nothing but practice and experience can produce expertness and facility in conducting this process. All that can be done for the learner in this case, is, to explain the manner in which the principles of this science may be made to bear upon questions in general; for as soon as they can be brought into the shape of equations, we have only to supply the foregoing Rules for finding the value of the unknown quantity or quantities. Before we proceed, therefore, to any actual examples, it may be proper to shew the relation which arithmetical and algebraic operations stand in to each other. 77. Suppose the following arithmetical question was proposed for solution; viz. "To divide the number 35 into two such "parts, that one part should exceed the other part by 9." A person unacquainted with Algebra might with no great difficulty solve this question in the following manner. 1. It appears, in the first place, that there must be a greater and a lesser part. II. The greater part must exceed the lesser by 9. 111. But |