Here our equations, cleared of fractions, will be 9x 5y = 90. And, if from the triple of the former the latter be sub tracted, we shall have 6y+5y = 288 90, that is, 96-24 = = 20. 3 2y 8= + 27. + 8/09 3 2y + 64, and 12x + 12y + 20x 480= 30y15x1620; which, contracted, become 4x-2y= 160, and = 47x-18y 2100: from the last of which subtract 9 times the former; so shall 11x=2100 x+x=14 ; to find x, y, and z. By subtracting the first equation from the second (in order to exterminate x) we have z-y= 1; to which the third equation being added, y will likewise be exterminated, there coming out 2z 16, or % = 8: whence y(x-1)= 7; and x ( 13 — y) = 6. Here the given equations, cleared of fractions, become 12x+8y+6x=1488 30x+24y=20x=4560 Now (to exterminate) let the second of these equations be subtracted from the double of the first; and also the triple of the third from the quintuple of the second whence is had To the double of the first, let the second equation be added; so shall the x's, on the contrary sides, destroy each other, and you will have 300+y=2y+4%, or 300= y+4z. Moreover, to the triple of the first, let the third equation be added, whence will be had +400=6y+3%, or 400=6y+2%. Now, if from the double of this last equation, the former, 300 y+4%, be subtracted, there will come out 500-11y; and, consequently, y =· 500 = 45; therefore Let xy-2, and xy+5x-6y=120; to exterminate x. By the former equation, x=y+2; which value being substituted in the latter (according to the second general method) it becomes y+2xy+5xy+2-6y=120, that is, y2+2y+5y+10—6y=120, or y2+y=110. EXAMPLE X. Let there be given x+y=a, and x+y=b; to extermi nate x. Here, by the first equation, x-a-y; and therefore x2=a-y12; which value being wrote in the other equation, we have a-y+y2b, that is, a2—2ay+y2+y2=b; and therefore y3—ay = 2 Given b-aa EXAMPLE XI. [axy + bx+cy=dy to exterminate y. Multiply the first equation by f, and the second by a, and subtract the latter product from the former; whence you will have bfx-agx+cfy-ahy=df-ak; which, by df-ak+agx-bfxc transposition and division, gives y of ah Let this value of y be now substituted in the first equation, and there will arise adfx — a2kx + a2gx2 — abfx2+cdf — cak+cagx— cbfx cf-ah + bx-d: which, multiplied by cf-ah, and contracted, gives ag-bf xax2+df—ak+cg—bh ×x=ck-hd. EXAMPLE XII. Supposing ax2 + bx + c = 0, and fx2 + gx + h = 0; to exterminate x. Proceeding here as in the last example, we have fbx =0; and, from thence, x == +fc― agx — ah = Whence, by substitution, x x ah-fc fb-ag ah — fc12 bxah-fc fb-ag + fb-ag + c = 0. This, by uniting the two last terms, and di viding the whole by a, gives: ah-fc12 bh-cg fb-ag fb-ag + consequently ah-ƒc]2 +ƒb—ag × bh —cg = 0. - = 0; After the same manner x may be expunged out of the equations ax + bx2 + cx + d= 0, and fx2+ gx + h = 0, &c. But, to show the use of the above example, suppose there to be given the equations x + yx — y2 = 0, and x2+3xy-100: then, by comparing the terms of these equations with those of the general ones, ax2 + bx + c = 0, and fx2 + gx + h 0; we have a = 1, y, c = y2, f = 1, g = 3y, and h = which values being substituted in the equation ah — fc +fb-ag x bh-cg=0, it thence becomes b = 10; 2 10+ yy]2 +y-Sy x-10y+ 3y3 =0, that is, 100-20y2+y+ 2012-6y=0; or, 100 5y; whence y may be found, and from thence the value of x also. = SECTION X. Of Proportion. QUANTITIES, of the same kind, may be compared together, either with regard to their differences, or according to the part or parts, that one is of the other, called their ratio. The comparison of quantities according to their differences, is called arithmetical; but according to their ratios, geometrical. When, of four quantities, 2, 6, 12, 16, the difference of the first and second is equal to the difference of the third and fourth, those quantities are said to be in arithmetical proportion. But, when the ratio of the first and second is the same with that of the third and fourth (as in 2, 6, 10, 30) then the quantities are said to be in geometrical proportion. Moreover, when the difference, or the ratio, of every two adjacent terms (as well of the second and third, as of the first and second, &c.) is the same, then the proportion is said to be continued: thus, 2, 4, 6, 8, &c. is a continued arithmetical proportion; and 2, 4, 8, 16, &c. a continued geometrical one. These kinds of proportions are also called progressions, being carried on according to the same law throughout. Arithmetical Proportion. THEOREM I. Of any four quantities, a, b, c, d, in arithmetical progression, the sum of the two means is equal to the sum of the two extremes. For since, by supposition, ba is dc, therefore is b + c =d+a, by transposition. THEOREM II. In any continued arithmetical progression (5, 7, 9, 11, 13, 15) the sum of the two extremes, and that of every other two terms equally distant from them, are equal. For since, by the nature of progressionals, the second term exceeds the first by just as much as its correspond *Although, in the comparison of quantities according to their differences, the term proportion is used; yet the word progression is frequently substituted in its room, and is, indeed, more proper; the former term being, in the common acceptation of it, synonymous with ratio, which is only used in the other kind of comparison. |