and therefore, (Art. 147), multiplying by 18, 2y-4x+1-18-24-6y+3x-3y; .. by transposition, 7-7x-11y. But from the second equation, 7x=12y. Substituting therefore this value in the preceding equation, it becomes and 3x+2y y-5 11x+152_3y+1 to find the values of x and y. Multiplying the first equation by 33, 33x-9y6-3x=33+15x+. 4y ; 3 multiplying again by 3, and transposing, we shall have 45x-31y=81. Multiplying the second equation by 12, .. by transposition, 19y-5x=131. Multiplying this by 9, 171y-45x=1179; but 45x-31y= 81; .. by addition, 140y=1280; and by division, y=9. Now, 5x-19y-131-171-131=40; .. by division, x=8. to find the values of x and y. Multiplying the first equation by 105, the least common multiple of 3, 7, and 15, 560+21=1925-60x-45y+120; .. by transposition, 81x+55y=1485; and dividing by 9, 9x+5y=165. From the second equation, 50y+6x-35=275+50x; .. by transposition, 50y-44x=310; and dividing by 2, 25y-22x=155; but multiplying the equation found above, by 5, 25y+45x=825; y Multiplying the first equation by c, and the se cond by a, we shall have ac bc + =mc, ... by subtraction, (bc-ad).=mc―na ; y Multiplying the first equation by 15y, ... 45y-21y-6x=75y-25x-45; and by transposition, 51y-19x=45. Multiplying the second equation by 2x+5, 8x+20+30xy+75y 2xy+5y 6x--2 107 =2xy· 8 6x-2 107_8x+20+30xy+75y, .. (Art. 186) 5y+ 8 and multiplying by 6x-2, we shall have and 321x-107=32x+340y+80; by transposition, 340y-289x187. The coefficients of y in this case, having aliquot parts; multiplying the first by 20, and the last by 3, 1020y-380x= 900, and 1020y-867x=-561; .. by subtraction, 487x=1461, consequently, 51y=45+19x=45+57=102; and x=3; 3x-2y+1 to find the values of x and y. Multiplying the first equation by 5+2y, 40x+16xy 80+300x+32y+120xy =16xy-107; ... trans" 40x+107= 80+300x+32y+120xy 3y-1 and multiplying by 3y-1, we shall have 120xy-40x+321y-107=80+300x+32y+ 120xy; .. by transposition, 289y-340x=187. And from the second equation, 27x-12y +15x+2y+2=27x2-12y2+38; ... by transposition, 15x+2y=36; whence, the coefficients of x having aliquot parts, multiplying the first equation by 3, and the second by 68, 867y-1020x=561, and 136y+1020x=2448; .. by addition, 1003y=3009, and y=3; consequently, 15x=36-2y=36-6=30; and.. by division, x=2. |