32. Given values of x. 3 xn = 4, to find the value 3 Ans. (32) 20 +10x3+54x2+145x=602, to find the 1. Given 2x-y=15, 2 and 2x-y=9, to find the values of x and y. From the second equation, x= 9+y2 2 ; substituting this value of x in the first equation, 9+ y2―y=15; by transposition, y2—y=6; completing the square, yo—y+—=6+1 = 20; From the first equation, y=2x-15; substituting this value of y in the second equa tion, 4x2-62x=-234; 2 62 4 completing the square, x2-6x+31 = 2. Given 4x+4y=56, 961 16 to find the values of a and 6x+3y=xy+12, and y. Dividing the first equation by 4, x+y=14;. y=14-x; substituting this value of y in the second equa tion, 6x+42-3x=14x-x2+12; by transposition, x2-11x=-30; completing the square, x2-11x+· 121 121 4 3. Given x2+ y2=50—x—? -y, to find the values of and y=10, Transposing the first equation, x2+y2+x+y=50; but, 2xy x and y. =40; (4 times 2nd. eq.) by addition, x2+2xy+y2+x+y=90; 1 or (x+y)2+(x+y)=90; completing the square, (x+y)2+(x+y)+ 1 19 extracting the root, x+y+2=± 22; by transposition, x+y=9, or —10; but 4xy =80, by subtraction, x2-2xy + y2=1, or 20; extracting the square root, x-y=1, or ±√20; but x+y= 9, or -10; by addition, 210, or 8, or -10±2/5; .. x = 5, or 4, or ➡5±√5; but by subtraction, 2y=8, or 10, or —10—2/5; .. y=4, or 5, or 5/5. 4. Given (x+y)2+2x=195—2y, to find the values and a2+y=97, of x and y. Transposing the first equation, (x+y)2+2.(x+y) =195; completing the square, (x+y)2+2.(x+y)+1=196; extracting the root, x+y+1=±14; by transposition, x+y=+14-1=13, or -15; squaring each side, x2+2xy+y2=169, or 225; but x2 + y2 = 97; by subtraction, 2xy 72, or 128; by subtraction, x2−2xy+y2=25, or —31; .. x—y=±5, or ±√√—31; but x+y= 13, or —15; by addition, 2x=18, or 8, or -15±√√—31; but by subtraction, 2y= 8, or 18, or ‡√—31—15; Dividing the first equation by 2, and completing the square, x1y2-2x2y+1=2208+1=2209; extracting the root, xy-1-47; by transposition, x2y=48, or -46; multiplying the second equation by y, x2y+y2=60—y; substituting the affirmative value of xy, into this equation, 48+y2=60−y ; by transposition, y2+y=12; completing the square, y2+y+1=12+ 1 = 4; But if the negative value of x2y be substituted, -46+y260-y; by transposition, y2+y=106; completing the square, y2+y+1=106+ EXAMPLES FOR PRACTICE In Adfected Quadratics of two unknown quantities. 1. Given x2-xy=60, to find the values of x and and x+y=19, 5 y. 2 y= 7, or 214. 2. Given xy+18x=1728, to find the values of x and 8y=9x, 3. Given 5x+4y=35, and 9y―xy=10x, 4. Given 2x+y=7, and x2+3y=13, } Ans. and y. to find the values of r and y. |