solve: y^3 = x^3 - 8x^2 + 6x + 8            for x,y are +ve integers?
826583
Yes, this is a difficult one.\nI'll keep trying, but don't hold your breath.\nAll I can say so far is that I could find\nonly one solution : (x, y) = (7,1).\nThere do not seem to be any more\nsolutions with x >= 8, and this is what\nneeds to be proved. So good luck with\nfinding someone who can do this.
763084
or solve equation for x and y components separately. \nfor x    0 +3+?=0 \nfor y    3+0 +?=0
795190
y = ax^3\n-5 = -8a, 5 = 8a, a = 5/8\n\nf(x) = (5/8)x^3
1050569
well...\n\n3(x-4)+8=x\n3x - 12 + 8 = x\n-4 = -2x\n2 = x\n\n...yes it is
701861
-5x+y=6 therefore y=5x+6\n2x-3y=60\n-----------\n2x-3(5x+6)=60\n2x-15x-18=60\n-13x=78\nx=-6\n\nWhen x=-6 is plugged into either of the original equations the value for y can be found by:\n\n-5(-6)+y=6\n30+y=6\ny=-24\n\n2(-6)-3y=60\n-12-3y=60\n-3y=72\ny=-24\n\nThe solution is in (x,y) form...(-6,-24)
905051
the equation is y=(x^3+5x+6)/6\nand the ordered pairs are\n(0,1)(1,2)(2,4)(3,8)(4,15)(5,26)\nsince it is a third degree equation the genera form is\nf(x)=ax^3+bx^2+cx+d\nf(1)=a+b+c+d...........=1....(1)\nf(2)=8a+4b+2c+d.....=2...(2)\nf(3)=27a+9b+3c+d...=4...(3)\nf(4)=64a+16b+4c+d.=8....(4)\nif you solve you will get  a=1/6,b=0,c=1\ngiving the equation as (1/6)x^3+(5/6)x+1
892609
y=2x-8\nx=2y-8\nadding 8\n2y=x+8\ndividing by 2\ny=(x+8)/2
917446
8x^2y\n\nas long as the x's and the y's are the same in both terms, you can just add them up...\n\nboth terms have x^2 and both terms have y and no other variables, so just add them up\n \n(3+5)x^2y = 8x^2y
831999
y = (x+3)(x-1)(x-2) = x^3 - 7x + 6
869535
5x+8+3x-5=6x-3\n8x+3=6x-3 (combine like terms)\n2x=-6 (subtract 6x, subtract 3)\nx=-3 (divide by 2)
605879
(x+3)(x+12)(x+2)(x+8)= [(x+3)(x+8)][(x+12)(x+2)]\n                     = (x^2+11x+24)*(x^2+14x+24)\nlet t=x^2+24+(25X)&#92;2 . \nSo we have: x^2+11x+24=t-(3x)&#92;2 and x^2+14x+24=t+(3x)&#92;2                                 \nso the equation becomes: [t-(3x)&#92;2][t+(3x)&#92;2] = 4x^2\n                    <-->t^2-(9x^2)&#92;4=4x^2  \n                    <-->t^2=(25x^2)&#92;4\n                    <-->t=(5x)&#92;2 or t=-(5x)&#92;2\n                    <-->x^2+10x+24=0 or x^2+15x+24=0\ncontinue by yourself
1082168
y=x^2+10x+8\ny=(x+5)^2 - 17\n\nVertex form: (5,-17)
795657
1. Move the constant to the right side of the equation:\n3x^2 - 8x = 6\n\n2. Divide out the x^2 coefficient:\nx^2 - 8/3x = 2\n\n3. Divide the coefficient of x (in this case, -8/3) by 2 and square it, then add it to both sides:\nx^2 - 8/3x + 16/9 = 34/9\n\n4. Factor the left side of the equation to (x + a)^2, where a is the number you just squared:\n(x - 4/3)^2 = 34/9\n\n5. Take the square root of both sides, remembering that this will produce both a positive and a negative root:\nx - 4/3 = +/-sqrt(34/9)\n\n6. Move the constant on the left to the right and simplify:\nx = 4/3 +/-sqrt(34/9) = 4/3 +/-(sqrt 34)/3 = (4 +/- sqrt 34)/3
532215
Your question is to factor in the following form y=a(x-p)^2+q for the question y=x^2-8x+8. Okay, let's roll.\n\ny=x^2-8x+8\n = (x-4)^2 -16 +8   [using (a-b)^2 = a^2 -2ab +b^2]\n = (x-4)^2 -8\n\nTherefore, in the form  y=a(x-p)^2+q, a=1, p=4 and q= -8.
681683
x-2y=8\nx=8+2y\n\n3x-2y=12\n3(8+2y)-2y=12\n24+6y-2y=12\n24+4y=12\n4y=12-24\n4y=-12\ny=-3\n\nx-2y=8\nx-2(-3)=8\nx+6=8\nx=8-6\nx=2
829339
x^2-6x+9 = (x-3)(x-3)\nx^2-9 = (x-3)(x+3)\n\nMultiple everything by (x-3)(x-3)(x+3)\n\n(x+3)(x+3) - (x+2)(x-3) - 5(x-3)(x+3)\n\nx^2+6x+9 - (x^2-x-6) - (5x^2-45)\n\nx^2+6x+9-x^2+x+6-5x^2+45\n\n-5x^2 + 7x + 60
835686
4x^2 + 3y^2 = 4\n8x x' + 6y y' = 0\n8 x '' + 6 y'' = 0\n6y'' = -8x ''\ny '' = -4x / 3