Solve x+y = 12 and 2xsquare + 3ysquare=7xy?
819984
restate the first equation in terms of y:\n\ny = 12 - x\n\nNow substitute this in for y in the second equation:\n\n2x² + 3(12 - x)² = 7x(12 - x)\n\nNow, multiply it out and solve for x:\n\n2x² + 3(144 - 24x + x²) = 84x - 7x²\n2x² + 3x² + 7x² - 72x - 84x + 432 = 0\n12x² - 156x + 432 = 0\n12(x² - 13x + 36) = 0\nx² - 13x + 36 = 0\n(x - 9)(x - 4) = 0\n\nx = 9 or 4\n\nSo, {x,y} = {9,3} or {4,8}
146086
2x+y=6 => y=6-2x \n3x+y=1 => 3x+(6-2x)=1 => x+6=1 => x=-5\ny=6-(2*-5) => y=16\n\nx=-5\ny=16
287407
3(16 y_squared - 24xy + 9 x_squared)
684399
if x + y = 12 and y = 2x, then in the original equation, you can substitute 2x for y.\n\nx + 2x = 12\n\ncombine like terms\n\n3x = 12\n\ndivide both sides by 3\n\nx = 4\n\nbecause y = 2x, substitute x for 4.\n\ny = 2(4)\n\ny = 8
469289
3x^2 = 12\n\nFirst, divide by three.\n\nx^2 = 4\n\nTake the square root.\n\nx = +2, -2
805022
let the numbers be x and y\nSo x + y = 7\nxy = 25\n\nNow 1/x + 1/y \n= x/xy + y/xy\n= (x + y)/xy\n= 7/25\n\nx + y = 7 So y = 7 - x\nxy = 25\n\nSo x(7 - x) = 25\n\n7x - x² = 25\n x² - 7x + 25 = 0\nx = [-b ± √(b² - 4ac)]/(2a)\n= [7 ± √(7² - 4.1.25)]/(2)\n= [7 ± √(- 51)]/(2)\n\nSo x = [7 - √(- 51)]/(2) and y = [7 + √(- 51)]/(2) \nThus 1/x = 1/[7 - √(- 51)]/(2)\n= 2/[7 - √(- 51)] * [7 + √(- 51)]/[7 + √(- 51)]\n= 2[7 + √(- 51)]/(49 - -51)\n= [7 + √(- 51)]/50\n\n1/y = [7 - √(- 51)]/50\nSo 1/x + 1/y = [7 + √(- 51)]/50 + [7 - √(- 51)]/50\n= 14/50\n= 7/25 (as above but with much more effort!!!!!!!!) (8^O)
783589
X & Y\n\n4XY=X+Y\n\nSolve for X or Y and then substitute your result in the above equation. \n\nX=4XY-Y\n1=3Y\nY=1/3\n\n4x/3=1/3+x\n4x=1+3x\nx=1\nProof:\n4(1/3)=1+1/3\n4/3=4/3
697620
x^2+y^2=(x+y)^2-2xy\nsubstituting\nx^2+y^2=3^2-2(1)\n            =9-2\n            =7\nx^2+y^2=7\nmy 10 pointsplease
753754
I presume you mean\n\n[(6x^2) / (4x^2y-12xy)] * [(x^2+x-12) / (3x^2+12x)]\n\nThere seems to be something wrong. Are you sure about those "y" terms?
1012196
3(x+12)=x+2\n3x+36 =x+2\n2x+36=2\n2x=34\nx= 17
225766
Given, x = 1/2y + 4 ............(i)\n and   2y +3 x = 12 ............(ii)\nSoln.: Multiply both side by 2 in eqn. (i)\n      2x = y + 8\ni.e.  2x - y = 8\ni. e. 4x - 2y = 16....(iii), multiplying both the side by 2\nNow, add eqn. (ii) and (iii) above,\n (2y+3x)+ (4x-2y)= 12 + 16\ni.e. 7x = 28\ni.e. x = 28/7 = 4.Ans.\nNow putting the value of x = 4 in either of the eqn. above, find out the value of y. Say, put value of 'x' in eqn.(ii),\n2y + 3x = 12\n2y + 3x4 = 12\n2y + 12 = 12\n2y = 12-12 = 0\nSo, y = 0/2 = 0 Ans.
773396
x+y-z=2---(1)\n2x-y+z=7-(2)\n-x-y+2z=3---(3)\nadding (1) and (2)\n3x=9\nx=3\nadding 1 and 3\nz=5\nsub x=3,z=5 in 1\n3+y-5=2\ny-2=2\ny=4\nthe solution set{3,4,5}\n\n2.x+2y-z=-3\n2x+y+z=6\nadding\n3x+3y=3\n=>x+y=1\neqn. 1*3\n=3x+6y-3z=-9\n-2x+2y+3z=3\nadding\nx+8y=-6\nx+y=1\nsubtracting 7y=-7\ny=-1\nsub in x+y=1,x=2\nsub in x+2y-z=-3\nz=3\nso solution set {2,-1,3}
1109407
-4(3-x) = 2(x+6) \n-12+4X=2X+12 use distributive property\n-12+4x-2x=2x-2x+12 move the variable x onto 1 side \n-12+12+2x=12+12 isolate the variable x\n2x=24\n(2x/2)=24/2\nx=12\n\nCheater!
1153750
7xe^y + 3x^3 = 5x^4\n\n=> 7e^y + 7xe^y (dy/dx) + 9x^2 = 20y^3 (dy/dx)\n\n=> 7xe^y (dy/dx) - 20y^3 (dy/dx) = -7e^y - 9x^2\n\n=> (dy/dx)(7xe^y - 20y^3) = -7e^y - 9x^2 \n\n=> dy/dx = (-7e^y - 9x^2)/(7xe^y - 20y^3).
932079
7P = (14x + 21y - 14)A + (21x + 14y + 35)B\n3Q = (-3x + 12y - 6)A + (9x - 12y + 21)B\nAs they are equal, then the coeficients of each A and B are equal:\n14x+21y-14 = 21x+14y+35    and\n3x+12y-6 = 9x-12y+21\nwhich gives you two equations in two unknowns.  The rest is easy.
851375
first subtract 2x from both sides to get:\n\n-7=3x+7  \n\nthen add 7 to both sides to get:\n-14=3x\nx=-14/3 or -4 2/3
695016
4x + 3y =12, find y = 4\n4x+3*4=12\n4x+12=12\n4x=0\nx=0