anyone know how to rewrite yx^2+6xy-x^2+2x+9y+3=0, into Y= form?
931054
Group all elements with y together:\n\ny(x^2 + 6x +9) -x^2 + 2x + 3 = 0\n\nMove the rest to the other side:\n\ny(x+3)^2 = x^2 - 2x + 3 (1)\n\nx = -3 is obviously not the solution of this equation therefore we can divide both sides of (1) by (x+3)^2.\n\n(1) is equivalent to:\n\ny = (x^2 - 2x + 3) / (x+3)^2
973495
Replace the y in the first equation with its value from the second:x+9.  You get x^2 + (x+9) = 9\n\nSubtract 9 from both sides to get x^2 + x = 0\n\nFactor out an x:  x(x+1) = 0\n\nSo x is 0 or x is -1\n\nPlug these into the original equations to get y.  \ny= 0 + 9; y=9\n\ny = -1 + 9; y=8\n\nSo the complete answers are (0,9) and (-1,8)Check them in the other equation: 0^2 + 0 + 9 = 9 check\n(-1)^2 + 8 = 9; 1+8=9 check
439090
((x^2-2x)/4y)/((x^2-4)/6)\n=(6x(x-2))/(4y*(x-2)(x+2))\n=3x/(2y(x+2))
844586
9x^2+y^2=9......(eqn no 1)\n3x-y=3 .............(eqn. no. 2)\nFrom equn. no. 2,we get y=3x-3=3(x-1)\nPutting the value of y in eqn.no.1,w get\n9x^2+{3(x-1)}^2=9\n=> 9x^2+9(x^2-2x+1)=9\n=>9x^2+9x^2-18x+9=9\n=>18x^2-18x=0\n=>18x(x-1)=0\nTherefore either 18x=0 or x-1=0\ntherefore x=0 or 1\nPutting the value of x in eqn no 2,we get 3X0-y=3 or y=-3  or 3X1-y=3,in which case y=0 \nTherefore x=0,y=-3 or x=1 and y=0
957977
5xy=(x+y)(x^2+2)\n5xy=x^3+2x+x^2y+2y\n5xy'+5y=3x^2+2+x^2y'+2xy+2y'\n5xy'-x^2y'-2y'=3x^2-5y+2xy\ny'(5x-x^2-2)=3x^2-5y+2xy)\ny'=(3x^2-5y+2xy)/(5x-x^2-2)
769967
removing the brackets\n18+3y=4y-16-y\nsimplifying\n18+3y=3y-16\nadding 16\n34+3y=3y\nadding -3y\n34=0\nwhich is absurd\nso the equation cannot be solved
981742
x² + 2x + y² - 6y - 6 = 0 --- Add 6 to both sides...\nx² + 2x + y² - 6y = 6 --- Add some parentheses...\n(x² + 2x) + (y² - 6y) = 6 --- Complete the Squares*...\n(x² + 2x + 1) + (y² - 6y + 9) = 6 + 1 + 9 --- Simplify the two expressions in the parentheses, and add the numbers to the right of the "=" sign...\n(x + 1)² + (y - 3)² = 16\n\nFormula: (x - h)² + (y - k)² = r² --- Where (h, k) is the center point of the circle, and r is the radius of the circle.\n\nUsing this formula, (-1, 3) would be the center, and 4 (= √16) is the radius.\n\n\n*Completing the square is an algebra technique, also used in many types of calculus. The essential objective is to reduce a quadratic polynomial in a variable in an equation or expression to a squared polynomial of linear order. This makes the equation easier to solve in many contexts. More information can be found here: http://en.wikipedia.org/wiki/Complete_the_square
145890
9x(2)-63xy-6x+42y which is (x-7) (9x-6)
11448
Substitution Method\nSolve this system of equations:\n3y - 2x = 11\ny + 2x = 9\n \n\n1.  Solve one of the equations for either "x =" or "y =".\nThis example solves the second equation for "y ="\n \n\n3y - 2x = 11\ny = 9 - 2x\n \n\n2.  Replace the "y" value in the first equation by what "y" now equals.  Grab the "y" value and plug it into the other equation. \n\n \n\n3(9 - 2x) - 2x = 11\n \n3.  Solve this new equation for "x".\n \n\n(27 - 6x) - 2x = 11\n27 - 6x - 2x = 11\n27 - 8x = 11\n-8x = -16\nx = 2\n4.  Place this new "x" value into either of the ORIGINAL equations in order to solve for "y".  Pick the easier one to work with!  \n\n \n \n\ny = 9 - 2(2)\ny = 9 - 4\ny = 5\n \n\n5.  Check:  substitute x = 2 and y = 5 into BOTH ORIGINAL equations.  If these answers are correct, BOTH equations will be TRUE!\n   \n\n3y - 2x = 11\n3(5) - 2(2) = 11\n15 - 4 = 11\n11 = 11\n\ny + 2x = 9\n5 + 2(2) = 9\n5 + 4 = 9\n9 = 9 \n\nElimination:\n x - 2y = 14\nx + 3y = 9\n \n\na.  First, be sure that the variables are "lined up" under one another.  In this problem, they are already "lined up".\n \n\nx - 2y = 14\nx + 3y = 9\n \nb.  Decide which variable ("x" or "y") will be easier to eliminate.  In order to eliminate a variable, the numbers in front of them (the coefficients) must be the same or negatives of one another.  Looks like "x" is the easier variable to eliminate in this problem.\n  \n\nx - 2y = 14\nx + 3y = 9\n \n\nc.  Now, subtract to eliminate the "x" variable.  \n(Remember:  when you subtract signed numbers, you change the signs and follow the rules for adding signed numbers.)\n \n x - 2y = 14\n-x - 3y = - 9    \n\n    - 5y = 5  \nd.  Solve this simple equation.   \n\n-5y = 5\n   y = -1\n \n\ne.  Plug "y = -1" into either of the ORIGINAL equations to get the value for "x".\n   \n\nx - 2y = 14\nx - 2(-1) = 14\nx + 2 = 14\nx = 12\n \n\nf.  Check:  substitute x = 12 and y = -1 into BOTH ORIGINAL equations.  If these answers are correct, BOTH equations will be TRUE!\n   \n\nx - 2y = 14\n12 - 2(-1) = 14\n12 + 2 = 14\n14 = 14  \n\nx + 3y = 9\n12 + 3(-1) = 9\n12 - 3 = 9\n9 = 9
1053403
Write it as:\n2x^2-4x+3x-6\n=2x[x-2]+3[x-2]\n=[2x+3][x-2]\nSimple middle-term factorisation!!!
1031270
x^2 + 5 + 5x + x y +5y\n\nx^2 + xy + 5x + 5y + 5\n\nx(x+y) + 5(x+y) + 5\n\nx(x+y) + 5(x+y+1)       OR     (x+5)(x+y)+5\n\nCynthia
1091081
64x^6 - 576x^5/y^2 + 2160x^4/y^4 - 4320x^3/y^6 + 4860x^2/y^8\n- 2916x/y^10 + 729/y^12\nIt's the same as mine except I put the y's in the denominator...but that doesn't matter.
232592
2x = 3y - 4y + 5 - 2\n2x = -y + 3\n2x + y = 3
1169467
y = x^2 + 5x - 3\n2y=3x - 2\n----------------\n2y = 2x^2 + 10x - 6\n2y = 3x - 2\n-----------------------\n    0 = 2x^2 + 7x - 4\n=>2x^2 + 7x - 4 =0\n=>2x^2 + 8x - x - 4 = 0\n=>2x(x + 4) -1(x+4) = 0\n=> (2x - 1)(x + 4) = 0\ntherefore 2x - 1 = 0 => x = 1/2 or \nx + 4 = 0 or x = -4\nx = - 4 or 1/2  \ngot it?
586962
(6y^2-3y)(y+7)-->(2y)(y-1)(y+7)\nif that helps
868484
-12x^2*3y / -21xy^4\n= -36x^2 y / -21xy^4\n= 12x / 7y^3
825955
it is really hard to do the math.. without actually seeing it bc im not sure what the problem actually is\n\n(2y^2)+((y-6)/(2y^2-9y+9xy^2 -2y -3 /y^2-1)\n\nThis is what you have and it doesnt make sense. you need to type it out better ashow perentaces and where the new sections start for example.\n\n((2x^2+3y)/2x) + ((3y-2x)/y^2)=