2y^2 + y-6 over 2y^2 - 9y + 9  *  y^2 -2y - 3 over y^2 - 1?
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)=
223041
uh..... no. Unless I am missing something, you substitute the 3 for 'x' and the 2 for 'y' and get 3+(3 x 2) = 3 + 6 = 9
273731
120\n\n(6y-10)^2 = 36y^2 -60y -60y +100
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
1182929
7x+27 over (x+9)(x-9)
167287
= (lnx + lnx)-lny = 6+6-2 = 10
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
791199
(6x^2)/4x^2y-12xy(x^2+x-12)/3x^2+12x\n3/(2y)-4y/x(x^2+x-12)+12x\n=(3/2y)-(4x^2y-4xy+48y)/x+12x\n=[3x-2y(4x^2y-4xy+48y)+24x^2y]\n/2xy\n=[3x-8x^2y^2+8xy^2-96y^2\n+24x^2y]/2xy
935267
3x+2y=12\n3x+y=9\n(2)*-2\n-6x-2y=-18\n3x+2y=12\nadding\n-3x=-6\ndividing by -3\nx=2\nsub x=2 in 3x+2y=12\n6+2y=12\nadding -6\n2y=6\ndividing by 2\ny=3
179549
(y + 9) * (y + 9)\n\n9 and 9 add to be 18 and multiply to be 81.
778763
1) (a - b)(x - y)\n2) (x^9 + y^3)(x^9 - y^3)\n3) (6x + 5)(x + 8)\n4) (x - 9)(x - 7)
815517
OK First you take out the common factor which is y. After you have done that.you have (y^2-12y+36). After this, it becomes simple factoring. The answer is\ny(y-6)(y-6)\nor also known as y(y-6)^2
766061
p^9-q^9/p^6-q^6\n=(p^3-q^3)(p^6+p^3q^3+q^6)/(p^3+q^3)(p^3-q^3)\n=(p^6+p^3q^3+q^6)/(p^3+q^3)
531417
(a+b)(b^6-ab^5+a^2b^4-a^3b^3+a^4b^2-a^5b+a^6)\nGood Night, and Good Luck!
916997
y+9=2x+12\n\nSolve for y:\n\n y+9=2x+12\n    -9=    - 9\n----------------\ny=2x+3\n\nsolve for x:\n\ny+9=2x+12\n -12=    -12\n--------------\ny-3=2x\n       ````\n        2\n\nx=y-3\n    ````\n      2\n\nx=y-3/2    y=2x+3
789649
We'll use the "addition" (or "elimination") method, but if we were to use the substitution method, we'd get the same result.\n\nIf you multiply both sides of the first equation by 3, you'll get:\n\n9x - 9y = 27\n\nAdding the two equations together then produces:\n\n9x - 9y - 9x + 9y = 27 - 25\n\nWhen you combine like terms, you get\n\n0 = 2\n\nNow, obviously, that's not true, which means this is an "inconsistent system" -- a system without any solutions.\n\nHope that helps!
643397
(8y^3 + 27)/(2xy - 10y + 3x - 15)\n((2y + 3)(4y^2 - 6y + 9))/((2xy - 10y) + (3x - 15))\n((2y + 3)(4y^2 - 6y + 9))/(2y(x - 5) + 3(x - 5))\n((2y + 3)(4y^2 - 6y + 9))/((2y + 3)(x - 5))\nthe (2y + 3)s cancel out\n(4y^2 - 6y + 9)/(x - 5)\n\nANS : (4y^2 - 6y + 9)/(x - 5)