solve: x^2 + 4y^2= 20   xy=4?
612878
xy = 4 then y = x/4\n\nx^2 +  4y^2 = 20\nx^2 + 4 x^2/16 = 20\nx^2 + x^2/4 = 20\n4x^2 + x^2 = 80\n5x^2 = 80\nx^2 = 80/5\nx^2 = 16\n\ntherefor\nx = 4\nthen \ny=1
859008
x+8y = 60, 4x+4y = 60\n\nsubtract: first equation minus twice second equation\n     x  +  8y  =    60\n  -8x  -  8y   = -120\n  -7x = -60\nx=60/7\n\nThen plug x into one of the equations, say the first one\n60/7 +8y = 60\n8y = 60 - 60/7\n8y = 360/7\ny=45/7
249211
Quickest way to solve it- take the first equation and square it, giving you \nx^2+y^2-2xy=25, while the other equation gives you\nx^2+y^2    =15\n\nSo xy = -5 (if that's not immediately obvious, take the difference of the two equations above and solve for xy).\n\nA word of advice on the side: there are certain quadratics that show up all the time on the SAT, such as (x+y)^2, (x-y)^2, and x^2-y^2. It's worth your while to become intimately familiar with these guys, so that you can expand and factor them immediately.
421777
3x + 4y = 12\n\n(0,3)\n(3,(3/4))\n(4,0)\n((8/3),1)
260457
20 - - 4 = 24\nminus and minus = plus\nie\n20 - - 4 = 20 + 4 = 24
1148686
(2y-8)(y-4)\n\n2y^2-8y-8y+32\n\n2y^2-8y^2+32
239058
You have to solve for one and substitue into the other...\n\nLets solve for x...\n\n2x + 3y = 40 \n2x = 40 - 3y\nx = (40 - 3y)/2\n\nThen Substitute in the other.\n-2x + 2y = 20\n-2(40 - 3y)/2 + 2y = 20  (the 2's in the first set cancel becaue 2/2 = 1 so...\n-(40-3y) + 2y = 20  (then distribute the -)\n-40 + 3y + 2y = 20 (combine like terms)\n5y = 60\ny =12\n\nThen sub y into the equation to find x\n\n 2x + 3y = 40 \n2x + 36 = 40\n2x = 4\nx=2\n\nDouble Check in the other equation...\n-2x + 2y = 20\n-2(2) + 2(12) = 20\n-4 + 24 = 20\n20 = 20 These equal so x=2 and y=12!
952496
Wasn't it 2/y^2-16 = 10/y^2+4y?\n\nIf so:\n\n2/y^2 - 16 = 10/y^2 + 4y\n10 (y^2 - 16) = 2 (y^2 + 4y)\n5 (y^2 - 16) = y^2 + 4y\n5y^2 - 80 = y^2 + 4y\n5y^2 - y^2 - 4y = 80\n4y^2 - 4y = 80\n4 (y^2 - y) = 80\ny^2 - y = 80/4\ny^2 - y = 20\ny^2 = 20 + y\n\nso now, we have this:\n\n2/y^2 - 16 = 10/y^2 + 4y (where we can replace y^2 with 20 + y)\n10 (y^2 - 16) = 2 (y^2 + 4y)\n10 (20 + y - 16) = 2 (20 + y + 4y)\n5 (20 + y - 16) = 20 + y + 4y\n100 + 5y - 80 = 20 + 5y\n20 + 5y = 20 + 5y
670230
x^2+4x+Y^2-8Y+16=0\nX^2+4x+Y^2-8Y=-16\nX^2+4x+4+Y^2-8Y+16=-16+16+4\nX^2+4x+4+Y^2-8Y+16=4\n(X+2)^2+(Y-4)^2=4\nR=Square root 4\nR=2
389436
5x4+4+4+5+4/2\n=20+4+4+5+2\n=35
264673
Let x = # of 20's\ny = # of 50's\n\nx + y = 32\n20x + 50y = 1000 or\n2x + 5y = 100\n\nfrom the first equation, we know that x = 32 - y. Simply substitute this to the remaining equation and solve for the value of y.\n\n2(32 - y) + 5y = 100\n64 - 2y + 5y = 100\n3y = 36\ny = 12\n\nThe answer is 12 ^_^
868484
-12x^2*3y / -21xy^4\n= -36x^2 y / -21xy^4\n= 12x / 7y^3
401465
(4x + 5y)(2y^2)\n8y^3 + 10y^3\n\na.)\n(4x + 2y^2)(5y)\n20xy + 10y^3\n\nb.)\n(4x) + (5y)(2y^2)\n4x + 10y^3\n\nc.)\n4x + (5y + 2y^2)\n4x + 5y + 2y^2\n\nd.)\n(2y^2)(4x + 5y)\n8xy^2 + 10y^3\n\nSo the answer is d.) (2y^2)(4x + 5y)\n\nI just did the math out.
90743
Well you must have two equations! If you wanted to say x+4y=16 then just add the two and you'll fing x (4x=10) and then find y from one of them.
846702
-x + 2y = 18\n x + 8y = -38\n\nadd up these two lines\n\n0x+ 10y = -20\ny= -20/10 = -2\n\nx + 8(-2) = -38\nx - 16 = -38\n\n\nx=-22\ny=-2
751075
Yes, you've got half the equation right. It should look like this:\nx = n+20
233924
48x^2y+20xz+20xy+5z\nI switched the position of the two middle terms.\n48x^2y+20xy+20xz+5z\n4xy(12x+5)+5z(4x+1)