Hi! I'm in algebra 1 and I just can't figure out this problem! Please help... thanks =) yx^2-30x+8?
364430
Did you mean to have the yx^2-30x+8 as your polynomial (a polynomial in two variables?)\nFor y=7   or 7x^2-30x+8\n\n7x^2-30x+8 = 7x^2-28x-2x+8\n\n7x^2-28x-2x+8= 7x(x-4)-2(x-4)\n\n7x(x-4)-2(x-4)= (7x-2)(x-4)
382994
So...\n\n2x + 35 = (x-2) + x + (x+2)?\n-2+2 makes zero, so you're left with \n2x + 35 = 3x.\nSubtract 2x from both sides.\n35 = x.\n\nSo, 33 + 35 + 37 equals 2(35) + 35. They are both 105. Is that the way your are learning it? I don't remember how I was taught to do that last year, but I think that's how.
120172
y-2 = -2(x-3)\ny-2 = -2x + 6 \ny+ 2x = 6+2 \ny+2x  = 8  ( linear equation)\n\nFINDING SOLUTIONS FOR THIS LINEAR EQUATION :\n\ny+2x = 8\n(a)  Assume, x=2 \n     subtitute the assumed value of x in the equation\n     y+ 2 (2) = 8\n     y+4 = 8\n     y = 8-4\n     y = 4\n     HENCE,  when -->   x= 2 then y= 4\n\n   Similarly you can find more solutions for any\n   given linear equation by assuming the value of any one unknown ( x or y ) and then finding the value of the other by substitution in the same given equation.
431382
let the 2 numbers be x and y :\n\nx - y=23----A\nxy = -120 ---B\nx = -120/y\n\nsubstituting x in A:\n-120/y - y = 23 \n-120/y - y2/y = 23\n-120/y - 23 = y2/y\n-120/y - 23/y = y2/y \n-120-23/y = y\n-120 - 23 = y2\n0 = y2 + 23 + 120\nsolving the quadratic equation by the formula;\ny = -b (sq.root) b2-4ac/ 2a\n\ny= -8 and x is 15\nand then 15-(-8) = 23 \n         15x(-8) = -120
899125
Q : 7x + 4 = 15x + 36\n\nfirst, move '+4' to the right side of the equation.\nand you get:\n7x = 15x + 36 - 4\n\nsecond, move '15x' to the left side of the equation.\nand you get:\n7x - 15x = 36 - 4\n\nthird, do the subtraction.\nand you get:\n-8x = 32\n\nforth, change the the left side of the equation to positive and the right side to negative.\nand you get:\n8x = -32\n\nthen, divide -32 by 8 so that you get 'x' equals to ___.\nand you get:\nx = -32/8\n   = -4 [the answer]
221619
x-4x+9+3(-2+1)+2=0\nx-4x+9+3*(-1)+2=0\nx-4x+9-3+2=0\n-3x+8=0\n-3x=-8\nx=8/3\nThis is if the equation was x-4x+9=-3(-2+1)-2, but if it was just -4x+9 = -3(-2+1)-2 than the answer is 2 (-4x=-8    x=2).
735754
just plug the numbers in.  if y=8, and x=2, \nits 8=a(2^2)\nso 8=a(4)\nso a=2.\nhope that helps!
793917
1)x+y-z=0.........(eqn. 1)\n2x-y+z=3...........(eqn. 2)\n4x+2y-2z=-2       (eqn.3)\nMultiplying eqn.1 by by 2 and subtracting it from eqn.2,we get\n-3y+3z=3...........(eqn. 4)\nMultiplying eqn.2 by 2 and subtracting it from eqn.3,we get\n4y-3z=-4...........(eqn 5)\nAdding eqn, 4 and 5 we get,y=7\nPutting the value of y in eqn 4 .we get\n-21+3z=3 or,3z=24  or z=8\nputting the value of y and z in eqn.1,we get\nx+7-8=0 \n=>x-1=0 or x=1\nTherefore,x=1.y=7 and z=8  ans\n2)x+2y=9....(eqn 1)\n 2x-3y=4     (eqn.2)\nmutiplying eqn.1 by 2 and subtracting it from eqn 2,we get\n-7y=-14 or y=2\nPutting the value of y in eqn 1,we get.\nx+4=9 or x=5\nTherefore x=5 and y=2 ans.
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.
453667
y=(x+8)^3-5\ny+5=(x+8)^3\n(y+5)^(1/3)=x+8\nx=(y+5)^(1/3)-8
810076
11^(0.35+7) = 3 this is a contradiction\n\nx=3 is a solution but the rest is just a mess.\n\nDo you have any idea of what you are doing?
586333
First, FUNNYPLAYA12'S answer is a partial of what I got, and my formula looked like this:\n\n-8X^3 + 10X^2 - 2X = 0\n-----------------------------------\nFactor out one X and you get:\n\n-8X^2 + 10X - 2 = 0\n\nFactor this and FOIL-check\n\n(-4X + 1)\n(2X - 2)\n___________\n-8X^2 (8X + 2X) -2\n\n-8X^2 +10X - 2\n\nX = {1/4, 1)\n\n\nTo verify:\n\n-8(1) + 10(1) -2(1) = 0\n\n-8 + 10 -2 = 0; and finally:\n\n10 - 10 = 0, which is very true!\n----------------------------------------\n\n-8(1/4) + 10(1/4) -2(1/4) = 0\n\n-2 + 2.5 - 0.5 = 0\n\n-2.5 + 2.5 = 0, which is also very true!\n\n\nOTBPOC
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
956578
x^2+3x-y=c\n2x +3 -y'=0\ny'= 2x +3\n\n(remember y^0=1)!
699742
-(3x+y)(3x-y)\n-(9x^2-y^2)\ny^2-9x^2
598932
Replace y in the first expression by the value from the second.  You get:\n8x - 4(2x - 4) = 16\n\nNow simplify:\n8x - 4*2x - 4*-4 = 16\nthat's four times minus four...\n8x - 8x - -16 =16\n16 = 16\nThis holds for any value of x.\nThis has occurred because the two expressions are the same!\nIf you divide the first expression by 4, you get:\n2x - y = 4\nnow make y the subject (add y to both sides, then take 4..)\n2x - 4 = y\n\nIf you were to substitute your solution into the second expression, you would have:\n-1 = 2*3 - 4   clearly this is not a solution...\n\nIf x were 3 then y would be 2*3-4 = 2\n\nI hope the explanation above is of some use!
636051
x=4\n\nplease solve...\n\n8-(1) :-) ;-(