solve for x when 0 = 1/(4 - x)?
971477
There is no real value for X. Here's a proof:\n\n0 = 1/(4-x)\n1 = 1/(4-x) + 1  (add 1 to both sides)\n1 = 1/(4-x) + (4-x)/(4-x)\n1 = (5-x)/(4-x)\n4-x = 5-x  (multiply both sides by 4-x)\n-x = 1-x\nx = x-1\n\nClearly not true. So the original equation was false.\n\n------------------------------\n\nBy the way, several others have mentioned that the limit as x approaches infinity (or -infinity) is 0, which is true. But you don't have limits mentioned...
1144396
you would take the 2x-4y=3 and solve for a variable, such as x=(3-4y)/3   you now have what x equals and you sub that into the other equation so you now have 2[(3-4y)/3]-3y=0 and you should be able to solve that equation
321475
*(2x/(x^2-4)) + (5/2-x) - (1/2+x)\n*(2x/(x^2-4)) + (5/2-x)((2+x)/(2+x)) - (1/2+x)((2-x)/(2-x))\n*(2x/(x^2-4)) + (10+5x)/(4-x^2) - (2-x)/(4-x^2)\n*(2x/(x^2-4)) + (8+6x)/(4-x^2)\n\nyou should be able to get it from there
814311
The issue lies in the fact that the original equation is 0=0.\n\nThen, if you put in any value for x, you are dividing by 0 which as we all know, is a "no no."
760030
okay well pretty much you have to solve for  x\nso inorder to get X by itslef you have to -4 from -8 like so:\n2x + 4-4 = –8-4\n2x=-12\nnow you still have 2X but now you can divide the 2 into -12\nx= \nI won't give you the answer but you should be able to find the answer!!!
1054506
given, F(x)= x-[x]\n=> F(1/x)= 1/x- [1/x]\n\nthen equation to be solved is :\n\nx-[x]+ 1/x -  [1/x]=1\n\nor x + 1/x = 1 + [x] + [1/x]\n\nnow when x>0, [x] = x \nthen from above equation,\nx+ 1/x= x+1/x+1  this is not possible for any positive x.\n\nwhen x<0,  [x]= -x\n\nthen,\nx+ 1/x = -x -1/x +1\nor, x+ 1/x = 1/2, this has no solution such that x is negative.\n\nHENCE the above equation has no solution.
665576
x^2<4x\n\ncase 1: x<=0\nno solution\n\ncase 2: x>0\ndivide by x\nx<4\n\ntherfore the solutions are 0<x<4
618060
this would multiply out to be 4x^2-9\n\nif you let the expression be equal to 0, then x would be positive or negative 3/2 because either (2x+3) = 0 or (2x-3) = 0, and just solve for x
785246
x^2-16=0\n(x+4)(x-4)=0\nx+4=0\nx=-4\n\nx-4=0\nx=-4\n\nx^3+14x^2+40x=0\nx(x^2-14x+40)=0\nx(x-10)(x-4)=0\nx=0\n\nx-10=0\nx=10\n\nx-4=0\nx=4\n\nx^4-5x^2+4=0\n(x^2-4)(x^2-1)=0\n(x-2)(x+2)(x-1)(x+1)=0\nxi2=0\nx=2\nx+2=0\nx=-2\nx-1=0\nx=1\nx+1=0\nx=-1
146214
D.    X (greater or equal to) 4\n      2(-2)+4 = 0\nso (-infin, -2) U [4, infin)
413913
x=0,2,3,4 and y=1,-1,0,3
894400
I'm not sure what you mean by ***, so I'll answer the second part of your question.\n\n-4x/-4x + 12/-4x = 1 - 3/x
363348
Are we solving for x?
703761
4x² -x = 3\n4x² - x - 3 = 0\n(4x + 3)(x - 1) = 0\nx = -3/4, x = 1
160474
I supose u mean 1/4(4x-8)+3x=0\nThe answer is then very simple:\n\nx=1/2\nCheers
491455
(x^2+4x+1)/(x-2) = x+2+3/(x-2)\n\nx  : x^2+4x+1-x^2-2x =2x+1\n2  : 2x+1 -2x-4 =3
60538
if [ -7x + x^2 =0 ]\nthe you will get y = 7/x