what is x in the algebraic equation x=x+9.1?
112971
There's a problem with equation. Go back and check it.\nAs it stands, no x would satisfy this.\nInfity is an abstraction, not a number.
208014
Factorising gives\n\n(x-4)(x+1) = 0\n\nThe solutions are therefore x = +4 and -1
16389
x+9 < 16\n\nadd -9 for all\n===============> x+9 -9 < 16 -9    (I)\nx+9 -9 =x (II)\n16 -9 = 7 (III)\n\n(I) & (II) & (III)\n===================> x < 7\n\nif x is a natural number => \nx= 1 or 2 or 3 or 4 or 5 or 6
660785
i don't fucking know.\n\nEDIT: waitaminnit i'm gonna give it a try!\n\nx^3-4x^2-7x+10 = 0\n\nfor x = -1\n-1 -1 +7+10 != 0\nso x != -1\n\nfor x = 1\n1-1-7+10 != 0\nso x != 1 either!! ...goshdarnit.\n\nx^3-4x^2-7x+10 = 0\nx(x^2-4x) - 7x = -10\nx^2-4x - 7 = -10/x (we divided both sides by x)\n..ok...i think i've lost the plot sorry.
672081
Answer: x = 1 or 4\n\nEXPLANATION:\nx^3 - 6x^2 + 9x - 4 = 0\nThis can be written as:\nx^3 - x^2 - 5x^2 + 5x + 4x - 4 = 0\nor x^2(x-1) - 5x(x-1) + 4(x-1) = 0\nor (x-1)(x^2 - 5x + 4) = 0\nor (x-1)(x^2 - 4x - x + 4) = 0\nor (x-1){x(x-4) - 1(x - 4)} = 0\nor {(x-1)^2}(x-4) = 0\nTherefore either x = 1 or x = 4
904672
let u = x^2 + 1, then you have du = 2xdx, => dx = 1/2x du,\n\nsubstitute back you are on the way  to solve it
175133
2x-1(xsquared-x+10)squared
572480
breaking it apart, we have:\n\n(x+1)^3=(x+1)(x+1)(x+1)=\n(x^2 + 2x + 1)(x+1) =\n(x^3 + 3x^2 + 3x + 1) SO\n6(x+1)^3 = 6(x^3 + 3x^2 + 3x + 1) = 6x^3 + 18x^2 + 18x + 6\n\n(x+1)^2 = (x+1)(x+1)=\nx^2 + 2x + 1   SO\n-3(x+1)^2 = -3(x^2 + 2x + 1)=\n-3x^2 - 6x - 3\n\n4(x + 1) = 4x + 4\nand -9 stays the same.\n\nSo putting everything together, you have\n(6x^3 + 18x^2 + 18x + 6) + (-3x^2 - 6x - 3) + (4x + 4) - 9 \ncombine like terms and you get:\n6x^3 + (18x^2 -3x^2) + (18x - 6x + 4x) + (6 - 3 + 4 - 9)\n6x^3 + 15x^2 +  16x - 2\n\nWHEW!  That took me back to college pre-calc!
681319
-18=x-9\n+9   +9\n-9=x\n\n27=x-11\n+11   +11\nx=38\n\n15=x-16\n+16  +16\nx=31
87719
x^2 + 9x + 10x + 90\n\n(x + 9)(x + 10)
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).
830946
c)\n\n= x^3 + x^2 + x - x^2 - x -1\n= x^3 -1
162293
On both equations, you have x^ but don't say what x was raised to. I'm assuming that it is x^2. Is this correct?
304206
the clue would be to simplify the equation  to \nlim {x +[-(x+1)+1]}/(x+1)\n x-->0 \nlim [x + 1 -x]/(x+1)\nx-->\nlim (1/(x+1)) =1\nx-->0
807939
= (1 + 8x)*(1 - 8x) = 1 - 64*x^2\n\nAnd what do you say?
518741
here by simple trial u can see that x=1 satisfies the equation so be sure x=1 is indeed a factor. \nnow when we know that x=1 is factor it means (x-1)=0 devides it compleately. now we can decide other values of x easily. here is a technique which i think will be helpfull:\nrewrite x^3+x^2-2 as (x^3 - x^2) + (2(x^2)-2x) + (2x-2)\nnow it is simple to take (x-1) common from all the terms\nie.\n(x-1)(x^2+2x+2)\n\n# this technique is very useful once u master the techniq \n\nfor more mail me at sush_smvdu@yahoo.co.in
1013726
There is no real value for x that satisfies that equation.  It simplifies to (x-3) = (x+1).  In other words, (4(x-3))/4 does not = x+1.