algebra solvinng equation qn3?
93711
First, distribute the one.  Now you have x+3.  Then, distribute the two.  Now you have -2+3x.  So, now your equation reads:  x+3-2+3x.  Combine like terms.  x+3x = 4x, 3-2=1.  Now, you have 4x + 1 = 0.  Subtract the one.  Now, you have 4x = 1.  Now, divide by 4 to get the x by itself.  You're left with x = 1/4.
681679
The solution involves complex numbers ('i' notation)\n\nThe factorized equation is:\n\n(b + 3 - i)(b + 3 + i)\n\nwhere 'i' is the imaginary complex number = square root of -1\n\n(TOOK ME WHOLE OF 15 MINS!!! MY MATH IS SLIPPING!!!)\n\neinsteinbrainiac - you came close to the solution
1100359
1)If it's an equation then solve it quadratically\n\n(X-3)= 0 or (X-3)= 0\n\ntherefore X= 3 (as the 3 moves to the other side making it positive)\n\n2)If you have to find the products ude the distributive law:\n\nXsquared -3X-3X+9\ntherefore the answer would be Xsquared -6X+9\n\nBut I'm sure you needed the Equation one so use my first answer and method.
886791
(px+2)(3x+q) = rx ^2 + x - 2\n\n(3p)*x^2 + (pq+6)*x + 2q =  rx ^2 + x - 2\n\nComparing coefficients on both side: we have\n\n2q = -2 , q = -1\n\npq+6 = 1, p(-1) + 6 =1 , p = 5\n\n3p = r , r = 3*5 = 15\n\nTherefore the solution is p = 5, q = -1 and r = 15\n\nHope it helps!
821200
(x²+4x+4) / (2x-3) / (x²-4) / (2x²-7x+6)\n\nYou have 4 terms and three divisions, so it's a little unclear if you mean (a/b) / (c/d) or what.  I will try to help to the extent I can.\n\nLet's assume you mean:\n\n[ (x²+4x+4) / (2x-3) ] / [ (x²-4) / (2x²-7x+6) ]\n\nFactor your x² equations to see if you can cancel terms:\n\nx²+4x+4 = (x+2)(x+2)\nx²-4 = (x+2)(x-2)\n2x²-7x+6 = (2x-3)(x-2)\n\nWoohoo - things are looking good!  Now we can rewrite our original problem using these factors:\n\n[ (x²+4x+4) / (2x-3) ] / [ (x²-4) / (2x²-7x+6) ] = \n\n[ (x+2)(x+2) / (2x-3) ] / [ (x+2)(x-2) / (2x-3)(x-2) ] =\n \n[ (x+2)(x+2) / (2x-3) ] * [ (2x-3)(x-2) / (x+2)(x-2) ] =\n\nnotice we can cancel out three terms: (2x-3), (x-2) and (x+2)\n\n[ (x+2)(x+2) / (2x-3) ] * [ (2x-3)(x-2) / (x+2)(x-2) ] =\n\n(x+2)\n\nHope that helped!\n\n~ ♥ ~
968026
I got   3(x² - 5x + 4) or 3( x - 4 )( x - 1 )\n\nLet me show you my work...\n\n3( x - 2 )² - 3(x - 2) - 6\n\nSimplify\n\n3( x -2 )( x - 2 ) - 3( x- 2 ) - 6\n\nDistribute\n\n3( x² - 4x + 4) - 3x + 6 - 6\n\nDistribute and Simplify\n\n3x² - 12x + 12 - 3x\n\nSimplify\n\n3x² - 15x + 12\n\nYou can now simplify by factoring out 3\n\n3(x² - 5x + 4)\nor\n3( x - 4 )( x - 1 )
25946
=3fg*(f^2g)+3fg*5f-3fg*4\n=3(f^(1+2g))*g+15(f^2)g-12fg
866906
q = (5r-3) / (4r-1) multiply by 4r-1\n4qr-q=5r-3 add q\n4qr=5r-3+q    subtract 5r\n4qr-5r=q-3 factor out r\nr(4q-5)=q-3   divide by 4q-5\nr=(q-3)/(4q-5)
501220
Are you sure it is x^3 and not x^2? If you're sure, then there is one relatively easy way to solve the equation:\n\n(x^3)-4x=24\n\nNow, this equation is of the form: x^3+px=q (where p=-4, q=24). The solution to this equation was given by the Italian mathematician "Cardan" to be:\n\nx=(sqrt((p^3)/27 + (q^2)/4) + q/2)^(1/3) - (sqrt((p^3)/27 + (q^2)/4) + q/2)^(1/3)\n\n\nFor your case, if you plug in the values for p,q you will get:\n\nx = (√(-64/27+144)+12)^(1/3) - (√(-64/27+144)-12)^(1/3)\n\nx = 2.888 - (-0.4629) = 3.3434\n\n\nThus, x=3.3434 is the real solution to the equation.
794971
3(3q-1)=-(9-q)\n9q-3=-9+q\n-3+9=q-9q\n6=-8q\n6/-8=q\nq=- 3/4
194716
x^2 + 13x + 52 + 160/(x-3)
685700
First convert into a matrix form which would be\n\n1    1    1\n1    2    3\n1    3    7\n\np\nq\nr\n\n= \n\n1\n4\n13\n\nTake the square matrix (3x3) on the other side so u ll need its inverse (i have directly calculated the inverse which is)\n\n5/2       -2      1/2                        \n-2         3      -1                           \n1/2       -1     1/2   \n\nMULTIPLY BY    \n\n1\n4\n13\n\n& the result is\n\n\n 1        p\n-3        q\n 3     =  r\n\nthis is the answer of the equation
234829
With factorisation you turn the equation x^2 + px + q = 0 into two smaller equations in brackets like this:\n\n(x+a)(x+b) = 0\n\nSo that a+b = p\nand a*b = q\n\nEG. Your first example you have p = 4 and q = 3\n\nSo you've got a+b = 4 so a could be 2 and b 2 or a 1 and b 3 etc\n\nand a*b = 3 so a could be 1 and b 3... Yes it matches.\n\n(x+3)(x+1) = 0\n\nThe solution is just x = -a and x = -b\n\n-> x = -1 or x = -3
850884
using chain rule and product rule\n\ny'= 3x^(2)*tan(4x)\n         +4x^(3)*sec^(2)(4x)\n  = x^2(3tan(4x)+4x*sec^(2)(4x) \n\ni hope that this helps
561262
u jammed the equation together.\ndo u mean\n6x^3 + ( 6/x - 2 * x^2 - x/3x2 - 3x + 3)\nits really hard to see ur question. can u write it out with spaces...\nhurry up i am waiting... lol
265295
q=(1/3j)/H or q=j/(3H)
892382
To solve the first one, divide each part of the first equation by two, so 2x-3y=-13; since -2x+3y = 13, adding the two equations gives 0x+0y=0.  No help.  Try substitution.\nFrom the first equation, 4x = 6y-26; x = 1.5y - 6.5\nsubstitute into the second equation -2(1.5y-6.5) + 3y = 13; -3y +13 + 3y = 13; 13=13.  Still no help.\nBoth equations are equivalent; the line that represents them is the solution.\n\nProblem two:\nSubtract equation 2 from equation 1.\n3x-3x+2y-3 = 6 - y\n2y-3 = 6-y\n3y = 9\ny = 3\n3x+6=6\nx=0\nchecking\n3x+3=y=3\n3x=0\nx=0