How do I solve 3rd order equations?
488850
Depress the cubic to eliminate the second order term.\n\nsee\nhttp://www.karlscalculus.org/cubic.html
929849
solve three body problems
224079
try this site... it shows step by step how to go about sloving these equations, good luck:
492288
x=\n3x-3+2x=5x-6-x\n3x+2x-5x+x=-6+3\n1x=-3\nx=-3&#92;1\nx=-3
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
201908
Ah, systems.\n\nthe first problem- there are many ways you can choose to do this, but I think we should multiply both sides of the first equation by two so that they positive and negative y's will cancel out.\n\nSo now we have:\n2(2x + y)=2(4)\n3x - 2y = -1\n\nSimplified:\n4x + 2y = 8\n3x - 2y = -1\n\nnow you add down, so you get:\n7x +0y = 7\n7x =7\nx=1\n\nsubstitute 1 for x in any equation:\n2(1)+y = 4   \n2 + y = 4\ny = 2\n\n\n\nfor the second one- lets cancel out the y's again.\n\n5x - 3y = 14\n-3(2x - y) = (6)-3\n\n5x - 3y = 14\n-6x +3y = -18\n\nAdd them together:\n-x = -4\nx = 4\n\nThen solve for y:\n5x - 3y = 14\n5(4) - 3y = 14\n20 -3y = 14\n-3y = -6\ny = 2
908426
Plug in each of the four answers.  The only one that works is (2), because you get\n\n | 3 (-2/3) - 3 | = 5\n | -2 - 3 | = 5\n | -5 | = 5\n 5 = 5
1160760
what you have is something like\na(b+c)\n\nthe solution is \nab+ac\n\n(x^3)(2x^-2)+(x^3)(4x)\nnow to multiply x^a * x^b you get x^(a+b)\n\n2x^((3+(-2))+4x^(3+1)\n\n2x+4x^4
763084
or solve equation for x and y components separately. \nfor x    0 +3+?=0 \nfor y    3+0 +?=0
765649
switch the variables.\n\nStart with\n\ny = sin(0.5x-3)\n\nand solve for x.\n\nto get rid of  the sin, we apply arcsin or sin^-1 (on calculators)\n\narcsin(y) = arcsin(sin(.5x - 3))\n\nthis gives\n\narcsin(y) = .5x - 3\n\nx = 2*[arcsin(y) + 3]
718048
x + y = 3\n\nSolve for a variable.  I chose X.\nx+y = 3\n -y    -y\nx = 3 - y\n\nSubstitute the value of X in the first equation.\n2(3-y) + y = 4\nDistribute.\n6-2y+y = 4\nAdd like terms.\n6-y = 4\n-6    -6\n-y = -2\nmultiply by -1.\ny=2\n\nSince you know Y = 2, substitute that back into the second equation that you solved for X with.\nx + y = 3\nX + 2 = 3\n  -2    -2\nX = 1\n\nX = 1\nY = 2
999333
There are several simple rules for re-arranging and solving equations.  It would be in your best interest to try and learn them, they aren't very hard.
826575
Use one equation to substitute into the other,then solve for a one variable problem.  Use your new information (the solved variable) to solve for the second variable.\n\nFor the first set:\n\n-3x - y = 2 rearranged becomes:\ny = -3x -2\n\nUsing this "definition" of y, substitute it into the second equation as such:\n\n-6x + 2(-3x -2) = -2\n-6x - 6x - 4 = -2\n-12x = 2\nx = -1/6\n\nSubstitute -(1/6) for x in either equation to solve for y:\n\n-3(-1/6) - y = 2\ny = -3/2\n\nDo the same for the second pair of equations to make sure you got it (the answers for x and y are x=9/7, y = 1/7).
904685
1)auxilary equation\n\nL^2+3L=0\n\n>>>>L=0,L= -3\n\ncomplementary function\n\nyc= C+De^(-3x)\n\n2)find a particular integral for\n                  d^2y/dx^2 +3dy/dx= x^2-2x\n\ntry y= ax^3+bx^2+cx...(1)\n\ndy/dx=3ax^2+2bx+c\n\nd^2y/dx^2= 6ax+2b\n\nsubstitute these into LHS of DE\n\n6ax+2b+3(3ax^2+2bx+c)=x^2-2x\n9ax^2+6(a+b)x +(2b+3c)=x^2-2x\n\nequating coefficients\n\n9a=1 >>>a= 1/9, 6(a+b)= -2\n         >>>b= -4/9 \n2b+3c=0 >>>>>>c=8/27\n\nsubstitute these values into (1)\n\nparticular integral\n\nyp=(1/9)x^3-(4/9)x^2+(8/27)x\n\n3)general solution\n\ny=yc+yp\n\n  =C+De^(-3x)\n    +(1/9)x^3-(4/9)x^2+(8/27)x\n\ni hope that this helps
145305
Use cubic equation \ny=ax3+bx2+cx+d\n\nSince you have four points data you can get four linear equations with four variables which is very easy to solve, get variables (a,b,c,d) and plug the values to cross check and see u will get answer. Use guass jordan or elimination method.
162979
From y=-3x-3 and y=x+5 you have -3x-3=x+5 so\n-3x-x=5+3\n-4x=8\nx=-2\nNow y=x+5=-2+5=3\nThe solution is (-2,3).
634747
What you want is express \n\nv = a*v1 + b*v2 + c*v3 \n\nWhere a, b and c are, of course, scalars. The coordinate vector will then be (a, b, c).\n\nNow how do we do this: \n\n(2, -1, 3) = a(1, 0, 0) + b(2, 2, 0) + c(3, 3, 3) \n\nEither you just look at it and try to guess what a b and c can be, or you write this up as a set of 3 equations in a,b and c: \n\n2 = 1a + 2b + 3c\n-1 = 0a + 2b + 3c\n3 = -0a + 0b + 3c \n\n...and then solve it: \n\nFrom the last equation we see that c = 1, then using the second equation, -1 = 2b + 3  <=>    -4 = 2b  <=>  -2 = b\n\nAnd substituting into the first, \n\n2 = 1a + 2(-2) + 3(1) = a - 4 + 3 = a - 1 \n\nthus 3 = a. \n\nSo the coordinate vector is (3, -2, 1) and with these numbers the criteria are (supposedly) satisfied. :)