solve the following system of equations: 2x+2y+z=0, 3x+2y+z=0, 2x+3y+z=0?
112444
Any method of solving will give x=0 y=0 and z=0 !
250421
Solutions do exist, it’s just possible you got your numbers mixed up (it happens to the best of us)\n\n2x-3y+z=5  --------(1)\n\nx+4y-2z= -3 --------(2)\n\n4x-2y+3z= 6 ---------(3)\n\nConsider (1) . let’s make z the subject of the formula to eliminate it\n\n2x-3y+z= 5\n\nTherefore z= 5+3y-2x    [by rearranging]\n\nYou can substitute z=5+3y-2x in (2) and (3), I’ll put it in (3) first, the order doesn’t matter\n\nTherefore 4x-2y+3(5+3y-2x)= 6\n\n4x-2y+15+9y-6x= 6\n\nSimplifying this you get 2x-7y= 9\n\nPut z=5+3y-2x in (2)\n\nSo  x+4y-2(5+3y-2x)= -3\n\nx+4y-10-6y+4x+-3\n\nsimplifying, you get 5x-2y= 7\n\nso now we have two new equations in terms of x and y, I’ll label them (4) and (5)\n\n2x-7y= 9 ---------(4)\n\n5x-2y= 7 ---------(5)\n\nWe have to solve these simultaneously now, consider (5)\n\n5x-2y= 7\n\nMaking x the subject of the formula we get\n\nx= (7+2y)/5\n\nput this new equation in (4)\n\n2* (7+2y)/5- 7y=9\n\nSimplifying this you get 14-45= 31y therefore y= -1\n\nNow we use ‘backward substitution’ to find x and z, first x\n\nFrom (4) we know 2x-7(-1)=9 therefore x=1\n\nNow put x and y in (1) to find z\n\n2(1)-3(-1)+z= 5\n\nFrom this you find z is 0\n\nTherefore x=1, y= -1 and z=0\n\nP.S: I'm not sure what you mean by the addition method, I used both elimination and substitution methods here, however I hope this helps!
567748
you are almost right - or maybe the end of your answer got cut off.\ndz/dx gives\n6xz+3x^2(dz/dx)\n-(yz^3+(3z^2)xy(dz/dx)=0\n\nthen just solve this equation for dz/dx
107281
Okay, I see your two equations are 5x + 5y = -30 and \n2x - y = 15.  I'm not certain if these are right so I'll explain how to solve systems of equations in general.\nThe way we solve a system of equations by substitution is to solve one of the equations for one of the variables.  We then take what that variable is equal to and substitute it into the equation we did not yet touch.  We then solve the resulting equation.  \nThe explanation is confusing so I will provide an example.\n4x + y = 9\n-x + 3y = 1\nSo I'll solve the second equation for x like so:\n-x + 3y = 1\n-x = 1 - 3y\nx = 3y -1\nThen, I substitute 3y - 1 into the first equation for x, since this expression is equal to x.\n4x + y = 9\n4(3y-1) + y = 9\n12y - 4 + y = 9\nAdding four to both sides and combining like terms I get:\n13y = 13\nTherefore y = 1.\nWe can then solve for x using one of the original equations, we'll pick 4x + y = 9.\n4x + y = 9, and we know y=1 so:\n4x + 1 = 9\nSubtracting 1 and solving gives us\n4x = 8, and therefore x = 2.\nYou can use this same process to solve your problem.  I suggest starting by solving the second equation for y.\nHope this helps, please remember to come back and choose a best answer!  :-)
258820
4x^2 -3x + 2 = 0 \nD = b^2 - 4ac = 9 - 4*2*4= 9- 32 = - 23<0\nThere are no real solutions, but there are 2 complex solutions.
164756
Easy!\n\n2x+2y=2 \n3x-y=1\n\nTake the second equation and isolate the single "y":\n\n3x - y = 1 becomes 3x - 1 = y (without going into how)\n\nSubstitute "3x - 1" in the first equation so it reads:\n\n2x+2(3x - 1)=2\n\nSimplify:\n\n2x + 6x - 2 = 2\n8x - 2 = 2\n8x = 4\nx = 1/2\n\nEarlier, we saw that y = 3x - 1, so:\n\ny = 3(1/2) - 1\ny = 3/2 - 1\ny = 1/2\n\nThere may be another way (less involved) that I've forgotten, but you can check each of those values for x and y in both equations to make sure they work.
695572
a) (x+4y)(x+y)=0    x=-y or -4y\n\nb) (2x+y)(x-3y)=0   x=-y/2 or 37\n\nc) (2x-y)(2x-y)=0  x=y/2\n\nd) 4x^2+7xy-2y^2=0\n    (4x-y)(x+2y)=0    x=y/4 or -2y\n\ne) x(5x+y)=0   x=0 or -y/5\n\nf) x(3x-7y)=0   x=0 or 7y/3
769967
removing the brackets\n18+3y=4y-16-y\nsimplifying\n18+3y=3y-16\nadding 16\n34+3y=3y\nadding -3y\n34=0\nwhich is absurd\nso the equation cannot be solved
702715
I know this can be done with matrix operations, but as I understand it, it represents the following system of equations which can be solved using junior algebra methods:\n\n-10x + 11y - z = -3\n14x  +  9y + 8z = 2\n12x  - 2y - 11z = 3\n\nAdding 8*first equation to second, gives \n-66x + 97y = -22\nSubtracting 11*first from third gives\n122x - 123y = 36, and I begin to see why one might try something different!\n\nThese last two give \nx = 393/1858\ny = -77/929  and then sub in the first to get\nz = -50/929\n\nPersonally, I don't believe it, but that's what I got.  How about you?
566070
0 = 2y + 2z\n\nz and y must be opposites and equal
11448
Substitution Method\nSolve this system of equations:\n3y - 2x = 11\ny + 2x = 9\n \n\n1.  Solve one of the equations for either "x =" or "y =".\nThis example solves the second equation for "y ="\n \n\n3y - 2x = 11\ny = 9 - 2x\n \n\n2.  Replace the "y" value in the first equation by what "y" now equals.  Grab the "y" value and plug it into the other equation. \n\n \n\n3(9 - 2x) - 2x = 11\n \n3.  Solve this new equation for "x".\n \n\n(27 - 6x) - 2x = 11\n27 - 6x - 2x = 11\n27 - 8x = 11\n-8x = -16\nx = 2\n4.  Place this new "x" value into either of the ORIGINAL equations in order to solve for "y".  Pick the easier one to work with!  \n\n \n \n\ny = 9 - 2(2)\ny = 9 - 4\ny = 5\n \n\n5.  Check:  substitute x = 2 and y = 5 into BOTH ORIGINAL equations.  If these answers are correct, BOTH equations will be TRUE!\n   \n\n3y - 2x = 11\n3(5) - 2(2) = 11\n15 - 4 = 11\n11 = 11\n\ny + 2x = 9\n5 + 2(2) = 9\n5 + 4 = 9\n9 = 9 \n\nElimination:\n x - 2y = 14\nx + 3y = 9\n \n\na.  First, be sure that the variables are "lined up" under one another.  In this problem, they are already "lined up".\n \n\nx - 2y = 14\nx + 3y = 9\n \nb.  Decide which variable ("x" or "y") will be easier to eliminate.  In order to eliminate a variable, the numbers in front of them (the coefficients) must be the same or negatives of one another.  Looks like "x" is the easier variable to eliminate in this problem.\n  \n\nx - 2y = 14\nx + 3y = 9\n \n\nc.  Now, subtract to eliminate the "x" variable.  \n(Remember:  when you subtract signed numbers, you change the signs and follow the rules for adding signed numbers.)\n \n x - 2y = 14\n-x - 3y = - 9    \n\n    - 5y = 5  \nd.  Solve this simple equation.   \n\n-5y = 5\n   y = -1\n \n\ne.  Plug "y = -1" into either of the ORIGINAL equations to get the value for "x".\n   \n\nx - 2y = 14\nx - 2(-1) = 14\nx + 2 = 14\nx = 12\n \n\nf.  Check:  substitute x = 12 and y = -1 into BOTH ORIGINAL equations.  If these answers are correct, BOTH equations will be TRUE!\n   \n\nx - 2y = 14\n12 - 2(-1) = 14\n12 + 2 = 14\n14 = 14  \n\nx + 3y = 9\n12 + 3(-1) = 9\n12 - 3 = 9\n9 = 9
850894
2x-3y=5 ….multiply by 1\n                           x+2y=-l .... multiply by -2\n\n                                       2x-3y=5 \n                                      -2x-4y=2 …..add two equations\n                                      ------------\n                                        -7y= 7\n                                          y=-1\n\n substitute y=-1 in the second equation coz it is easier in order to get the x value\n\nx+2y=-1\nx-2=-1\nx=1\n\nSo the solution is (1,-1)
275027
Solve for y in the first equation and then substitute into the second equation.\n4x + 2y = 0  Subtract 4x from each side\n2y = -4x  Divide both sides by 2\ny = -2x\nSubstitute -2x for y in the equation 6x + 3y = 0\n6x + 3(-2x) = 0\n6x - 6x = 0\n0 = 0\nThis means there are infinitely many solutions of the form (x,y) where x is any real number and y = -2x.\nTherefore, the answer is\n(x,-2x), where x is any real number.
406322
q*(x,y,z)X(a,b,c) = q(yc - zb, az - xc, xb - ya). In your case, B is (0, 0, -1.2) so you have -5nC(-1.2y, 1.2x, 0) = (-3.6*10^(-7), 7.6*10^(-7), 0) which is a trivial system of equations.
783856
First, you should know that you can multiply any one of these equations through by a nonzero constant, without changing the solution. You can also add or subtract any two of the equations term by term without changing the solution.\n\nWe begin by adding equations 1 and 3 together\nx+y+z = -5\nx-y+4z=12\n---------------\n2x + 5z = 7  (notice that the ys cancel out)\n\nNow we triple equation 1, and subtract equation 2 from it\n\n3x+3y+3z= -15\n2x+3y-2z = 8\n--------------------\nx +5z = -23\n\nNow we have two equations, in which we have eliminated y.\nWe subtract them, and solve for x.\n\n2x + 5z = 7 \nx    +5z = -23\n-----------------\nx = 7 - (-23) = 30\n\nIt is easy to find y and z, now that we know x is 30/\n\nx + 5z = -23\n30 + 5z = -23\n5z = -53\nz = -(53/5) or -10.6\n\nx + y + z = -5\n30 + y + (-53/5) = -5\ny = 53/5 - 30 - 5 = -24.4\n\nSo (x,y,z) = (30,-24.4,-10.6)
166772
2x - 3y = 5        \n2x^2 + 3y^2 = 35 \n\n2x = 5 + 3y\nx = (5 + 3y)/2\n\n\n2 [(5 + 3y)/2]^2 = 5 + 3y\n2 [(25+30y+9y^2)/4] = 5 + 3y\n25+30y+9y^2 = 10+6y\n9y^2-24y+15=0\n(3y-3)(3y-5)=0\n\ny=1; y=5/3\n\nif y=1\n2x - 3(1) = 5\nx = 4\n\nif y = 5/3\n2x - 3(5/3) = 5\nx = 5
496037
-3x + 5y = 8\nx + 2y = 1\n\n(-3x + 5y) + (x + 2y) = 8 + 1\n-3x + 5y + x + 2y = 9\n-2x + 7y = 9\n\n7y = 2x + 9\ny = (2/7)x + (9/7)\n\nx + 2y = 1\nx + 2((2/7)x + (9/7)) = 1\nx + (4/7)x + (18/7) = 1\n7x + 4x + 18 = 7\n11x + 18 = 7\n11x = -11\nx = -1\n\ny = (2/7)x + (9/7)\ny = (2/7)(-1) + (9/7)\ny = (-2 + 9)/7\ny = (7/7)\ny = 1\n\nx = -1\ny = 1