Find x,y,z, and w by the GAUSE- JORDAN method?
952689
5x+3y+z+5w=-2\n3x+2y+2z+4w=2\n2x+4y+3z-3w=-11\n4x-3y-2z+2w=-3\n\n=>\n5     3     1     5   I  -2\n3     2     2     4   I   2\n2     4     3    -3   I  -11 \n4    -3    -2     2   I  -3\n\nR4 - (4/5)*R1 , R3 - (2/5)*R1 and R2 - (3/5)*R1 will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000    1.0000    3.2000\n         0    2.8000    2.6000   -5.0000  -10.2000\n         0   -5.4000   -2.8000   -2.0000   -1.4000\n\nR4 - (-5.4/0.2)*R2 , R3 - (-5.2/0.2)*R2 will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000    1.0000    3.2000\n         0    0.0000  -17.0000  -19.0000  -55.0000\n         0    0.0000   35.0000   25.0000   85.0000\n\nR4 - (-37/17)R3 will give\n\n5.0000   3.0000    1.0000    5.0000   -2.0000\n         0   0.2000    1.4000    1.0000    3.2000\n         0   0.0000  -17.0000  -19.0000  -55.0000\n         0   0.0000    0.0000  -14.1176  -28.2353\n\nR4 / -14.1176 will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000    1.0000    3.2000\n         0    0.0000  -17.0000  -19.0000  -55.0000\n         0    0.0000    0.0000    1.0000    2.0000\n\nR3 - (-19)R4 will give\n\n    5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000    1.0000    3.2000\n         0   -0.0000  -17.0000         0  -17.0000\n         0    0.0000   -0.0000    1.0000    2.0000\n\nR3 /(-17) will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000    1.0000    3.2000\n         0    0.0000    1.0000    0.0000    1.0000\n         0    0.0000    0.0000    1.0000    2.0000\n\nR2- R4 will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000             0    1.2000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000    0.0000    1.0000    2.0000\n\nR2- (1.4)R3 will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    0.0000             0   -0.2000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000    0.0000    1.0000    2.0000\n\nR3 /(0.2) will give \n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    1.0000    0.0000             0   -1.0000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000    0.0000    1.0000    2.0000\n\nR1- 5R4 will give\n\n5.0000    3.0000    1.0000             0  -12.0000\n         0    1.0000    0.0000             0   -1.0000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000   -0.0000    1.0000    2.0000\n\nR1-R3 will give \n\n5.0000    3.0000             0             0  -13.0000\n         0    1.0000    0.0000             0   -1.0000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000   -0.0000    1.0000    2.0000\n\nR1-3R2 will give \n\n5.0000             0   -0.0000             0  -10.0000\n         0    1.0000    0.0000             0   -1.0000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000   -0.0000    1.0000    2.0000\n\nR1/5 will give \n1.0000             0   -0.0000             0   -2.0000\n         0    1.0000    0.0000             0   -1.0000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000   -0.0000    1.0000    2.0000\n\nAnd thats the final answer.\n\nSo, your answer is x=-2, y=-1,z=1, w=2\n\nVerification\na=\n     5     3     1     5\n     3     2     2     4\n     2     4     3    -3\n     4    -3    -2     2\nb= \n    -2\n     2\n   -11\n    -3\ninverse of a is\n\n    0.0250   -0.0250    0.1250    0.1750\n    0.4000   -0.4000    0.0000   -0.2000\n   -0.5125    0.6792    0.1042    0.0792\n    0.0375    0.1292   -0.1458   -0.0708\n\nThus,\ninv(a)*b = [-2 -1 1 2]'\n\nI hope i answered your question.\n\nAll the best
965376
z=w+whx\nz=w(1+hx)\nz/(1+hx) = w &#92;
752077
y=ax^2\n\nx=4\ny=24\n\n24=a*4^2\n24=a*16\na=24/16\na=3/2\na=1.5
951523
vertical: x=3\nhorizontal: y=5\n`
803558
1.\nc=(xy^2-1)/y\ndifrentiate w.r.t x\n(y(y^2 + 2xyp)-p(xy^2-1))/y^2=0\ny^3 + 2xy^2p - pxy^2 + p=0\ny^3 +xy^2p +p=0\nwhere p=(dy/dx)\n2.\nsimilarly diffrentiate \n(x+y^2)/x^2=-c\nto get\n2(x^2)yp - x^2 - 2xy^2=0\nthanks
44938
Let \nBlue = x\nYellow = y\n\nThen:\n2x + y = Green   [Given]
547629
assuming u is an unknown variable y=0
387063
X=[-(30y-433)]/1232\nY=[-(1232x-433)]/30
861275
x(y-1) = 2y + 3\n\nxy - x = 2y + 3\nxy - 2y = x + 3\n(x - 2) = (x + 3) / y\n y = (x + 3) / (x -2)
511948
There are an infinite number of ordered pairs that would satisfy this equation. If you want to find a few of them, assign a value to x and then solve for y.
793917
1)x+y-z=0.........(eqn. 1)\n2x-y+z=3...........(eqn. 2)\n4x+2y-2z=-2       (eqn.3)\nMultiplying eqn.1 by by 2 and subtracting it from eqn.2,we get\n-3y+3z=3...........(eqn. 4)\nMultiplying eqn.2 by 2 and subtracting it from eqn.3,we get\n4y-3z=-4...........(eqn 5)\nAdding eqn, 4 and 5 we get,y=7\nPutting the value of y in eqn 4 .we get\n-21+3z=3 or,3z=24  or z=8\nputting the value of y and z in eqn.1,we get\nx+7-8=0 \n=>x-1=0 or x=1\nTherefore,x=1.y=7 and z=8  ans\n2)x+2y=9....(eqn 1)\n 2x-3y=4     (eqn.2)\nmutiplying eqn.1 by 2 and subtracting it from eqn 2,we get\n-7y=-14 or y=2\nPutting the value of y in eqn 1,we get.\nx+4=9 or x=5\nTherefore x=5 and y=2 ans.
1159782
you can use linear algebra, look up gauss jordan elimination. You will want to put each equation in its own row inside a matrix. Then you will have a coefficient matrix and an augmented matrix. Row reduce the left side and you will have your desired answers on the right. If you have a TI-83+ type in your matrix and hit RREF([A]) and it will do this for you. Go to the link I provided and scroll down a bit and it will show you.
773396
x+y-z=2---(1)\n2x-y+z=7-(2)\n-x-y+2z=3---(3)\nadding (1) and (2)\n3x=9\nx=3\nadding 1 and 3\nz=5\nsub x=3,z=5 in 1\n3+y-5=2\ny-2=2\ny=4\nthe solution set{3,4,5}\n\n2.x+2y-z=-3\n2x+y+z=6\nadding\n3x+3y=3\n=>x+y=1\neqn. 1*3\n=3x+6y-3z=-9\n-2x+2y+3z=3\nadding\nx+8y=-6\nx+y=1\nsubtracting 7y=-7\ny=-1\nsub in x+y=1,x=2\nsub in x+2y-z=-3\nz=3\nso solution set {2,-1,3}
586962
(6y^2-3y)(y+7)-->(2y)(y-1)(y+7)\nif that helps
823237
There is several way to solve the problem, but I think it is easies, to find the parametric representation, and then use, that the scalar-product of the orientations-vectors is zero, if the line are perpendicular, and the orientations-vectors is linear depended if parallel.\n\nL1:  y = -½x-3 => (x,y)= (1,-½)*t+(0,-3)\n4x-2y+5 = 0 <=> y=2x+5/2=>\nL2: (x,y) = (1,2)*t + (0+5/2)\n\nbecause (1,-½)*(1,2)=1-1=0 the lines are perpendicular
633125
P, Q, R, S, T, U, V, W, X, Y, Z!!! Did I get it right???
714493
This *is* the way to write it so that x is the input and w is the output. If you want to solve for x in terms of w (i.e. have w the input and x the output), you are better off doing numerical approximations than using the cubic formula (it is really, really nasty).