Using the Gauss jordan Method to solve this problem?
449432
Gauss Jordan elimination has to do with matrix operations.  Take your coefficients and constants and put them in an augmented matrix (basically a matrix with an extra column for the constants):\n\n| -2 -5 | -15 |\n| -6 -15 | -45 |\n\nUsing elementary row operations, try to get this into row reduced echelon form (all 1's on the diagonal):\n\nR1/(-2) -> R1\n| 1 5/2 | 15/2 |\n| -6 -15 | -45 |\n\nR2 - (-6)R1 -> R2\n| 1 5/2 | 15/2 |\n| 0 0 | 0 |\n\nFrom here, we see that we can't get this into RRE, and from this we can say that there is no solution.\n\nThis was seen in the previous answers in solving it as a system of equations versus using a matrix.  From inspecting the equations as a system, you can see that these are actually the same line - the second is the first multiplied by 3.  Since they are the same line, there are actually infinite solutions.\n\n(If these were parallel lines, then you would have no solution.  Remember that a "solution" is a point of intersection in this case.  If they are parallel, they never intersect, and thus the system would have no solution.  But if they overlap, they always intersect, and thus the system would have infinite solutions.)\n\nsomeguy's explanation in turning it into a parameter is a legitimate approach if you _have_ to have an answer.  The answer he came up with works, regardless of which equation you started with.\n\nSo from this, we have three answers:  no solution (from G-J elimination), infinite solutions (from inspecting the lines graphically/analytically), and (15/2 - 5y/2, y) (from turning it into a parameter solution).  Take your pick based on whatever method you are required to use for this particular problem.
1142827
You don't give a feasible region. However, only one of your answers actually satisfies z=6x+10y. That should solve your problem.
510279
Jimbo, you gave up too soon......\n\nV = L*W*H = L*W*x in this problem\n\nV = (6-2x)*(4-2x)*x\n\nV  oila!
908083
Tried to do it algebraically, that was a booger.\n\nI plugged that into mathematica and it gave me an error normally generated when the inverse of a function is needed to represent a solution and the solution variable occurs in more than one argument of that function.\n\nThat one might not have a solution?
846527
You'll have to do this in more than one step.  Basically it will sort of have to be a program.  I would start with the restart; command and then  define the function (I would look this up in the help menu since it is dependent on 2 variables).  Then use a command to find the derivative.  Then a command to solve for the pairs (x,y) when the derivative is 0.  That will give you 1.  For 2 and 3 you will also have to do a series of commands.  Use your calc book to refresh your memory on the procedure of finding these, it should be a similar procedure in maple.
1082511
I think your question have a typo.  There are three variables and I am not sure which one are you trying to solve for.
929849
solve three body problems
929836
v = b - u/k\nv - b = u/k\nk(v-b) = u
224079
try this site... it shows step by step how to go about sloving these equations, good luck:
441742
Assume there is a pair a_i, b_j with a non zero product.\nWrite fg= (a_i x^i + other terms) (b_j x^j + other terms) =\na_i b_j x^(i+j) is non zero.\n\nContradiction.
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
865560
A. Just multiply CF and then subract the resulting matrix from C. Is there something specific about the required operations you don't understand?
920579
use synthetic  divsion method \ni think this link would help\nhttp://www.purplemath.com/modules/synthdiv.htm
578062
use the properties of equalities and inequalities.\n(subraction,additon,multiplication,division).
1132233
this question will require calcs using logarithm.\nhence will take time
473829
The solution set is (-infinity, -7) U (-3/2, infinity)
837061
it can be the original way you had it (c=mx +y) but it is alot simpler in the long run to rearrange it to (y=mx+c) (y being y, x being x, m being gradient, c being a constant). this way it is easyer to solve.