If I do not solve these equations but am trying to write a augmented matrix how can I do it?
438919
I dont know how to put this up here. I tried to put it in a nice format, but it would accept.\nBut you put the following in one matrix \n-2  6\n+2  2\n\nThen you put a * (multiply)\n\nThen you put the following in one more matrix\n\nx\ny\n\n\nThen put an ' = ' symbol\n\nThen put the following in a matrix\n\n40\n8
1126894
This normally means that you are going to be using Gaussian (or Gauss-Jordan) reduction, where ther eare three "moves" that are allowed.\n\n1) You can switch any two rows in the matrix.\n2) You can multiply any row in the matrix by a non-zero scalar.\n3) You can add a multiple of one row to another.\n\nUsing these three "moves," the goal is to reduce the matrix into a usable form (the echelon form, for example).
844357
Gauss-Jordan will be longer.. Matrix solution will be simpler n faster\n\nDet matrix   \n                    5    3    1    5\n                    3    2    2    4\n                    2    4    3   -3\n                    4   -3   -2    2\n\nPlug the RHS colum in the column u want to solve,, and divide that by the Det...
888367
You need to give some directions.  What is the problem asking you to do?\nNote: The person below is incorrect.  (4+3x)^5 does NOT equal\n4^5 + (3x)^5
916972
When all the variables cancel out, then you have an infinite number of solutions, or no solutions.  Here, you have none.\n\nThis problem shows you that you need to check for linear independence.  That is, none of your equations should be linear combinations of any others.  What is a linear combination?  It's when you add equations which may or may not be multiplied by constants.  If you get something like this, then you can't solve your system, as you'll have fewer equations than unknowns.\n\nIn your case, 3[equation1]-[equation3] yields 0=-1.  This is false, so your system has no solution.\n\nIf there was a 3 instead of a 4, you would have an infinite number of solutions, where they satisfied the first and second equations.
291866
The same equation can be written as \ny = 2x^2-3x+2\n\nNow, for the new equation, let it be written in terms of the new shifted coordinates (X,Y). In the new system, what was the origin (0,0) earlier, now becomes (-1,-2). Hence the equations relating the two systems would be\nX = x - (-1) and Y = y - (-2)\nor X = x + 1 and Y = y + 2\n\nWritten in another way,\nx = X - 1 and y = Y - 2\n\nSubsituting in the main equation,\n\n(Y-2) = 2(X-1)^2 - 3(X-1) + 2\nOn simplifying,\nY = 2X^2 - 7X + 9\n\nHence, the general equation that you need is\nf1(x) = 2x^2 - 7x + 9
429488
If you have two unknowns (i.e. x, y) then you must have two equations to solve for x and y. You only have one equation, unless I missed something.
802863
| A - I(lamda)| = 0 ==> eigenvalue ... \n\nlamda is an eigenvalue...\n\nsolve the matrix first
236017
sorry i don't know mathematica but you can solve it easily using matlab. Enter rthe coefficient matrix as follows for example,\nA=[1 2 3 4; 2 3 3 4;........]\nand the right hand side\nb=[ 2 3 4 4]\nonly for illustration. ";"indicates columns.\nA/b solves this system and gives the solution vector x.
77049
first you have to tell us what you're solving for.(either x or y).\nIf you want to combine like terms u have to add 2y+2y you get 4y,then you try 4y+4x = ?.\nBut you forgot what is equal to if you're solving \nfor the letter x or y
283652
When you try to see if your vector u is a linear combination of a basis of that subspace, you will be lead to write a system of linear equations, with the components of u in the right side of the equations. If that system has no solution, then u is not in that subspace, otherwise it is in it.
550482
-2a(b+2)-6b=\nor\n-2(ab+2a+3b)
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.
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
241815
If this matrix is A, we want to solve an equation AX=B, to solve this equation we have to cancel A, so multiply both sides by an inverse, A^(-1) and obtain X=BA^(-1).\nIf the matrix in non invertible, we cannot cancel it as easily. At least I assume that was the cancellation you requested.
658708
this one isn't possible..wait...i think i got it....\n\nxa+yb+xz+yz\nxa+yb+z(x+y)\n\nif a typo, and a = b then:\nxa+ya+z(x+y)\na(x+y)+z(x+y)\n(a+z)(x+y)\n\notherwise, not possible unless, use the math trick:\nxa+yb+z(x+y)+ya-ya+xb-xb\nxa+xb+ya+yb+z(x+y)-xb-ya\n(x+y)(a+b)+z(x+y)-xb-ya\n(x+y)(a+b+z)-xb-ya
533429
well..write them down ...try to derive them (if their derivation is simple enough or else skip it )......\n\ntry to connect each formula with other one with a hint ....a small hint will help u to recollect other one ....\n\ntry to do at least one problem based on each formula