Given\n x-y+ z=2\n5x+   z=-1\n  3y+az=3\nwhere a is some real number.\nFind all values of a for which this system has a unique solution
124528
Put them in matrix form, invert the matrix and you should get the answer.
525993
**************\nQuestion 1\n**************\n\n2x - 4y = 8\nx - 2y = 4\n\nx - 2y = 4\nMultiply both sides of this equation by -2\n(-2)(x - 2y) = -2*4\n-2x + 4y = -8\n\nNow we will add this equation to the other\n  2x - 4y =  8\n-2x + 4y = -8\n____________\n0 + 0 = 0\n0 = 0\n\nThere are an infinite number of solutions to this problem because both sides of the equal sign end up the same.\n\n**************\nQuestion 2\n**************\nWe can add these together right away because the x and -x will cancel each other out.\n-x - y =  4\n x - y = -8\n_________\n0 -2y = -4\n\nNow we will solve for y.\n-2y = -4\nDivide each side by -2\ny = 2\nNow we plug that into one of the equations to get x\n x - y = -8\n x - 2 = -8\nAdd 2 to both sides\nx = -6\n\nSo, x = -6 and y = 2
147634
1.y+2\n2. x+3\n3. z+6\n4. i wasnt sure what it was equal to, so i assued 0... making the answer (x+2)(x+5) however,if it is x2+7x=10...then this answer's wrong...
200336
I don't condone getting Homework answers here but I can help you. first you can eliminate 1 of the variables for example x by multiplying the 3rd eq. by -2 and adding eq. 2 and  eq.3 \n\n2x-y+ 2z = 11 + (-2x - 4y - 4z = -12)\nthis will leave you with 0x -5y - 2z = -1\n\nthen you can do the same thing with eq. 1 and eq. 2\nnamely (4x + 5y + z = 6) +(-2)(2x-y+2z = 11)\nthis simplifies to \n\n0x + 7y -3z = -16\n\nyou have to repeat the process and eventually you get a solution to 1 of the variable which you can then put back into one of the equations to find the other answers
376134
3x+2y-z=-11\n-10x-2y+4z=34     (3)*-2\nadding \n-7x+3z=23     (4)\n2x-3y+4z=11     (2)\n15x+3y-6z=-51  (3)*3\nadding\n17x-2z=-40    (5)\n(4)*-2     14x-6z=-46\n(5) *-3    -51x+6z=120\nadding\n-37x=74\nx=-2\nsub14+3z=23\nz=3\nsub y=-1\nsolution set{-2,-1,3}
239424
Solving a system by addition involves simply adding the two equations together. First, however, you have to do a little prep work. Either the x's or the y's must be of the opposite sign. I'll show you how to do the y's first, because it is easier.\n\n5x-3y=13\n4x-3y=11\n\nThe y's need to be opposite. What is the simplest way to do so? Make one of the -3y's into a positive 3y. Thus, multiple one of the equations by -1.\n\n-1(5x-3y=13)\nbecomes\n-5x+3y=-13\n\nNow you have\n-5x+3y=-13\n4x-3y=11\n\n+3y + -3y = 0\n\nYou have just canceled out the y's.\n\nNow you have:\n-5x=-13\n4x=11\n\nAdd these together and you get:\n-x=-2\nThus, x=2\n\nNow you know x. Take either of the original equations, and substitute in your new x.\n\n5(2)-3y=13\n10-3y=13\n-3y=3\ny=-1\n\nThat's it! Hope I helped.
227060
-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
626104
5  3  4     *   x                        19\n3 -1 1          y            =           1\n                   z\nthere are infinite solutions:\n\nx = (22-7z)/14\ny = (52-11z)/14
333049
No solution.\n\nDo (equation 3) minus 2*(equation 1)...\nyou get\ny + z = 13\nwhich contradicts equation 2.  Either you have a mistake, or the answer is no solution.\n\nHope this helps,\nGod bless,\nShane
340047
D\n\nEither is sufficient.  \n1) will lead to 2 equations,2 unknowns which yield a unique solution for x and y:  x=y+1, x+y=77, therefore x=39, y=38.  xy=1482 \n\n2) If the sum is 77 and x and y have the same tens digit, then there are two solutions for {x,y}, namely {38,39} and {39,38}, however the product xy is the same. xy=1482
318989
25x + 5y = 25........(1)\ny = 5 – 5x...........(2)\n\n25x+5y=25............(1)\n5x+y=5...............(2)\n\nOn rearranging u find both equations are the same.\ni.e equation1=5 * equation(2)\nso solution is x=0,y=0;\n\nif u r looking for values\nu can use trial and error methods\n\none solution would be x=-1 and y=10\ni.e 5 * -1+y=5\ny=10\n\nlike wise u can put values for x and find the corresponding value for y.
179907
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.
10918
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.
345366
Rearranging equation 2:  x = y + z + 1\n\nSubstitute this in for x in each of the other two equations:\n\ny + z + 1 + y + z = 1   simplifies to  y + z = 0\n\n2y + 2z + 2 + 2y - z = -4  simplifies to 4y + z = -6\n\nSubtracting these simplified equations gives 3y = -6\n\ntherefore y = -2\n\nNow substitute this into y + z = 0, giving z = 2\n\nNow substitute these values of y and z into x = y + z + 1, to get x = 1\n\nSo the solution is:\n\nx = 1\ny = -2\nz = 2
534029
in order to solve the system of equations you can do one of two things.  The easiest thing to do is to set the two equations to be equal. i.e. \n\nsince y=2x and y = -3x+5, then 2x=-3x+5.  In solving for x you will find that 5x=5 or x=1.  Then you can plug your resulting x back into the equation to get y=2(1) =2.  The main thing you are trying to do is get one of your variables to be equal to some combination of your other variables.  For example, if you are trying to solve 3y+2x-1=5x+3 and 6y=2x+7.  The first thing you want to do is to get a single variable by itself on one side of an equation.\nLet's start with the first equation, 3y+2x-1=5x+3. You would try to group the y terms by themselves on one side of the equation, an group the x terms and constants on the other side of the equation.  In this example, in order to group the terms correctly, you would have 3y=3x+4.  Then solve for y by itself by dividing both sides by 3.  The result is y=x+4/3.  You take this result and substitute for y in the second equation, 6y=2x+7.   You will get 6(x+4/3)=2x+7.  Then you solve for x by itself.  6x+8=2x+7\n4x=-1;  x=-1/4.  Then you can substitute this result back into either of your original equations to produce your solution for this system.  6y=2(-1/4)+7 => 6y=13/2 =>y=13/12.  In order to check your answer, you can graph the two linear equations.  Where the two lines intersect is the solution to your system.\n\n\nIf you are looking for a more advanced way to solve a system of equations, then you may want to consider using an augmented matrix.  Rearrange your equations so that you have only variables on one side of your equations, and only constants on the other side of your equations, then sole the system of Ax=b.  The matrix A is the coefficients of your variables, and b is a vector of your constants.  x is a vector of your variables.  Augment the matrix of your coefficients with your constant vector and do Gaussian Elimination until the matrix is in reduced row echelon form.  This will provide the exact solution to the system of equations.
530101
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
411922
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.