im doing math matrix's & need hella help:  5x-3y=-2 &4x+2y=5 another problem: 3x+2y+2z=3 &x+2y-z=5 &2x-4y+z=0
459985
Find the inverse of the matrix\n5 -3\n4  2\nA 2x2 matrix has a formula for the inverse.\nThe inverse of the matrix\na b\nc d\nis \nd/(ad-bc) -b/(ad-bc)\n-c/(ad-bc) a/(ad-bc)\nThus, for the above problem\na = 5, b = -3, c = 4 and d = 2\nad-bc = 5(2) - (-3)(4) = 10+12 = 22\nThe inverse is\n1/11  3/22\n-2/11 5/22\nMultiply this times the  matrix\n-2\n5\n= \n(1/11)(-2) + (3/22)(5)\n(-2/11)(-2) + (5/22)(5)\n= \n1/2\n3/2\nThus, x = 1/2 and y = 3/2 or (1/2,3/2)\nFor the second one the matrix is\n3 2 2\n1 2 -1\n2 -4 1\nThe inverse of this is\n1/14 5/14 3/14\n3/28 1/28 -5/28 \n2/7 -4/7  -1/7\nTake the inverse times the matirix\n3\n5\n0\nto get the matrix\n2\n1/2\n-2\nAnswer: x = 2, y = 1/2 and z =-2 or (2,1/2,-2)
677039
(1) { -2x-y-z=0\n(2) { -x-2y-z=0\n(3) { -x-y-2z=0\n\nLet's look at (2):\n-x-2y-z=0 => x=-2y-z\nNow at (3):\n-x-y-2z=0 => x=-y-2z\nNow we can see that\n-2y-z=-y-2z => *** y=z ***\n\nNow let's put these values into (1):\n-2x-y-z=0 => -2x-y-y=0 => -2x-2y=0 => 2x=-2y => *** x=-y ***\n\nFor now, we know that x=-y=-z\nLet's put all of what we know into (2), and express everything using 'x':\n-x-2y-z=0 => -x-(-2x)-(-x)=0 => -x+2x+x=0 => 2x=0 => *** x=0 ***\n\nThus, x=0 => y=-0=0 => z=0\nSo the only solution is (0,0,0).\n\nGood luck!
137368
2x+3 = 5\nsubtract 3 from both sides\n2x+3-3 = 5-3\n2x = 2\nDivide both sides by 2\nx=2/2\nx=1
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}
1021939
2x+y=3..........eq 1         -2x+5y=-9..............eq 2\nAdd eq. 1 and 2\n6y=-6\ny=-1\nPutting this value in eq 1\nx=2\nS.S={ 2, -1}\n\n3x+5y=0............eq 3           2x-5y=-25.................4\nMultiply eq 3 by 2 and eq 4 by 3\nFrom eq 3: 6x+10y=0................eq 5\nFrom eq 4: 6x-15y=-75..............eq 6\nSubtract eq 6 from 5\ny=3\nPut the value of y in eq 3\nx=-5\nS.S={ -5, 3}
790308
1.  5x + 7= 3x + 21\n     -3x       -3x\n     2x + 7 = 21\n           -7    -7\n     2x = 14\n     /2     /2\n        x = 7\n\n2.  3x - 2 (5-x) = -3(x-10) + 3x\n     3x - 10 - 2x = -3x + 30 + 3x\n     x - 10=30\n     x=40\n\n3.  5( x+1) = 3(2x +3) + 5\n     5x + 5 = 6x+9 + 5\n    5x+5=6x+14\n    5=x+14\n    x=-19
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!
720107
4x -12y =5...........(1)\n-x +3y = -1..........(2)\n\nMultiplying eq.(2) by 4 and add with eq.(1)\n(4x -12y =5) + (-4x +12y = -4)=\n0 = 1\nwhich is invalid.
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!  :-)
450302
3y+2x=12\n2x=-3y+12\nx= -3/2y+6\nx=-3/2(x-1)+6\n =-3/2x-(-3/2)+6\n = -3/2x+15/2\n5/2x=15/2\nx= 3\n\ny= -3/2y+6-1\n5/2y= 5\ny= 2
1165093
3x + 5 = -4(x-6)\nFirst, you need to combine the second part.\n\n3x + 5 = -4x + 24\nNow you switch.\n\n3x + 4x = 24 - 5\nGroup.\n\n7x = 19\nDivide.\n\nx = 19/7
704941
The second equation can be rewritten as:\n3x + 2y = -1\nMultiply both sides by 2\n6x + 4y = -2\n6x + 5y = -7 (this is the first equation)\nSubtract the first equation from the second one.\ny = -5\nNow, plug the y = -5 into the second equation:\n3x +2(-5) = -1\n3x - 10 = -1\n3x = 9\nx = 3\nSo your solution is x = 3 and y = -5
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?
905206
8x - 2 = 3x - 2\n8x = 3x\n5x = 0\nx = 0
940849
2x + 4y = -6\nX - 3y = 7 multiply by two\n\n2x + 4y = -6\n2x - 6y = 14\n{subtract equation 2 from equation 1}\n\n2x - 2x + 4y - (-6y) = -6 - 14\n10 y = -20 divide by 10\ny = -2\n\n\nplug into a equation \n2x - 6(-2)  = 14\n2x + 12 = 14\n2x = 2\nx = 1\n\nx = 1, y = 2
86920
5x - 13 =3y , 4x - 11 = 3y\nSubstitue\n5x - 13 = 4x -11\nsubtact 4x\nx - 13 = -11\nadd 13\nx = 2\n\nplug 2 in\n5 * 2 -3y = 13\n10 - 3y = 13\n-3y = 3\ny = -1
631059
3y+5=y-25\n2y+5=-25\n2y=-30\ny=-15\n\nSee, you just use inverse operations.\n\n4n-5=2n+3\n2n-5=3\n2n=8\nn=4\n\nSo your answers are y=-15 and n=4. I hope that helped!