how do you solve the system of equations  3x+4y+z=7; 2y+z=3; -5x+3y+8z=-31   not using matrices?
817577
you are going to have to use SUBSTITUTION. meaning place the equations one above the other. See what you can cross out... i.e.\n\n1) TAKE OUT THE X'S BY MULTIPLYING THE FIRST EQUATION BY 5 AND THE SECOND EQUATION BY 3. YOU WILL GET...\n\n15x+20y+4z=35.\n-15x+9y+24z= -93\n\nthus, you can cross out the x's and get the equations:\n\n20y+4z=35\n9y+24z= -93, and now you must cross out the z's by multiplying the top equation by -6...\n\n-120y-24z=-210\n9y+24z= -93\n\nand, so on....
272177
multiply both sides by 3 and you'll get\n3x-3*4=-2*(y-9)\n\n3x-12=-2y+18\n-2y+18=3x-12\n-2y=3x-12-18\n-2y=3x-30  divide both sides by -2\n\ny = (-3/2)x + 15
438685
from the last equation: Z = 8 - 2X\nthen replace Z in the second equation = -3X + 2Y -(8 - 2X) = -4\nthis equals to : Y = X/2 +2\nNow that u have Z and Y replace them in the first equation:\nX + 2 (x/2 +2) + 6(8 - 2X) = 4\nthat's X = 24/5\nZ = -8/5\nY = 22/5
984033
2x+3y=4\n4x+3y=-36\nsubtracting\n-2x=40\ndividing by -2\nx=-20\nsobstituting\n-40+3y=4\nadding 40\n3y=44\ny=44/3
145305
Use cubic equation \ny=ax3+bx2+cx+d\n\nSince you have four points data you can get four linear equations with four variables which is very easy to solve, get variables (a,b,c,d) and plug the values to cross check and see u will get answer. Use guass jordan or elimination method.
408462
to solve such a system in general, what you do is this: take the first equation, solve it for x treating y as you would a constant. In your case, this is x = 4 + y.\n\nThen, take your expression for x and substitute it everywhere you see x in the second equation; in your case you get 3 *(4+y) + 2y = 7.\n\nSolve this now 1-variable equation for y, once you find y (in this case, -1).\n\nSubstitute the value of y into either equation, and you get the value of x (3 in this case).
865074
To start off, you are given a simple equation to find x with respect to y and z. \n\nx = 2y - 3z +5\n\nThe next step is to substitute into the other two equations.\n\n3(2y - 3z +5) -3y + z = 9\n6y - 9z + 15 - 3y + z = 9\n3y - 8z = -6\n\n5(2y - 3z + 5) + y - 3z = 3\n10y - 15z + 25 +y - 3z = 3\n11y - 12z = -22\n\nNow what I would do is take the first new equation and solve for y:\ny = (8/3)z - 2\n\nAnd then substitute it into the second equation:\n\n11((8/3)z - 2) - 12z = -22\n(88/3)z - 22 - 12z = -22\nz = 0\n\nInsert this into the previous equation:\ny = (8/3)(0) - 2\ny = -2\n\nAnd then finally into the very first equation:\nx - 2(-2) + 3(0) = 5\nx + 4 = 5\nx = 1\n\nNow, to be sure, we check our answer with the other two equations:\n\n3(1) -3(-2) + 0 = 9\n3 + 6 = 9\n\n5(1) + (-2) - 3(0) = 3\n5 - 2 = 3\n\nEverything checks out, so:\nx = 1\ny = -2\nz = 0
201908
Ah, systems.\n\nthe first problem- there are many ways you can choose to do this, but I think we should multiply both sides of the first equation by two so that they positive and negative y's will cancel out.\n\nSo now we have:\n2(2x + y)=2(4)\n3x - 2y = -1\n\nSimplified:\n4x + 2y = 8\n3x - 2y = -1\n\nnow you add down, so you get:\n7x +0y = 7\n7x =7\nx=1\n\nsubstitute 1 for x in any equation:\n2(1)+y = 4   \n2 + y = 4\ny = 2\n\n\n\nfor the second one- lets cancel out the y's again.\n\n5x - 3y = 14\n-3(2x - y) = (6)-3\n\n5x - 3y = 14\n-6x +3y = -18\n\nAdd them together:\n-x = -4\nx = 4\n\nThen solve for y:\n5x - 3y = 14\n5(4) - 3y = 14\n20 -3y = 14\n-3y = -6\ny = 2
835073
okay well\n\n1) combine two to figure out the first variable\nx-2y+z=4\n-x+2y+3z=-4\nso 4z=0; z=0\nthen figure out the next variable with a different set of equations\nlike and multiply so the ys will cancel [take Zs out cause zero]\nx-2y+z=4\n2(2x+y)=6\n\ntherefore\nx-2y=4\n4x+2y=6\n5x=10\nx=2\n\nthen substitute for y\n\n-x+2y=-4\n-2+2y=-4\n2y=-2\ny=-1\n\nso y=-1, x=2, z=0\n\n\ncontinue the process with the rest.
772255
Note that you have a 7c in both equations, so write them both down like is shown below, then subtract the second equation from the first equation to get:\n\n7c + 3d = -16\n7c + 5d = -8\n----------------------\n0c - 2d = -8\n\nNow divide both sides by -2 to get d=4\n\nNow substitute d=4 into either one of your original equations to solve for c:\n\n7c+3(4) = -16\n7c+12 = -16\n7c = -28\nc = -4\n\nANSWER:  c = -4,  d = 4
887085
First, I multiplied the first equation by 2 and the second one by -5 in order to make it easy for me to cancel out both x's.\n2(-5x+7y=-40)\n-5(-2x+4y=-16)\n\n-10x+14y=-80\n10x+-20y=80\n\nNow, you can actually cancel out the x's and the constants at the end.\n\n-6y=0\ny=0\n\nNow you plug y=0 into one of the equations to find x.\n\n-5x + 7(0) = -40\n-5x +0 = -40\n-5x = -40\nx = 8\n\n(0,8) is the answer.
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!  :-)
753426
X2   5x - 4y = 1\n10x -8y =2\n-10x + 8y = -3\n-------------------------\n  0=-1                NO SOLUTION
913862
3x + 2y - 2z\n\n3(5) + 2(2) - 2(1)\n\n15 + 4 - 2\n\n19 - 2\n\n17\n\nThe answer is 17\n\n- - - - - - -s-
627033
6y + 3 = -4y^2\n4y^2 + 6y + 3 = 0\n\na=4, b=6, c=3\n\ny = -b+-SQRT(b2-4ac) / 2a\ny = -6+-SQRT[36-(4)(4)(3)] / [(2)(4)]\ny = -6+-SQRT[-12] / 8\n\nBut SQRT[-12] is not possible, unless we are looking for imaginary values, then:\n\ny = -6/8 +- SQRT[-12]/8\ny = -3/4 +- [2SQRT3]i/ 8\ny = -3/4 + [(SQRT3)/4]i or -3/4 - [(SQRT3)/4]i
424288
1. y2+6y+8 = (y+4)(y+2)\n2. x2+7x+12 = (x+4)(x+3)\n3. z2+2z-24 = (z-4)(z+6)\n4. x2+7x+10 = (x+2)(x+5)
709836
2x -  4y = -5\n3x + 4y = -15\n\n5x = -20\n\nx = -20/5                                       \n\nx = -4\n\n*************************\n\n2(-4) - 4y = -5\n\n-8 - 4y = -5\n\n- 4y = -5 + 8\n\n- 4y = 3              \n\ny = -3/4\n\n**************************\n2 (-4) - 4 (-3/4) = -5                                        \n\n-8 - (-3) = -5           \n\n&\n\n3 (-4) + 4 (-3/4) = -15                           \n\n-12 + (-3) = -15