Can anyone explain how to solve x, y, and z in the following systems of three equations?
830984
You have to first eliminate one variable from two of the equations which would leave two equations with two variables.\nThen eliminate one of those two variables to get one value.  Plug this in to get a second value; plug these in to get the third.\n\nFor example in your problem b, if you multiply the second row by -1 and add it to the first row, that would eliminate z\n\n(2x - 1y = 1)\n\nThen divide the third row by 2 and add it to the (original) first row, which also eliminates z\n\n(20x - 4y = 16)\n\nNow take those 2 rows, eliminate either x or y, and start plugging in
134530
First, generate 3 equations by inserting the values of x and y for each of the coordinates.\n\nEqn 1. For (-1,1): 1=a*-1^2+b*-1+c yields: 1=a-b+c\nEqn 2. For (1,3): 3=a*1^2+b*1+c yields: 3=a+b+c\nEqn 3. For (3,2): 2=a*3^2+b*3+c yields: 2=9a+3b+c\n\nStep 1: solve Eqn 1. for a:  a=1+b-c\nStep 2: insert a from Step 1 into Eqn 2.: 3=(1+b-c)+b+c ==> 2=2b ==> b=1\nStep 3: insert b=1 back into Step 1 and re-solve for a: a=1+1-c ==> a=2-c\nStep 4: insert the results from Step 2 for b and Step 3 for a into Eqn 3 to find c.\n  2=9*(2-c)+3*1+c ==> 2=18-9c+3+c ==> 19=8c ==> c=19/8\nStep 5. Insert the results from Step 4 and Step 2 into Eqn 1 to find a.\n  1=a-1+19/8 ==> a=-3/8\n\nTherefore:\n  a=-3/8\n  b=1\n  c=19/8\n\nStep 6: Verify for -3/8*x^2 + 1*x +19/8\n (-1,1): 1=-3/8*(-1^2)+1*-1+19/8\n (1,3): 3=-3/8*(1^2)+1*1+19/8\n (3,2): 2=-3/8*(3^2)+1*3+19/8
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?
956578
x^2+3x-y=c\n2x +3 -y'=0\ny'= 2x +3\n\n(remember y^0=1)!
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
805693
#1\n4x+1+3=0\n4x+4=0\n4x=-4\nx=-1\n\n#2\n2y+7+4=y\n2y+11=y\ny=-11
406322
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.
239864
Solve for y in the second equation, then plug that into the first equation and solve for z.  After you find the value of z, plug that into the other equation.\nMake sure you always check your answer!  To do that just plug in the values for y and z into each equation to see if you get a true statement.  If you don't then your answer is wrong or else you made a mistake when checking the answer.
786186
2x + y + z = 4- -- - - - - Equation 1\n-x - 3y + 3z = 3 - - - - - Equation 2\n3x + 4y - 4z = 1- - - - - Equation 3\n- - - - - - - - - - - \nMultiply equation 2 by 2\n2(-x) -2(3y) + 2(3z) = 2(3)\n- 2x -6y + 6x = 5 New equation 2\n- - - - - - - - - - -\nElimination of x\n2x + y - z = 4\n- 2x - 6y + 6z = 6\n- - - - - - --\n- 5y- 5z = 10- - - - -Equation 4\n\n- - - - - - - - -\nMultiply equation 1 by -3 and equation 3 by 2\n\n2x + y - z = 4\n-3(2x + (-3)(y) - (-3)(z)= -3(4)\n-6z - 3y - 3z = -12 - - New equation 1\n- - - - - - - - -\n\n3x + 4y - 4z = 1\n2(3x + 2(4y) - 2(4z) = 4(1)\n6x + 8y - 8z = 4- - -New equation 3\n\n- - - - - - - - - - \nElimination of x\n\n- 6x - 3y - 3z = - 12\n6x + 8y - 8z = 4\n- - - - - - - - - -\n5y - 6z = - 8 - - Equation 5\n\n- - - - - - - - - - - - -\nElimination of y from equation 4 and 5. \n\nSolving for z.\n\n- 5y + 5z = 10\n5y - 6z = -8\n- - - - - - - -\n-z = 2\n-1(-z) = -1(2)\n\nz = - 2\n\nThe answer is z = -2\nInsert the z value into equation 4\n\n- - - - - - - - - - -\n\nSolving for y\n\n- 5y + 5z = 10\n- 5y + 5(-2) = 10\n-5y + (- 10) = 10\n- 5y - 10 = 10\n- 5y - 10 + 10 = 10 + 10\n- 5y = 20\n\n-5y/- 5 = 20/-5\n\ny = - 4\n\nThe answer is y = - 4\n\nInsert thy y value into equation 1\n\n- - - - - - - - -\nSolving for x\n\n2x + y - z = 4\n2x + (- 4) - (-2) = 4\n2x - 4 + 2 = 4\n2x - 2 = 4\n2x - 2 + 2 = 4 + 2\n2x = 6\n\n2x/2 = 6/2\n\nx = 3\n\nThe answer is x = 3\n\nInsert the x value into equation 1\n\n- - - - - - - - - - - - -\nCheck For equation 1\n\n2x + y - z = 4\n\n2(3) + (- 4) -(-2) = 4\n\n6 - 4 + 2 = 4\n\n2 + 2 = 4\n\n4 = 4\n- - - - - - - - - - - -\n\n Check for equation 2\n\n-x - 3y + 3z = 3\n\n-(3) -  3(- 4) + 3(-2) = 3\n\n-3+  12 + (- 6) = 3\n\n-3  + 12 - 6 = 3\n\n9 - 6 = 3 \n\n3 = 3 \n\n- - - - - - - - - - - - - - - - - \n\nCheck for equation 3\n\n3x + 4yy - 4x = 1\n\n3(3) + 4(- 4) - 4(- 2)  = 1\n\n9+ (-16)  - (- 8) = 1\n\n9 - 16 + 8 = 1\n\n- 7 + 8 = 1\n\n1 = 1\n\nThe solution set { 3, -4, - 2 }\n\n- - - - - - - -s-\n\nFollow this procedure of elimination\n\n\n- - - - - - - -s-
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!
994889
Start with a general cubic equation:\nx^3 +ax^2 +bx+c=0.\nFirst, we want to get rid of the x^2 term, so let \nx=y-(a/3).\nIf you plug this in and expand, the y^2 terms will drop out and you will get an equation of the form\ny^3 +Cy+D=0.\nNow, if you let y=u+v (to get an extra bit of freedom), and expand, you get\nu^3 +v^3 +(3uv+C)(u+v)+D=0.\nChoose v so that 3uv+C=0, i.e. v=-C/(3u). Plug this into\nu^3 +v^3 +D=0\nand multiply through by u^3 to get a quadratic in u^3. Solve\nthis by the quadratic formula and take a cube root to find\nu. There will be three cube roots in the complex plane, so find all of them. For each u, find v=-C/(3u), then y=u+v, and finally x=y-(a/3).\n\nThis is often a rather cumbersome proceedure, even if you do some of it ahead of time. Also, the roots may not 'look' like what you expect. If you apply this to an equation that has a root of, say, 4, you will obtain 4 as a sum of two cube roots of quadratic expressions. To recognize the answer as 4 is a challenge in and of itself.
1162328
1) -  Y-2x-8 = 0 - so Y=-2x-8\n2) -  Y^2 +8x=0 - (-2x-8)^2+8x=0 \n\n\n4x^2+32x+64+8x=0\n\n4x^2+40x+64=0 - so (4x+8)(x+8)=0 so x=-8 or 4x=-8, x=-2\n\nWhen x=-8, Y-2(-8)-8=0 so Y=-8    A=(-8,-8)\nWhen x=-2, Y-2(-2)-8=0 so Y=4     B=(-2,4)
344034
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...
712820
write them so they can be added top to bottom (i.e. line up the x's, y's, and constants):\n\n4x + 2y = -4\nx - 2y = -3  (we need to rearrange the original equation here to line things up)\n\nnow add top to bottom:\n\n(4x)+(x)  +  (2y)+(-2y)  =  (-4)+(-3)\n5x + 0 = -7\n5x = -7\nx = -7/5\n\nnow plug x in to either equation to solve for y:\n\n2y = -7/5 + 3\n2y = -22/3\ny = -22/6 = -11/3\n\nHope this helps :)
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!  :-)
11448
Substitution Method\nSolve this system of equations:\n3y - 2x = 11\ny + 2x = 9\n \n\n1.  Solve one of the equations for either "x =" or "y =".\nThis example solves the second equation for "y ="\n \n\n3y - 2x = 11\ny = 9 - 2x\n \n\n2.  Replace the "y" value in the first equation by what "y" now equals.  Grab the "y" value and plug it into the other equation. \n\n \n\n3(9 - 2x) - 2x = 11\n \n3.  Solve this new equation for "x".\n \n\n(27 - 6x) - 2x = 11\n27 - 6x - 2x = 11\n27 - 8x = 11\n-8x = -16\nx = 2\n4.  Place this new "x" value into either of the ORIGINAL equations in order to solve for "y".  Pick the easier one to work with!  \n\n \n \n\ny = 9 - 2(2)\ny = 9 - 4\ny = 5\n \n\n5.  Check:  substitute x = 2 and y = 5 into BOTH ORIGINAL equations.  If these answers are correct, BOTH equations will be TRUE!\n   \n\n3y - 2x = 11\n3(5) - 2(2) = 11\n15 - 4 = 11\n11 = 11\n\ny + 2x = 9\n5 + 2(2) = 9\n5 + 4 = 9\n9 = 9 \n\nElimination:\n x - 2y = 14\nx + 3y = 9\n \n\na.  First, be sure that the variables are "lined up" under one another.  In this problem, they are already "lined up".\n \n\nx - 2y = 14\nx + 3y = 9\n \nb.  Decide which variable ("x" or "y") will be easier to eliminate.  In order to eliminate a variable, the numbers in front of them (the coefficients) must be the same or negatives of one another.  Looks like "x" is the easier variable to eliminate in this problem.\n  \n\nx - 2y = 14\nx + 3y = 9\n \n\nc.  Now, subtract to eliminate the "x" variable.  \n(Remember:  when you subtract signed numbers, you change the signs and follow the rules for adding signed numbers.)\n \n x - 2y = 14\n-x - 3y = - 9    \n\n    - 5y = 5  \nd.  Solve this simple equation.   \n\n-5y = 5\n   y = -1\n \n\ne.  Plug "y = -1" into either of the ORIGINAL equations to get the value for "x".\n   \n\nx - 2y = 14\nx - 2(-1) = 14\nx + 2 = 14\nx = 12\n \n\nf.  Check:  substitute x = 12 and y = -1 into BOTH ORIGINAL equations.  If these answers are correct, BOTH equations will be TRUE!\n   \n\nx - 2y = 14\n12 - 2(-1) = 14\n12 + 2 = 14\n14 = 14  \n\nx + 3y = 9\n12 + 3(-1) = 9\n12 - 3 = 9\n9 = 9
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)