systems of equations 3?
924170
Not really sure what you want but I'm just going to solve x and y.\ny= x-4       -1\nx^2 -2y -11 = 0    -2\nreplace 1 in 2\nx^2 -2x -3 = 0\n(x-3)(x+1) = 0\nthus x = 3 or x = -1\nreplace in 1\nthus if x = 3 ; y = -1\nand if x = -1 ; y = -5
1137447
2 linear equations\n\n X  + 3Y - 1 = 0\n4X + 7Y + 5 = 0\n\nFROM (1)   4X + 12Y - 4 = 0\nFROM (2)   4X +  7Y + 5 = 0\n\nSUBTRACT\n                          5Y - 9 = 0\n\n                          5Y = 9\n\n                            Y = 1.8\n \n  FROM (1)      X + 5.4 - 1 = 0\n\n                        X + 4.4 =   0\n\n                        X =  -4.4
1144396
you would take the 2x-4y=3 and solve for a variable, such as x=(3-4y)/3   you now have what x equals and you sub that into the other equation so you now have 2[(3-4y)/3]-3y=0 and you should be able to solve that equation
27501
Start out by multiplying the\n            first equation by -1 and add\n            it to the second equation to \n            eliminate x from the second\n            equation.\n\n            -x  - y - z = -4\n             x - 2y - z = 1\n            ----------------\n               -3y - 2z = -3\n\n            Now eliminate x from the third\n            equation by multiplying the first\n            equation by -2 and add it to\n            the third equation.\n\n            -2x - 2y - 2z = -8\n             2x -  y - 2z = -1\n            ------------------\n                 -3y - 4z = -9\n\n            Next, eliminate y from the third\n            equation by multiplying the second\n            equation by -1 and adding it to\n            the third equation.\n\n             3y +  2z = 3 \n            -3y -  4z = -9\n            --------------\n                  -2z = -6\n\n            Solve the third equation for z.\n                 \n            -2z = -6\n              z = 3\n\n            Substitute 3 for z in the\n            second equation and solve for y.\n\n            -3y - 2z = -3\n            -3y - 2(3) = -3\n            -3y - 6 = -3\n            -3y = 3\n            y = -1\n\n            Lastly, substitute -1 for y and\n            3 for z in the first equation\n            and solve for x.\n\n            x + (-1) + 3 = 4\n            x + 2 = 4\n            x = 2\n\n            The answer is (2, -1, 3).\n\n\nnext time, do your homework on your own.
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).
791030
3x – 2y = 6\n3x = 6 + 2y\nx = 2 + 2/3y
518063
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.
86546
(3y+1)(-2)+y\n-6y-3+y\n-5y-3\n\n12s+(7-s)2\n12s+14-2s\n10s+14\n\n7x-3x(x+1)\n7x-3x^2-3x\n-3x^2+4x\n\n3t(t-5)+6t^2\n3t^2-15t+6t^2\n9t^2-15t\n\n4w^2-w(2w-3)\n4w^2-2w^2+3w\n2w^2+3w
492288
x=\n3x-3+2x=5x-6-x\n3x+2x-5x+x=-6+3\n1x=-3\nx=-3&#92;1\nx=-3
843483
using (x+y)^3 = x^3+3x^2y+3xy^2+y^3\n\nwe get (2a+b)^3 = 8a^3+12a^2b+6ab+b^3\n\n(ab+c)^3 = a^3b^3+3a^2b^2c+3abc^2 + c^3\n\n(6ab-c)^3 = 216a^3b^3-108a^2b^2c+18 abc^2 - c^3
768805
Factor out the 3:\n3(x^2+3x+2)\nUnfoil:\n3(x+1)(x+2)\nTake the roots:\nx=-1 or -2
924424
The equations:\n(a+b)^2=a^2+2ab+b^2\nA^2-B^2=(A+B)(A-B)\nJames Chan's solutions were right.
379982
For the second equation, move the 9x to the left side and the -21 to the right side\n3y - 21 = 9x becomes\n-9x + 3y = 21\nRight the equations, so that one is right below the other one.\n3x - 2y = -8\n-9x + 3y = 21\nYou can cancel out the x's by multiplying the top equation by 3\n3(3x - 2y) = 3(-8) becomes\n9x - 6y = -24\nNow you have\n9x - 6y = -24\n-9x + 3y = 21\nAdd the two equations together\n-3y = -3\nThus, y = -1\nTo find x, replace y with -1 in either one of the original equations.\nYou will find that x = -2\nAnswer: (-2,-1)
824262
well... imagine 2 functions\n\n2x + 3y = 5\n-2x + 2y = 10\n\nadd the two togather\n\n5y = 15\ny=3\n\nplug y into one of the first equations\n\n2x + 3y = 5\n2x + 3(3) = 5\n2x + 9 = 5\n2x = -4\nx = -2 and y = 3\n\nyou can check it with the other equation too\n\n-2x + 2y = 10\n-2(-2) + 2(3) = 10\n4 + 6 = 10\n10 = 10\n\nand the system is solved... \n\nx = -2 and y = 3
763084
or solve equation for x and y components separately. \nfor x    0 +3+?=0 \nfor y    3+0 +?=0
732343
(1) 5x-2y+z=11\n(2) 3x+4y-2z=-9\n(3) -2x+2y-z=-4\nadd (1) 5x-2y+z=11\nand (3) -2x+2y-z=-4\n3x=7\nx=7/3\n\nnow the equations become\n(4) 5*7/3+2y-z=11\nor 2y-z=11-35/3=-2/3\n\n(5) 3*7/3+4y-2z=-9\nor 4y-2z=-9-7=-16\n\n(6) -2*7/3+2y-z=-4\nor 2y-z=-4+14/3=2/3\n\n4 & 6 are the same, just multiplied by -1\n5 is parallel to them. There is no solution
811856
Okay multiply the top equation by 7 and the bottom equation by negative 3 to obtain 2 new equations with the x variable to eliminate:\n\n7(3x + 7y = -43)\n-3(7x + 6y = -59)\n\nthis becomes \n\n21x + 49y = -301\n-21x - 18y = 177\n\nNow cancel out the 21x's and add down.  Doing that, the equation becomes\n\n31y = -124   this can easily be solved for y\n\ny = -4\n\nNow plug -4 back into one of the original equations\n\n3x + 7(-4) = -43\n\n3x - 28 = -43\n\nadd 28 to both sides to yield\n\n3x = -15   and solve for x\n\nx = -5\n\nNow check x = -5 and y = -4 into both and see that you have solved this system correctly.