What would be the answer to x - 3y= 9     -x + 2y = 1 using elimination?
103832
Place one equation below the other and add:\nx-x elimates x\n-3y+2y = -y\n9+1 = 10\nso -y = 10\ny= -10\n\nso x --3*10 =9\nx=9-30\nx = -21
309995
do this\n9x-3/4y=1  move 9x to other side\n3/4y=-9x+1 get rid of 3/4, by7 multiplying everything by 4/3\ny=-9x*4/3 +1*4/3\ny=-12x+4/3
753930
The most intuitive way to answer is to use Gauss-Jordan elimination, as described below.\n\n\nw + x + y + z = 24\n3w + 3x + 3y + 3z = 72\n3w - 2x - 3y + 3z  = -9\n5x + 6y= 81\n\nw + x + y + z = 24\n5w + 5x + 5y + 5z = 120\n5w + 3x - 2y - 2z  = 4\n2x + 7y + 7z = 116\n\nw + x + y + z = 24\n2w + 2x + 2y + 2z = 48\n2w + 4x + 4y - 5z  = 22\n-2x - 2y + 7z = 26\n\n2x + 7y + 7z = 116\n-2x - 2y + 7z = 26\n\n4x + 9y = 90\n5x + 6y= 81\n\n20x + 45y = 450\n20x + 24y = 324\n21y = 126\n\ny = 6\n\n4x + 9y = 90\n4x + 54= 90\n4x  = 36\nx = 9\n\n2x + 7y + 7z = 116\n18 + 42 + 7z = 116\n7z = 56\nz = 8\n\nw + x + y + z = 24\nw + 9 + 6 + 8 = 24\nw = 1\n{w,x,y,z} = {1,9,6,8}\n\nAnother method is to put the coefficiants into a matrix, and put it into reduced-row echelon form.\n\n|1 1   1  1  24|\n|3  -2 -3 3 -9|\n|5  3  -2 -2 4|\n|2  4  4  -5 22|\n\nIn RREF...\n|1 0 0 0 1|\n|0 1 0 0 9|\n|0 0 1 0 6|\n|0 0 0 1 8|\n\nThese are all new valid equations, just in a much nicer form. Specially, 1w+0x+0y+0z=1, etc.\nThis tells us that w=1, x=9, y=6, and z=8.\n\nAlso, you could use Cramer's rule with these equations and get the same answer, but I will not demonstrate that one.
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)
892382
To solve the first one, divide each part of the first equation by two, so 2x-3y=-13; since -2x+3y = 13, adding the two equations gives 0x+0y=0.  No help.  Try substitution.\nFrom the first equation, 4x = 6y-26; x = 1.5y - 6.5\nsubstitute into the second equation -2(1.5y-6.5) + 3y = 13; -3y +13 + 3y = 13; 13=13.  Still no help.\nBoth equations are equivalent; the line that represents them is the solution.\n\nProblem two:\nSubtract equation 2 from equation 1.\n3x-3x+2y-3 = 6 - y\n2y-3 = 6-y\n3y = 9\ny = 3\n3x+6=6\nx=0\nchecking\n3x+3=y=3\n3x=0\nx=0
706243
-2y+3=7y-42\n-2y+3-3=7y-42-3\n-2y=7y-45\n-2y-7y=7y-7y-45\n-9y=-45\n-9y/-9=-45/-9\ny=5
240934
Solution: y = 9 - x or y = -x + 9\n\nHere are the steps I took:\nThe equation must equal y = mx + b, so we start with:\n1. x + 2y = 18\n2. x + y = 9 (Divide both sides by 2 to remove it from the "y")\n3. y = 9 - x (Subtract "x" from both sides)\n\nIn this type of equation in algebra, you have to produce steps to isolate the "y".
164756
Easy!\n\n2x+2y=2 \n3x-y=1\n\nTake the second equation and isolate the single "y":\n\n3x - y = 1 becomes 3x - 1 = y (without going into how)\n\nSubstitute "3x - 1" in the first equation so it reads:\n\n2x+2(3x - 1)=2\n\nSimplify:\n\n2x + 6x - 2 = 2\n8x - 2 = 2\n8x = 4\nx = 1/2\n\nEarlier, we saw that y = 3x - 1, so:\n\ny = 3(1/2) - 1\ny = 3/2 - 1\ny = 1/2\n\nThere may be another way (less involved) that I've forgotten, but you can check each of those values for x and y in both equations to make sure they work.
973495
Replace the y in the first equation with its value from the second:x+9.  You get x^2 + (x+9) = 9\n\nSubtract 9 from both sides to get x^2 + x = 0\n\nFactor out an x:  x(x+1) = 0\n\nSo x is 0 or x is -1\n\nPlug these into the original equations to get y.  \ny= 0 + 9; y=9\n\ny = -1 + 9; y=8\n\nSo the complete answers are (0,9) and (-1,8)Check them in the other equation: 0^2 + 0 + 9 = 9 check\n(-1)^2 + 8 = 9; 1+8=9 check
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}
861275
x(y-1) = 2y + 3\n\nxy - x = 2y + 3\nxy - 2y = x + 3\n(x - 2) = (x + 3) / y\n y = (x + 3) / (x -2)
1137023
(2x-9)(2x+9)\n\na guess... its been a long time since I did that stuff
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).
433587
This is a simple linear equation. Sole for y.\n\ny=(11/9)-x.\nThis is quickly put in the calculator and graphed.\nThere are infinetly many solutions to this equation since there are no limits on the domai.\n\nTo prove this:\n\nYou use substitution.\nSolve for x    x=(11/9)-y\nThen, substitute this for x in the original equation.\n9((11/9)-y)+9y=11\nDistribute the 9\n11-9y+9y=11\nAn identity 11=11 demonstrates that there are infinetly many solutions.
840475
x^9- y^3z^3=(x³-yz)³ \n                  =x^9-3*x^6*yz+3x³y²z²-y³z³
696505
3y-2x=14\n3y= 2x+14\ny= 2/3x+14/3\n\n7-6 over 3-4 = 1/ -1\nm= -1\n\ny=mx+b\n6=-1(4)+b\n6=-4+b\nb=10\n\ny=x+10
829339
x^2-6x+9 = (x-3)(x-3)\nx^2-9 = (x-3)(x+3)\n\nMultiple everything by (x-3)(x-3)(x+3)\n\n(x+3)(x+3) - (x+2)(x-3) - 5(x-3)(x+3)\n\nx^2+6x+9 - (x^2-x-6) - (5x^2-45)\n\nx^2+6x+9-x^2+x+6-5x^2+45\n\n-5x^2 + 7x + 60