Algebra Linear equations?
814480
The slope-intercept form of a line is\ny = mx + b, where m is the slope and b is the y-intercept.
469735
The system of equations:\n[a11 a12 a13; a21 a22 a23]\n[1 -3 4; 2 3 1]\n\nPerforming indicated row operation:\n[a11 a12 a13; a21' a22' a23']\n[1 -3 4; -1 -6 3]\n\na21' = a11 - a21 = 1 - 2 = -1\na22' = a12 - a22 = -3 - 3 = -6\na23' = a13 - a23 = 4 - 1 = 3
588758
a.  ax + bx\nb.  21x^3 + 24x^2y\nc.  24x^2 + 31x + 10\nd.  24x^2 - 8x - 6\ne.  40x^3 + 66x^2 + 38x + 6  (I assumed that you meant 7x in the first half of the equation.  If not, this is not correct.)\nf.  x^2 -1\ng.  x^3 - 8
589722
Step 1: Solve one equation for one variable in terms of the other.\nStep 2: Plug that into the other equation and solve for the second variable.\nStep 3: Plug the second variable into the first equation and solve for the first variable.
978722
You can't, algebraically.  You have two equations and three variables.  Sorry.  You need one more equation, such as a = b.
576570
algebra...usually (well i do) set up a equation for everything, and solve for the variable. If u need help u can always get help in www.algebra.com .....
926227
Well Aq = b. where A is the matrix of coefficients, q is [x; y; z; w], and b is the answer [-2; 2; -11; -3] So we are looking for the Rx = d, that corresponds to the original equation. The aumented matrix is \n     5     3     1     5    -2\n     3     2     2     4     2\n     2     4     3    -3   -11\n     4    -3    -2     2    -3\nThe first step is to create a 1 in 1,1 so we divide Row1 by 5.\nThen subtract 3xRow1 from row 2, 2xRow1 from Row3, 4xRow1 from Row4\nThis is the result\n         1    0.6     0.2    1       -0.4\n         0    0.2     1.4    1        3.2\n         0    2.8     2.6   -5    -10\n         0   -5.4    -2.8   -2      -1.4\nContinue down and to the right with row operations in the same vein until the matrix is in row echelon form then go up and to the left in order to create 0s in all the non-step positions and get the matrix in reduced row echelon form.\n\nThe reduced row echlon form of  the augmented matrix [A b] is \n     1     0     0     0    -2\n     0     1     0     0    -1\n     0     0     1     0     1\n     0     0     0     1     2\n(Heh look at the nice numbers. You got this out of a textbook didn't ya?)\nThis was of course gotten by the Gauss-Jordan Method.\nSo now it's just a matter of reading off the coefficients\n1q1 + 0q2 + 0q3 + 0q4 = -2\n0q1 + 1q2 + 0q3 + 0q4 = -1\n0q1 + 0q2 + 1q3 + 0q4 =  1\n0q1 + 0q2 + 0q3 + 1q4 =  2\n\nSo q = [-2; -1; 1; 2] and x = -2, y = -1, z = 1, w =2.\n\nIf you want all the row operations, well there are an awful lot of them and they're all simple and straightforward. It's simply a matter of carefully working through the algorithm, so you are better off working them out for yourself. You'll probably get them before I can type them all in.
704540
Add the two equations together: (2x+y)+(x-y) = 1+8, so 3x = 9, x = 3.  Then 2*3+y = 1 so y = -5.\n\nSteve
695176
6x^2-3x-4x+2\n3x(2x-1)-2(2x-1)\n(2x-1)(3x-2)
931883
Ummm...\nThe equation of algebra, eh?\nWhat on earth are teachers teaching kids these days?!
540264
a linear equation is just x+y=a number, or y=x+a number.  It could also have a coefficient in front of x or y in either equation.  To graph it, either use the slope-intercept form, where you take y=x+a number, the coefficient in front of x is the slope, and the number after x is the y-intercept, so, you plot (0, the number) and use the slope to find another point, then draw your line.  The second is to find two or three points that fit the equation and plot them.  Just substitute a number in for x two or three times, find y each time, and plot those numbers, then draw a line through them.
1160760
what you have is something like\na(b+c)\n\nthe solution is \nab+ac\n\n(x^3)(2x^-2)+(x^3)(4x)\nnow to multiply x^a * x^b you get x^(a+b)\n\n2x^((3+(-2))+4x^(3+1)\n\n2x+4x^4
222015
5x - 2y = 4  (1)\n2x + 3y = 13 (2)\nmul. eq.(1) by 2: 2 (5x - 2y = 4)\n                    10x - 4y = 8\n\nmul. eq.(2) by 5: 5 (2x +  3y = 13)\n                    10x + 15y = 65\n\nsubtract: 10x -  4y =  8\n      (-) 10x + 15y =  65\n          ________________\n               -19y =  -57 \n                  y =  -57/-19\n                  y =  3\n\nsubstitute value of y to any of the eqs.\n\nSubstituting to eq. (1):\n\n    5x - 2y   = 4   \n    5x - 2(3) = 4\n    5x - 6    = 4\n    5x              = 4 + 6\n    5x              = 10\n     x              = 10/5\n     x              = 2\n\nso, x = 2\n    y = 3\n\nto check, substitute values of x and y to any of the two equations:\n    for eq. (1),\n\n    5x   - 2y   = 4   \n    5(2) - 2(3) = 4\n      10 - 6    = 4\n\n    for eq. (2),\n\n    2x   + 3y   = 13\n    2(2) + 3(3) = 13\n      4  + 9    = 13
1051508
Your equation would look like x^2-ax+b.  Which means when you solve the equation you would two answers one of which will contain a plus sign other will contain a minus sign since the middle term is negative.  Thus the answer is C.
202779
This is quadratic. A quadratic equation looks like:\nax^2+bx+c = 0\n\nSo a=4, b=-1, and c=3\n\nSo by using the quadratic equation: \n\nx= (-b+sqrt(b^2-4ac))/2a\nand\nx= (-b-sqrt(b^2-4ac))/2a\n\nTry to get the answer from that.... note that imaginary numbers may be involved!
796486
a)\n4x + 4y = -12\n-2x + 3y = 14\n\nDouble the second equation and add:\n4x + 4y = -12\n-4x + 6y = 28\n10y = 16\ny = 16/10\ny = 8/5\n\nThen solve for x:\nx + (8/5) = -3\nx = -3 - 8/5\nx = -4 3/5\n\nb) \n3x + 2y = 5\n2x - 3y = 3\n\nMultiply the top by 2 and the bottom by 3, then subtract:\n6x + 4y = 10\n6x - 9y = 9\n13y = 1\ny = 1/13\n\nThen solve for x:\n3x + 2(1/13) = 5\n3x + 2/13 = 5\n3x = 63/13\nx = 21/13\nx = 1 8/13
9550
aX +b =c\n=> aX = c-b\n=> X = (c-b)/a\n\nin this case:\n1/3(X-4) =1\n=> 1/3X - 4/3 = 1\na=1/3\nb=-4/3\nc=1\n\n=> X = (c-b)/a\n=> X = (1+4/3)/(1/3)\n=> X = 3(7/3)\n=> X = 7