I got -1 -1 When I sovled this equation using the Gauss Jordan method?
356990
Gauss Jordan method for this:\n\n| 6 -5 | 4 |\n| -6 5 | -2 |\n\nR1/6 -> R1\n| 1 -5/6 | 2/3 |\n| -6 5 | -2 |\n\n6R1 + R2 -> R2\n| 1 -5/6 | 2/3 |\n| 0 0 | 2 |\n\nFrom this, we see that there is no solution, since we have a row of 0's on the left of the matrix.\n\nIf you try to solve this by substitution or by any other method you know, you'll find that you can't.  It turns out that these are parallel lines, and parallel lines don't intersect, so you won't get an answer for this particular set of equations.
316686
the process u follow is correct, but actually it is wrong. fundamentally wat u did is ok, but technically it is wrong. because this becomes a quadratic equation which has more than 1 root. so this equation has more than 1 answers for which it is true.
510352
The solution by a_ebnlhaitham is not quite correct in that he omitted the 2 from the power of r (i.e.(teta)).The answer should be 1/(1 - r^2) not 1/(1 - r). The series converges because r lies in the open interval (-1,1).
35268
slope = (-4-(-6))/(-2-0)\nslope = -1
889697
well gauss jordan is a easy method for matrix manipulations. if u use reverse triangulation also in conjunction with it . it wud be much more simpler.\n\n2x-y+3z=8\n-x+2y+z=4\n3x+y-z=0\n\nthis is set of equations given.write them in matrix form\n\n|| 2  -1  3 ||  ||x||     ||8||\n|| -1  2  1 ||  ||y||  = ||4|| \n|| 3   1  -1||  ||z||     ||0||\n\nA    X= B\nnow the equation is in the form of AX=B\n\nnow our concentration will be on matrices A and B\n\nour aim is to make the lower triangle elements equal to 0.\n\nnow follow the steps one by one. apply the calculations on both A and B  matrices.\n\nwe have 3 rows in matrix A and let us name them as R1 , R2 and R3.\n\nreplace R2 by R2/-1 and R3 by R3/3\n\nnow the matrices will be\n\n||2   -1    3    || ||x||   ||8 ||\n||1   -2   -1    || ||y||= ||-4||\n||1   1/3  -1/3 || ||z||   ||0 ||\nreplace R2=R1-2xR2\n||2   -1    3     || ||x||    ||8 ||\n||0    3    5     || ||y||=  ||16||\n||1   1/3  -1/3 ||  ||z||   ||0 ||\ncheck once what we have done here. \nevery element of R2 is calculated from the preceding element in that colum by the formula R1-2xR2.\n\nthe elelment of colum one is\n2-2x1=0\nthe element of colum 2 is -1-(2x-2)=3\nthe element of colum 3 is 3-(2x-1)=5\n\nlike wise \nreplace R3=R1-2xR3\n||2   -1    3      || ||x||     ||8 ||\n||0    3    5      || ||y|| =  ||16||\n||0   -5/3  11/3 || ||z||     ||8 ||\n\nnow we got the first colum zero in the lower triangle.\n\nnow the second element in the third row needs to be zeroed.\nreplace R3=R3x(-3/5) or  R3=R3/(-5/3)\n||2   -1    3      || ||x||   ||  8   ||\n||0    3    5      || ||y||= ||  16  ||\n||0    1  11/5   || ||z||   ||-24/5 ||\nreplace R2=R2/3\n||2   -1    3      || ||x||   ||  8   ||\n||0    1    5/3   || ||y||= || 16/3 ||\n||0    1  11/5   || ||z||   ||-24/5 ||\n\nnow replace R3=R3-R2\n||2   -1    3      || ||x||   ||  8     ||\n||0    1    5/3   || ||y||= || 16/3   ||\n||0    0    8/15 || ||z||   ||-152/15 ||\n\nnow apply the final steps of reverse triangulation technique.\nfrom the matrix equation, if again converted in to normal equations \n\n2x-y+3z   =8\n  y+5z/3  =16/3\n    8z/15 =-152/15\nnow z= (-152/15) divided by (8/15)\n= (-152/15)/(8/15) =-152/8=-19\nthere fore z=-19\nnow substitute z=-19 in equation 2\n\ny+5x(-19)/3=16/3\ny=(16/3)+(95/3)=111/3=37\nsubstitute the values of y and Z in equation 1\n\n2x-37+3x-19=8\n2x=8+37+57=102\nX=102/2=51\nfinally we got the values of x = 51, y=37 and z=-19\n\nactually u can continue the guass jordan method without applying the reverse triangulation method to get the values of x, y \n\nand z directy from the matrix for that after making the lower triangle zero we have to make the upper triangle elements also \n\nequal to zero.\n\nlet us see how to do that.\n||2   -1    3      || ||x||   ||  8     ||\n||0    1    5/3   || ||y||= || 16/3   ||\n||0    0    8/15 || ||z||   ||-152/15 ||\nreplace R3=R2/(8/15)\n||2   -1    3     || ||x||   ||  8     ||\n||0    1    5/3  || ||y||= || 16/3   ||\n||0    0    1     || ||z||   || -19    ||\nR2=R2-5xR3/3\n||2   -1    3    || ||x||  ||  8     ||\n||0    1    0    || ||y||= ||  37    ||\n||0    0    1    || ||z||  || -19    ||\n\nR1=R1/3\n||2/3   -1/3    1    || ||x||   ||  8/3   ||\n||0       1     0     || ||y||= ||  37    ||\n||0       0     1     || ||z||   || -19    ||\n\nR1=R1-R3\n||2/3   -1/3    0    || ||x||   ||  65/3  ||\n||0       1     0     || ||y||= ||  37    ||\n||0       0     1     || ||z||   || -19    ||\nR1=R1x-3\n||-2    1     0    || ||x||   ||  -65   ||\n||0       1     0    || ||y||= ||  37    ||\n||0       0     1    || ||z||   || -19    ||\nR1=R1-R2\n||-2    0     0    || ||x||   ||  -102  ||\n||0       1     0    || ||y||= ||  37    ||\n||0       0     1    || ||z||   || -19    ||\nR1=R1/(-2)\n||1    0     0    || ||x||  ||  51    ||\n||0       1     0    || ||y||=||  37    ||\n||0       0     1    || ||z||  || -19    ||\nnow we can diretly read the matrix into equations as\nx=51, y=37 and z=-19\n\n\nhope und
187139
Yes (-1,0) and (1,-2) share the tangent line y=-x-1\n\nwhich y=-x-1 is tangent to the function y= f(x)=x^4-2x^2-x\n\nNote that (-1,0) and (1,-2) satisfy y=x^4-2x^2-x
424681
substituiting your values you get (the values you gave are not very clear, can't make out whether positive or negative.\nAssuming all values are positive...)\ny-2 = 2(x-5)\ny-2 = 2x -10\n2x -y - 8 =0  is of the form ax + by + c =0
1117062
if ur just solving that equation...u dont need to switch the sign.....\nwatch (im not gonna have the equal sign on the bottom takes too long). \n\nx-1<-3\n(add one to both sides now)\nx<-2\n\nthers ur answer
291468
Oh yes i have cracked it at last....\n\nsolution...\n\nLet x=tan-1 (y) (i mean tan inverse)\nthen tan x = y,and cot x = 1/y (coz tan-1 (x) = cot-1 (1/x))\n\nthen ur given equation's Left hand side part (LHS) will be\n\n(1/y)/(1-y) + (y)/(1-(1/y))\n\n= 1/(y-y2)+ y/(1-(1/y))\n\n= 1/(y-y2) + y2/(y-1)\n\n= y2(y-1) - 1/y(y-1)\n\n= y3/y(y-1)\n\nI hope u understood till this step.\n\nNow we have (a3-b3) = (a-b) (a2+ab+b2)\n\nthen the LHS becomes...\n\n(y-1)(y2+y+1)/y(y-1)\n\n= (y2+y+1)/y\n\n= y+1+(1/y)\n\nsubstitute back y=tanx, 1/y = cot x and leave 1 as it is...\n\nU have LHS = 1 + tan x + cot x...Thats ur RHS !!!
657596
the answer doesnt work out correctly because you have a negative under the square root of the quadratic equation.  but it can be solved using i.  then the answer would be 2.33i and -1.73i
868180
the roots are (2a-1)x(2a-1)
781847
(x²+1)/(x+1).\nf(x) = x²+1, g(x) = x+1\n[ 2x*(x+1) - x² * 1 ] / (x+1)²\n(2x²+2x - x²) / (x+1)²\nx²+2x / (x+1)²\n\nf(x) = x²+2x, g(x) = (x+1)²\n(( [ (2x+2)*(x+1)² ] - [ x²+2x * 2(x+1) * 1 ]  )) / ((x+1)²)²\n[ (2x+2)(x²+2x+1) - (x²+2x)(2x+2) ] / (x+1)^4\n[ 2x³+4x²+2x+2x²+4x+2 - (2x³+2x²+4x²+4x) ] / ((x+1)²)²\n[ 2x³+4x²+2x+2x²+4x+2-2x³-2x²-4x²-4x) ] / ((x+1)²)²\n[ 2x+4x+2-4x) ] / ((x+1)²)²\n(2x+2) / ((x+1)²)²\n2(x+1) / (x+1)^4\n2 / (x+1)^3  <----dy²/dx²  Yours was incorrect.\n\nAnd yes, this will never have an inflection point.  It will only approach y=0 as x approaches +/- ∞ and approach x=0 as x approaches 0
458222
0 1 0 0\n0 0 1 0\n1 0 0 0\n0 0 0 1\nPut the identity matrix behind your matrix.\nTurn you matrix into the identity matrix using row operations.  The matrix in the back will be the inverse.\n0 1 0 0  1 0 0 0 \n0 0 1 0  0 1 0 0 \n1 0 0 0  0 0 1 0 \n0 0 0 1  0 0 0 1\nSwap row 1 with row 3\n1 0 0 0  0 0 1 0 \n0 0 1 0  0 1 0 0 \n0 1 0 0  1 0 0 0 \n0 0 0 1  0 0 0 1\nSwap row 2 with row 3\n1 0 0 0  0 0 1 0 \n0 1 0 0  1 0 0 0\n0 0 1 0  0 1 0 0 \n0 0 0 1  0 0 0 1\nThus, the inverse is\n0 0 1 0\n1 0 0 0\n0 1 0 0 \n0 0 0 1
699429
c'mon dude, these are easy.\n\nAdd second to first.\n\n2x = 2, x = 1, therefore y = -2.
325892
= 1-{1-[1-(0)]}\n= 1-{1-[1]}\n= 1-{0}\n= 1
399118
y-int= (0,-5)\nx-int= (-1.2,0)
802801
i think -1; y=mx+b, "b" is the 3 and "m"(slope) is negative one.