How do I solve these equations:?
821835
Problem 1.\nu = sin x\nu^2 + 3u + 2 = 0\nu = -3/2 +/- 1/2\nu = -1, -2\nsin x = -1  [because the sine is never -2]\nx = 270 degrees\n\nProblem 2.\nu = cos x\nu^2 + 2 u + 1 = 0\n(u+1)(u+1) = 0\nu = -1\ncos x = -1\nx = 180 degrees
537199
got to math.com, click on calculators and click on solve equations.....it'll do the math for ya
81494
I'm getting some funky answers on these equations\n\nx= -48/7 and y= -156/7
551906
Add each equation\nA + B = 3x^2 + 2x - 2\nA - B = -x^2 + 4x - 8\n2A = 2x^2 + 6x - 10\nDivide each side by 2\nA = x^2 + 3x - 5\nNow plug into one of the original equations to get B\nB = 3x^2 + 2x - 2 - (x^2 + 3x - 5)\nB = 2x^2 - x + 3
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
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!
803558
1.\nc=(xy^2-1)/y\ndifrentiate w.r.t x\n(y(y^2 + 2xyp)-p(xy^2-1))/y^2=0\ny^3 + 2xy^2p - pxy^2 + p=0\ny^3 +xy^2p +p=0\nwhere p=(dy/dx)\n2.\nsimilarly diffrentiate \n(x+y^2)/x^2=-c\nto get\n2(x^2)yp - x^2 - 2xy^2=0\nthanks
695176
6x^2-3x-4x+2\n3x(2x-1)-2(2x-1)\n(2x-1)(3x-2)
366751
make x the width, and solve the two equations (the one given, and the one assumed about the rectangle where l*w=A)
741586
set y=0\nsolve quadratic eqn\nquadratic eqn\n\n-7 +- (49-16)^.5\n-------------------\n2\n= (-7 +- 33^.5)/2
975552
x = -5/2\ny = 6\n\nSolve both equations for y.  Set them = to each other and solve for x.  That will give you the x value -5/2.  Then plug -5/2 into either one of the equations for x, and solve for y.
758594
2. The equation is (y-y1)/(y1-y2)=(x-x1)/(x1-x2)\n=>(y+2)(-2-0)=(x+1)(-1-0)\n=>(y+2)(-2)=(x+1)(-1)\n=>-y-2=-2x-2\n=>-y+2x=0=> 2x-y=0 which is the required equn.\n3.The equation is (y-4)(4+4)=(x-0)/(0-4)\n=>y-4/8 =x/-4\n=>8x=-4y+16  or,8x+4y-16=0  o4 2x+y-4=0 is the reqd.equation\n4.the eqn. is (y+3)/(-3-3)= (x+2)/(-2-0)\n=>(y+3)/-6=(x+2)/-2\n=>-6x-12=-2y-6\n=>-8x+2y-6=0 or,-4x+y-4=0 which is the equired equation
741999
whats the equation or formula u need help with?
801397
equation (2)*2\n6x-4y+8z=12\n9x+4y-z=0\nadding\n15x+7z=12    (4)\n(1)*2\n18x+8y-2z=0\n6x-8y-3z=3\nadding\n24x-5z=3        (5)\n(4)*5        75x+35z=60\n(5)*7      168x+35z=21\nadding     243x=81\nx=1/3\nsubstituting\n8-5z=3\n-5z=-5\nz=1\nsub\n3+4y-1=0\n4y=-2\ny=-1/2\nsolution set{1/3,-1/2,1}
708644
Use the quadratic formula.\nx1 = 1\nx2 = -1/2
782468
1.ac\n\n2.-3r\n\n3.3m/-2r\n\n4.(x-4)\n\n5.(a-b)(b-c)
498659
1) y=x+4\n   3x+y=16 (I guess you made a mistake by writing +16)\n--------------\nsubstitute x+4 for y in second equation and you'll get\n\n   3x+x+4=16\n    4x+4=16\n    4x=12\n     x=3\n\nnow y=x+4=3+4=7\n\nsolution (x,y)=(3,7)\n\n2) 4x+5y=48\n    y=3x+2\n--------------\nsubstitute 3x+2 for y in first eq.\n\n4x+5(3x+2)=48\n\n4x+15x+10=48\n\n19x=38\n\nx=2\n\nnow y=3x+2=3*2+2=6+2=8\n\nsolution (x,y)=\n(2,8)\n\n3)3x-y=0\ny=6\nin first equation instead of y write 6\n\n3x-6=0\n3x=6\nx=2\n\nso, the solution is (2,6)\n\n4) x-2y=-10\n3x-y=0\n\nfrom the first equation you have\nx=2y-10\nnow substitute 2y-10 for x in second one\n\n3(2y-10)-y=0\n6y-30-y=0\n\n5y=30\ny=6 \nnow x=2y-10=2*6-10=12-10=2\nso x=2 and y=6