Need DETAILED help w/a literal equation?
447259
You've done 90% of work.\n\n(x - 3m)(x - 2m) = 0\nIt can take place only if either x - 3m = 0 or x - 2m = 0.\nFirst equation means that x = 3m.\nSecond means that x = 2m.\n\nSo, there are two roots of equation: 3m and 2m.\nFor m = 3 we have roots: x = 9 and x = 6.
224079
try this site... it shows step by step how to go about sloving these equations, good luck:
909284
((1/x) + (1/y)) / ((1/x) - (1/y))\n\n((y + x)/xy) / ((y - x)/xy)\n\n((y + x)/xy) * ((xy)/(y - x))\n\n(xy(y + x))/(xy(y - x))\n\n(y + x)/(y - x) = (x + y)/(-x + y) = (x + y)/(-(x - y)) = -(x + y)/(x - y)
167128
There's an infinite number of answers.\n\nIf you do it all out, they both equal the same line...\nI multiplied the second equation by a -4 and got 0=0.\n\nIf you just multiply it by a 4 you'd see that it's the same exact line.
240687
4m+H-(4m-h) =\n4m+H-4m+h=\n4m-4m+H+h=\n0+H+h=\nH+h
724411
Sure...\n\nfirst we want to combine like terms on each side of the equals sign.  Think of the equals like the balance point on a scale... you can do anything on one side as long as you match it (keep it balanced) on the other!\n\nso we want to combine the variables and the constants...\n\n8a - 3a + 4 = -a -4 + 2\n5a + 4 = -a - 2\n\nnow we want all the variables on one side, and constants on the other:\n\n5a -(-a) + 4 - 4 = -a-(-a) - 2 - 4\n5a + a = -2 - 4\n6a = -6\n\nnow we just divide by 6 to get a alone:\n\n6a/6 = -6/6\na = -1\n\nHope this helps! :)
902613
I just plugged this into my calculator.\n\n[1, 0, 2, 0, 2, 0 | 0\n0, 1, -3, 0, -2, 0 | 0\n0, 0, 0, 1, -1, 0 | 0\n0, 0, 0, 0, 0, 1 | 0\n0, 0, 0, 0, 0, 0 | 1 ]\n\nTake it however you want.
1154468
make a common denominator\n\na/b - c/d = a*d/b*d - c*b/d*b = (a*d-c*b)/b*d
931316
(-0.242+2.18-1.94)*10^-18\n=(-0.002)*10^-18
1046507
[2x/(x+3)(x-3)]*[(x-3)/4(x)]\ncancelling 2 and x-3 we will get\n=x/(x+3)2(x)\nsimplify we will get\n=x/(2x^2+6x)
971390
{3XY^2M/1}-{2M/[X^-1Y^-1]^2\n+{[4X^2Y^2]/M^-1}-\n(3X^(3Y)^(3M)^-1/(XYM)^-2
736379
for this just substitute for x and y in eqn and see if LHS=RHS\nLHS=4\nRHS=(-1/2)(-3)-5/2\n=3/2-5/2\n=-2/2\n=-1\nnot = RHS\nso pt. not on line
934968
Okay, assuming you mean:\n∫e^(2x)/√(1-e^(2x)) dx\nLet u=1-e^(2x), du=-2e^(2x) dx, dx = -1/2 e^(-2x)\n-1/2 ∫1/√u du\n-1/2 ∫u^(-1/2) du\n-u^(1/2) + C\n-√(1-e^(2x)) +C
703752
simply subtract 7 from both sides. This leave you with\n\nx<_3
613986
=IF(E12=0,0,\nIF(AND(E12>0,E12<4),\n5.9,IF(E12>=4,12.9,\n"negative value")))\n\nAll on one line!\n\nEdit:  The >= (greater than or equal to) is required else the value of exactly 4 would return "negative value"
898860
First one: d= - 1\n\nEDIT: the rest I'm too lazy to do. Do your own homework! =D
29264
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.