What is the solution to this equation?
921168
is there a typo in the first equation?\n---\n\nthe idea is to find a common multiple for one of the variables x and y so that we can "get rid" of it. one way to do that is the following:\n\nmultiply first equation by 2 to obtain \n -14x + 12y = 84     --- (3)\n\nmultiply second equation by 3 to obtain\n -9x + 12y = 84       --- (4)\n\nnow the multiples of y in both equations are the same. we can proceed to "get rid" of y\n\ndifference of equations (3) and (4):\n       -14x + 12 y = 84\n     - (-9x  + 12y = 84 )\n-----------------------------\n        -5x             = 0\n\nthus x = 0\n\nnow substitute x = 0 into either the first or second equation and you'll get y = 7
295616
D (the discriminant) = b^2 - 4ac = 25 - 4*(-2)*0 = 25\nSince D>0  there are two real solutions.
413333
let suppose that \nr(@) = 4/ sqrt( 4cos^2(@) +sin^2(@) )\nthen x(@(t) ) = r(@) . cos (@) emplies\n\nx^2 (t) = r^2 (@) . cos^2(@)\n\nx^2 = 16 / (4 cos^2 + sin^2)  . cos^2\n\nso x^2(t) = 16/ (4+ tan^2(t) )\n\nBut tan( @(t) ) = y(t) / x(t)\n\nThus: x^2 = 16 / (4+ y^2 / x^2 )\nThus: \nEQ : x^2/4 + y^2/16 =  1\n\nCheck @=0  --> r(@=0) = 2\nAnd (x=2 , y=0 ) is solution of EQ
897159
x + y = 2- - - - - - -Equation 1\ny - x = 2- - - - - - -Equation 2\n- - - - - - - -\n\nRearrange equation 2 to read\n\n-x + y = 2\n\n- - - - - - - -\n\nx + y = 2\n-x + y = 2\n- - - - - - - -\n\n2y = 4\n\n2y/2 = 4/2\n\ny = 2\n\nThe answer is y = 2\n\nInsert the y value into equation \n\n- - - - - - - - - - - - - - - - - - - - - \n\nx + y = 2\n\nx + 2 = 2\n\nx + 2 - 2 = 2 - 2\n\nx = 0\n\nThe answer is x = 0\n\nInsert the x value into equation 1\n\n- - - - - - - - - - - - - - - - - - - - - -\n\nCheck for equation 1\n\nx + y = 2\n\n0 + 2 = 2\n\n2 = 2\n\n- - - - - - - - -\n\nCheck for equation 2\n\ny - x = 2\n\n2 - 0 = 2\n\n2 = 2\n- - - - - - -\n\nThe solution set is { 0 , 2 }\n\n- - - - - - -  -s-
741586
set y=0\nsolve quadratic eqn\nquadratic eqn\n\n-7 +- (49-16)^.5\n-------------------\n2\n= (-7 +- 33^.5)/2
69673
The Question is reversed \n\ny =4e^(-x) sin3x     ---------    (eqn1) \n\ny'=[4e^(-x) 3cos3x] + [-4e^(-x) sin3x]\n\ny'=[4e^(-x) 3cos3x] - [y] ----- From(eqn1)y =4e^(-x).sin3x \n\ny'=[4e^(-x) 3cos3x] - [y] ------(eqn2)\n\ny"=[4e^(-x)(-9)sin3x] + [ -4e^(-x)3cos3x ] - y' \n\ny"+y'=[-9y] - [4e^(-x) 3cos3x]   ----From(eqn1)y=4e^(-x)sin3x \n\ny"+y'=[-9y]- [y'+y] ------ From(eqn 2) 4e^(-x).3cos3x =y'+y\n\ny"+2y'+10y=0\n\nSo this differential equation has  y=4e^(-x)sin3x as one of it's solutions
517123
It's hard to show it clearly here but the solution is actually:\n\n[-b±√(b²-4ac)]/2a\n\nthere will usually be two solutions!\nThe site below has it laid out more clearly...\n\n("invited" to leave school early.. 1974)
929231
(a^2-b^2)=(a-b)(a+b)\nHence,\nsec^4y-tan^4y = (sec^2y-tan^2y)(sec^2y+tan^2y)\n\nSubstitute that back into the initial equation...\n\n(sec^2y-tan^2y)(sec^2y+tan^2y)\n------------------------------------------\n           (sec^2y+tan^2y)\n\nand cancel common terms\n\n= (sec^2y-tan^2y) \n= 1 (a standard trig identity)\n\nhope that solves it.
279194
It easily follows that the general solution to this differential equation is y=Ae^{kt} for some constant A [You may differentiate to double check your answers.] \n\nSo given the initial condition, i.e. y(t=60)=0.5*A, we obtain k=(ln(0.5))/(60).
417188
y=Asin(2x)+Bcos(2x)+(x/4)\n\nNow you get to determine why.
879495
so easy :\nx^2 - Sx + P = 0\n\nwhere, S is the sum of the roots and P is the Product of the Roots :\n\nS = 2+(sqrt)2 + 2-(sqrt)2 = 4\n\nP = [2+(sqrt)2][2-(sqrt)2]= 4-(2) = 2\n\n==> the equation is x^2 - 4x + 2 = 0\n\nhope this helps enjoy !!??!!
797913
That's the quadratic formula. It is the solution to equations of the form\nax^2+bx+c=0
979866
If I am reading this as you have submitted, then, the solution of x=2secx is transcendental, and needs to be solved graphically giving a set of values that satisfy it; you plug that set of values for x that satisfy it into y=3+2tanx, and that gives the solution set for y.
741999
whats the equation or formula u need help with?
531201
the solution is 1. see 6^1,5^2,4^3,3^4,2^5,1^6.
402617
No solutions.\nFrom the first equation you have\ny=x^2+2x+9/2\n\nUsing the second:\nx^2+2x+3=x^2+2x+9/2\n3=9/2 , so there is no solution
196036
If you divide 2 to the second equation, you will notice that it is identical to the first one, so they are dependent to each other, and there are infinitely many solutions to this problem.\n\n^_^