I need help solving these equations for the unknown?
825343
Answer 1:\n(4y + 2)/2 = 21\n4y + 2 = 21*2\n4y + 2 = 42\n4y = 42 - 2\n4y = 40\ny = 40/4\ny = 10\n\nAnswer 2:\na + 4 = 3a - 81\na = 3a - 81 - 4\na = 3a - 85\na - 3a = -85\n-2a = -85\n2a = 85 (The minus signs cancel each other out)\na = 85/2 (or) 42.5\n\nAnswer 3:\n146 = (36y + 2)/4 = 236.5?\nThis means that 146 = 236.5 which is never possible\nThis implies that the equation has no solution.
408462
to solve such a system in general, what you do is this: take the first equation, solve it for x treating y as you would a constant. In your case, this is x = 4 + y.\n\nThen, take your expression for x and substitute it everywhere you see x in the second equation; in your case you get 3 *(4+y) + 2y = 7.\n\nSolve this now 1-variable equation for y, once you find y (in this case, -1).\n\nSubstitute the value of y into either equation, and you get the value of x (3 in this case).
663777
SOrry, I can't help you. If I had the formulas I could do it, but there is no way I remember that off the top of my head! Good luck
610726
there is a solution, all x,y that satisfy this equation:\ny = 4 - 2x\n\nwhich is a straight line with intercepts (0,4) & (2,0)\n\nIt has an infinite amount of solutions. To be able to solve for actual numbers, you need as many equations as you have variables.
555500
here are the answers:\n-17x-19 for the first one \n22x+1 for the second one\n-6x^2+3x+15 for the third one, which can also be written as -3(2x^2-x-5)
983612
First swap the variables, then solve for R again.\n\nv = (21/2) - (1/64)R^2\nR^2 = -64v + 672\nR = ±√(-64v + 672)
967876
1) 2a^3 - 13a^2b + 11ab^2 + 14b^3\n\n2) (2a-4b) (4a^2+8ab+16b^2)\n\n3)  (b-8)(b-1)=0\n      b=8   or    b=1\n\n4) 4m^3 - 26m^2 - 14m=0\n2m(2m^2 - 13m - 7)=0\n2m=0  or  2m^2 - 13m -7=0\nm=0  or    (2m+1)(m-7)=0\n                  m=-0.5   or  m=7\n\n so the final answer is m=0 and m=-0.5 and m=7
423853
[2x^{2}+ 2x2y]+[2x2y+2x^{2}]+\n[6yz^{9}+3y^{2}9z^{8}]+\n[3y9z^{8}+y^{3}72z^{7}]\n\nthis is using z as the last term.  Based on what I used, I think a 3D parabola of sorts will emerge
930340
PROBLEM 1:\n\nsqrt(x^3) = 8\n\nSquare both sides:\nx^3 = 64\n\nNow take the cuberoot of both sides:\nx = cuberoot(64)\n\nSolve:\nx = 4\n\nPROBLEM 2:\n\ncuberoot(x^2) =4 \n\nCube both sides:\nx^2 = 4^3\n\nSimplify:\nx^2 = 64\n\nTake the square root of both sides:\nx = ±8
620893
The best way to work this out would be to plot a speed time graph for the information.\n\nOr you could express it as a simultaneous equation.
725212
Yeah, make them graphable on your calc.\n\ny = 4-x\ny = 2x-2\n\nIf you want to solve, just set 4-x and 2x-2 equal to each other (since y must also be equal in both equations)\n\n4-x=2x-2\n3x=6\nx=2\n\n\n2+y=4\ny=2\n\nTheir intersection is (2,2)\n\nThe y-intercept of the first is 4, and the second is 2.\nThe slope for the first is -1, and the second is 2.\nThe x-intercept of the first is 4, and the second is 1.
502102
well chinaren got it started but I believe his explianation is weird.\nLets assume x=3^t\nthen log3(x)+log2(x+1)=3\n=>log3(3^t)+log2(3^t+1)=3\n=>tlog3(3)+log2(3^t+1)=3\n=>log2(3^t+1)=3-t\n=>3^t+1=2^(3-t)\n=>3^t+1=2^3/2^t\n=>(3^t+1)*2^t=2^3\n=>3^t*2^t+2^t=8\n=>6^t+2^t=8\nfrom here there is one one solution possible as t=1\nso x=3^1=3
654517
x^2-2x+1\n(x-1)(x-1)\nso zero at x=1\n\nx^2-x=6\nx^2-x-6=0\n(x-3)(x+2)=0\nzero at x= 3, and x=-2\n\nx^2-3x-4\n(x-4)(x+1)\nzero at x=4  or x=-1\n\nx^2+8x+15\n(x+3)(x+5)\nzero at x=-5 or x=-3\n\nx^2-6x+5\n(x-5)(x-1)\nzero ar x=5, or x=1\n\nx^2+9x+20\n(x+4)(x+5)\nso zero at x=-4 or x=-5
101802
I don't think there is a unique solution for it. You have more unknown variables than the equations.
925536
To solve, first, you need to know certain Laplace Transforms.  You can find them from a table or solve for them using the definition:\n\nL[f'(x)] ---------> sF(s) - f(0)\nL[f''(x)] --------> s^2F(s) - sf(0) - f'(0)\nL[1] ------------> 1/s\nL[x] ------------> 1/s^2\n\nPlugging this into the equation, you get:\n\n[s^2F(s) - sf(0) - f'(0)] - [sF(s) - f(0)] = 1/s - 1/s^2\n\nPlugging in the initial conditions and simplifying the right side, you get:\n\ns(s-1)*F(s) = (s-1)/s^2 ---------------> Now solve for F(s)\n\nF(s) = 1/s^3\n\nNow use the general inverse Laplace Transform property:\n\nL-1[1/s^n] ---------> x^(n-1)/(n-1)!\n\nIn our case, n = 3, pluggin in you get:\n\nf(x) = x^2/2\n\n--------------\n\nHope this helps
974937
please complete the equation
741149
1.f(x)=x^2+4x-7\nx=[-4+/-rt(16+28)]/2\n=[-4+/-2rt(11)]/2\n=-2+/-rt11\n\n2.3x^2-11x-20=0\n=3x^2-15x+4x-20=0\n=(3x+4)(x-5)=0\nx=-4/3 or 5