solve algreticly for the unknown 3a-2=6a+4?
398441
a= do your homework boy
967763
Equivalent to [H+]^2/([HCH3CO2] - [H+]).  You can solve for [H+] knowing Ka and [HCH3CO2].
966048
(3ab^2)(3ab)^2\n(3ab^2)(9a^2b^2)\n27a^3b^4
222653
x+6(x^2)= 12\nx+6x^2  = 12\n6x^2+x-12=0\n(2x+3)*(3x-4)=0\n2x+3=0 3x-4=0\n2x=-3  3x=4\nx=-3/2 x=4/3
483173
((6p - 18)/(9p)) / ((3p - 9)/(p^2 + 2p))\n((6(p - 3))/3(3p)) / ((3(p - 3))/(p(p + 2)))\n((2(p - 3))/(3p)) / ((3(p - 3))/(p(p + 2)))\n((2(p - 3))/(3p)) * ((p(p + 2))/(3(p - 3))\n(2p(p - 3)(p + 2))/((9p)(p - 3))\n(2p(p + 2))/(9p)\n(2(p + 2))/9\n(2p + 4)/9
1060834
(5x+2)/3 -(2x+4)/6 = -4/3\n\nMultiply everything by 6:\n\n10x + 4 - (2x+4) = -8\n10x + 4 - 2x - 4 = -8\n8x = -8\nx = -1
1098481
6(2x + 8) - 6 =0\n12x + 48 - 6 = 0\n12x + 42 = 0\n12x = -42\nx = -42/12\nx = -3.5
390719
so you have (1/2)a+a-(6/3)=4+(1/3)a\n\nadding like terms: (3/2)a-2=4+(1/3)a\n\nsubtract (1/3)a from (3/2)a and add two to both sides:\n(3/2-1/3)a=4+2\n(9/6-2/6)a=6\n(7/6)a=6\n\nnow solve and what do you get? :)
769967
removing the brackets\n18+3y=4y-16-y\nsimplifying\n18+3y=3y-16\nadding 16\n34+3y=3y\nadding -3y\n34=0\nwhich is absurd\nso the equation cannot be solved
227387
D = 36 - (4)*(-2)*(-3)= 36- 24 = 12\nTwo real solutions.
1026680
3(-4 + 8a) = 25a -15\n-12 + 24a = 25a - 15\n25a-24a = -12 + 15\na = 3
810099
Use distribution to simplify...\n\n6f + 27 = -7 + f\n\nNow isolate the variable and solve.
197473
This is not an equation unless it is equal to something.  It is an expression.\n\nFirst, you can simplify the middle: a-5a = -4a\n\nThen, multiply the first 2 together: 3a(-4a)=-12a^2\n\nNow multiply this by the last part: -12a^2(a-35)= -12a^3 + 420a^2.\n\nIf you were supposed to set this equal to 0 and solve for a, you would not need to multiply it out.  You would just set each part = 0 and solve\n\n3a = 0\n-4a = 0\na-35 = 0
695176
6x^2-3x-4x+2\n3x(2x-1)-2(2x-1)\n(2x-1)(3x-2)
627033
6y + 3 = -4y^2\n4y^2 + 6y + 3 = 0\n\na=4, b=6, c=3\n\ny = -b+-SQRT(b2-4ac) / 2a\ny = -6+-SQRT[36-(4)(4)(3)] / [(2)(4)]\ny = -6+-SQRT[-12] / 8\n\nBut SQRT[-12] is not possible, unless we are looking for imaginary values, then:\n\ny = -6/8 +- SQRT[-12]/8\ny = -3/4 +- [2SQRT3]i/ 8\ny = -3/4 + [(SQRT3)/4]i or -3/4 - [(SQRT3)/4]i
744241
3/(x + 6) - 1/(x - 2) = -8/(x^2 + 4x -12)\n\nStep 1: Create a common denominator for the left side. This is convenient, because (x + 6) * (x - 2)  = (x^2 + 4x - 12)\n3/(x + 6) * (x - 2) / (x - 2) - 1/(x - 2) * (x + 6) / (x + 6)\n3 (x - 2) / ((x + 6) * (x - 2)) - 1 (x + 6) / ((x + 6) * (x - 2))\n\nFinally:\n(3 (x - 2) - 1 (x + 6)) / (x^2 + 4x -12) = -8/(x^2 + 4x -12)\n\nStep 2: Cancel the x^2 + 4x -12 denominator on both sides, then solve.\n(3 (x - 2) - 1 (x + 6)) / (x^2 + 4x -12) = -8/(x^2 + 4x -12)\n3 (x - 2) - 1 (x + 6) = -8\n3x - 6 - x - 6 + 8 = 0\n2x - 4 = 0\n2x = 4\nx = 2 (solution!)
892382
To solve the first one, divide each part of the first equation by two, so 2x-3y=-13; since -2x+3y = 13, adding the two equations gives 0x+0y=0.  No help.  Try substitution.\nFrom the first equation, 4x = 6y-26; x = 1.5y - 6.5\nsubstitute into the second equation -2(1.5y-6.5) + 3y = 13; -3y +13 + 3y = 13; 13=13.  Still no help.\nBoth equations are equivalent; the line that represents them is the solution.\n\nProblem two:\nSubtract equation 2 from equation 1.\n3x-3x+2y-3 = 6 - y\n2y-3 = 6-y\n3y = 9\ny = 3\n3x+6=6\nx=0\nchecking\n3x+3=y=3\n3x=0\nx=0