algebra!!! how to solve?
106861
How about:\nx ^2/(x^2-4)+1/(x+2) = 0\nx^2/(x^2-4) = -1/(x+2)\nx^2/(x+2)(x-2) = -1/(x+2)\nx^2/(x-2) = -1\nx^2 = -x + 2\nx^2 + x – 2 = 0\n(x+2)(x-1) = 0\ncan you take it from here?
706334
To solve #1, you  have to use partial fractions.  \n\nYou have to convert \n\n** x / [(3x-1)(x+2)] = A/(3x-1) + B/(x+2) for some constants A,B.\n\nTaking the integral of x / [(3x-1)(x+2)] would be the same as taking the integral of A/(3x-1) + B/(x+2), so let's do that.\n\nIntegral (A/(3x-1) + B/(x+2))dx = \nIntegral (A/(3x-1))dx + Integral (B/(x+2))dx\n\nAlways pull out constants.\n\nA*Integral(1/(3x-1))dx + B*Integral(1/(x+2))dx\nA * 1/3 * ln |3x - 1| + B * ln|x+2| + C\n\nNow, solve for A and B using the equation marked as **.\n\nx / [(3x-1)(x+2)] = A/(3x-1) + B/(x+2)\n\nFirst, multiply both sides by (3x-1)(x+2) to get\n\nx = A(x+2) + B(3x -1)\n\nRemember that the above is a statement that is __true for all x___, so in order to solve for A and B, we can let x be ANY value.\n\nFor simplicity purposes, we're going to let x=-2 and then x=1/3.\n\nx = -2:  \n\n-2 = A(-2 + 2) + B(3(-2) - 1)\n-2 = B(-7)\nB = 2/7\n\nx = 1/3:\n\n1/3 = A(1/3 + 2) + B(3(1/3) -1)\n1/3 = A(7/3)\nA = 1/7\n\nTherefore, substitute A=1/7 and B=2/7 back into here:\n\nA * 1/3 * ln |3x - 1| + B * ln|x+2| + C\n\nto get\n\n1/7 * 1/3 * ln |3x - 1| + 2/7 * ln|x+2| + C\n7/3 * ln |3x - 1| + 2/7 * ln|x+2| + C\n\nTo solve #2:\n\nUse integration by parts.\n\nlet u = arcsin(3x)                  dv = dx\ndu = 1/sqrt(1-(3x)^2) * 3         v = x\n\nSo our new integral becomes\n\nx*arcsin(3x) - Integral [ (3x)/sqrt(1 - (3x)^2) ] dx\n\nAfter that it becomes substitution.
705864
Alright, don't listen to the above post...\n\nGrade 12 here, answering questions all day for mere fun.\n\n12)\ny - 3 = -1/3 (x + 4)\n\nEXPLAINED\n\nadd 3 to both sides\ny = -1/3 (x + 4) + 3\n\nexpand 1/3(x + 4) \n= x * 1/3 + 4 * -1/3\n= x/3 + 4/3\ny = -1/3x - 4/3 + 3\n\nTo add -4/3 and 3, use 3 = 9/3\ny = -1/3x - 4/3 + 9/3\ny = -1/3x + 5/3\n\nTrue, my bad. I originally added 4/3 instread of -4/3\n\n\n16)\n(-2, -1) (4, -4)\n\nm = Dy / Dx\nm = ((-1) - (-4)) / ((-2) - (4))\nm = 3 / -6\nm = -1/2\n\ny = mx + b\n(-1) = -1/2(-2) + b\n-1 = 1 + b\n-1 - 1 = b\n-2 = b\n\ny = -1/2x - 2\n\nCHECK\ny = -1/2(4) - 2\ny = -2 - 2\ny = -4\nTherefore it works.\n\n18)\n(0, 0) (2, 3)\n\nm = Dy / Dx\nm = ((3) - (0)) / ((2) - (0))\nm = 3/2\n\nb = 0\n\ny = mx + b\ny = (3/2)x + 0\ny = (3/2)x\n\nNote, the lower post copied my answers
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? :)
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!)
368125
if this is the problem:\n((x+3)/2)*((x^2-x)/(x+3))=21\nthen you solve by\ncancelling out (x+3)\n(1/2)*(x^2-x)=21\nmultiplying by 2\nx^2-x=42\nmaking it quadratic\nx^2-x-42=0\nand using x=(-b(+-)srt(b^2-4ac))/2a\nx=(1(+-)srt(1-4*1*(-42)))/2\nx=(1(+-)srt(169))/2\nx=(1+13)/2  or x=(1-13)/2\nx=7   or x= -6
842934
Ok, first you must get the -8 and the x/-2 out of parentheses, so u just multiply them both by one. So then is -8 * x / -2. First, you do -8 * x, which is -8x. The problem then becomes -8x / 2. In this step, just don't mind the negative or the x and just put them back in at the end. So you're really just doing 8 / 2, which is 4. You must put the - and the x back in, so the answer is -4x. Ta Da!
982146
Just add and substract terms in both sides.\n\nSince \ny-x^2+8x\nis equal to 7, \nyou can add x^2 and substract 8x to both sides.\n(y-x^2+8x) +x^2-8x = y\n7 +x^2-8x = x^2-8x+7\n\nso y=x^2-8x+7\n\nby the way, if you factorize in terms of x then you'd get\ny=(x-1)(x-7)
1169576
Try:\n\nhttp://www.sosmath.com/cyberexam/algebra/q001/1/1.html\n\nhttp://www.math.com/practice/Algebra.html\n\nhttp://www.glencoe.com/sec/math/studytools/prealglblr.shtml \n\nhttp://www.syvum.com/squizzes/algebra/
1053403
Write it as:\n2x^2-4x+3x-6\n=2x[x-2]+3[x-2]\n=[2x+3][x-2]\nSimple middle-term factorisation!!!
831674
1. replace x with -3, calculate and see if you get 0\n2. divide by (x+3) and solve the remaining quadratic equation
790277
ill give it a shot - im not sure , anyways i got 2 answers..\n\n(1st)\n\n(3-x)x+2 = 2 (x^2-2)\n\n3x-x^2+6-2x = 2x^2-4\n\nx-x^2+6 = 2x^2-4\n\nx-1/x+6 = 2/x-4\n\nx-1+6x = 2-4x\n\nx+4x+6x = 2+1\n\n11x = 3\n\nx = 3/11\n\n\n(2nd possible answer)\n\n\n(3-x)x+2 = 2 (x^2-2)\n\n3x-x^2+6-2x = 2x^2-4\n\nx-x^2+6 = 2x^2-4\n\nx-1/x+6 = 2/x-4\n\nx-1+6/x = 2-4/x\n\n-5 = -2/x\n\n-5x = -2\n\nx = -2+5\n\nx = 3
795656
The first thing you should remember is the phrase PEMDAS.  This is shorthand for parenthesis, exponents, multiplication/division, and addition/subtraction.  This is the order of operations in math.  You should solve all equations in this order.  So for the first one, you should first distribute (multiply) -7 with (3-3x).  After that combine like terms, i got 25x -30; you can also factor that to be 5(5x-6).  The same steps apply for the rest of the problems.  I hope it helps.
439883
Use the following four steps to solve a quadratic equation, in this order:\n\nStep 1: Guess the answer\n\nIf x^2=0, can you guess the answer?\nIf x^2=1, you can guess, but remember that there may be two answers. (1 and -1)\n\nYou can tell if you are guessing right by plugging back into the original equation. If it worked, you're done.\n\nIf you can't guess in 30 seconds, you are done with Step 1. Go to Step 2.\n\n\nStep 2: Try factoring\n\nx^2 + 4x + 4 = (x+2)(x+2)\n\nRemember that the constant term of your factors has to multiply to the constant term of the quadratic. \n\nFor example, if you have x squared plus something plus 6, remember that your constant terms must be multiples of 6. \n\nTry:\n\n(x + 1)(x + 6)\n(x - 1)(x - 6)\n(x + 2)(x + 3)\n(x - 2)(x - 3)\n\nYou just ran out of factors of 6. If this didn't work, go to step 3.\n\n\nStep 3 -- Complete the Square\n\nOne important thing to remember about completing the square is that it does exactly the same thing as the quadratic equation. Pick either Step 3 or Step 4 as your last step. Guessing and factoring don't always work because you can't always guess and you can't always factor. But completing the square and using the quadratic formula will always find the answer if there is one.\n\nSuppose you have \n\nx^2 - 6x - 16 = 0\n\nHere's how to complete the square:\n\n#1. Move the constant (whatever it is) to the right side\n\nx^2 - 6x = 16\n\n#2. Find half of the coeficient of the x term and add the square of that number to both sides of the equation:\n\nThe x term here is -6x\nThe coefficient is -6\nHalf of -6 is -3 (Remember this number -3)\n-3 sqared is 9. Add 9 to both sides of the equation:\n\nx^2 - 6x + 9 = 16 + 9\n\n#3. If you did #2 correctly, then the left side is a perfect square and the constant term is -3 (remember -3?)\n\n(x - 3)^2 = 25\n\n(If you're not sure, multiply it out and see if you get what you had before.)\n\n#4. Take the square root of both sides. Remember, it could be negative or positive.\n\nx - 3 = 5\nx - 3 = -5\n\n#5. Solve your two equations\n\nx = 8\nx = -2\n\n#6. Put your answers back into the original equation and see if you got the right answer. \n\n\nStep 4. \n\nIf you did Step 3, you already have the answer, and you're already done. If you didn't do Step 3, use the quadratic formula to get the answer. You can find the quadratic formula here:\n\nhttp://en.wikipedia.org/wiki/Quadratic_formula\n\nMemorize the quadratic formula and never forget it.  Write it on your forhead until you have it memorized.
822539
first multiple everythingby x, then simplify and solve:\n\n1/x-2 = 3/x+3 =>\n\nx/x - 2x = 3x/x +3x =>\n\n1 - 2x = 3 + 3x =>\n\n-2  = x
1160760
what you have is something like\na(b+c)\n\nthe solution is \nab+ac\n\n(x^3)(2x^-2)+(x^3)(4x)\nnow to multiply x^a * x^b you get x^(a+b)\n\n2x^((3+(-2))+4x^(3+1)\n\n2x+4x^4
156840
Look up "Algebra Solved 2005" or "algebrator", these two programs not only solve problems, but they give step by step solutions for understanding. they have tutorials and practice test. I went from knowing none to fairly advanced in a few weeks.