how do i do this Algebra 2 problem?
916972
When all the variables cancel out, then you have an infinite number of solutions, or no solutions.  Here, you have none.\n\nThis problem shows you that you need to check for linear independence.  That is, none of your equations should be linear combinations of any others.  What is a linear combination?  It's when you add equations which may or may not be multiplied by constants.  If you get something like this, then you can't solve your system, as you'll have fewer equations than unknowns.\n\nIn your case, 3[equation1]-[equation3] yields 0=-1.  This is false, so your system has no solution.\n\nIf there was a 3 instead of a 4, you would have an infinite number of solutions, where they satisfied the first and second equations.
777699
4a^2-4a+1\nUse the FOIL method (multiply First terms, then Outside terms, then Inside terms, then Last terms).\n\nSo you have:\n(2a-1)*(2a-1)\n(2a*2a)+(2a*-1)+(-1*2a)+(-1*-1)\n4a^2-2a-2a+1\n4a^2-4a+1
672754
Take out the five so you have:\nx^2 + 8 + 16\nthen you can change that to\n(x + 4)^2 b/c you have to find a number that is the product of the a and c and the sum of b\nwhere you have: ax^2 + bx + c
863644
take out 4x first\n4x(x2-6x+8)\n4x(x-4)(x-2)
780960
I also have difficulty typing this thypes of math problems.  First, simplify the ax^2+bx+c parts in the first half to (x+c)(x+c) looking things first.  So...\n\n[(x-3)(x-1)]} is the first part.  With the bottom of the second half, divide by 3, so that you get 3(x+1).  Now line everything up like so...\n\n(x+1)(x+3)(x-3)\n_____________\n3(x-3)(x-1)(x+1)\n\nNow do the cancellations and you get the right answer...\n(x+3)\n____\n3\n\nSorry for not working to convention, but it was easier to type it that way.
201432
x=2 and y =2,\n\n2^2=2^2 and 2*2 = 2 + 2 =4
22859
Subract the two terms on the left giving:\n-2t < 16\n\nAdd 2t to both sides giving\n0 < 16 + 2t\n\nSubtract 16 from both sides\n\n-16 < 2t\n\nDivide both sides by 2\n\n-8 < t
716170
since 2 is a root you take (x-2)*(x-2) and -3 is one also so you multiply (x+3) now you know it is a fifth degree so you need two more so you can take (4x^2+1) and multiply\n(x-2)(x-2)(x+3)(4X^2+1) you do the algebra
87883
First write y as a product instead of a quotient:\n\n(2+x)*(2-3x)^(-1)\n\nNow you should be able to do this. \nUse 2+x as your f and (2-3x)^(-1) as your g. \nYou will have to use the chain rule to find g'.
1063248
To solve correctly, distribute two to x and -2, add four to both sides, and divide both sides by x:\n\n2(x - 2) = x + 3\n2x - 4 = x + 3\n2x = x + 7\nx = 7\n\nThe error that was made was trying to simply remove the parentheses instead of actually distributing 2 to both of its contents.\n\n-----\n\nIf this is not what you meant and I have read it backwards, this is the other interpretation:\n\n-----\n\nTo solve correctly, add two to both sides, then divide both sides by x:\n\n2x - 2 = x + 3\n2x = x + 5\nx = 5\n\nThe error that was made was trying to change 2x [2 times x] into 2(x - 2) [2 times the difference of x and 2]. This is not allowed.
942253
the answer is -3x square root of 2
1108240
2(x-5)^2 = 3\ndivide by 2:  (x - 5)^2 = 3/2\nsquare root: (x - 5) = +/- root(6)/2\nadd 5: x = 5 +/- root(6)/2
142621
x² + (a-3 )x – 2a² + 6a – 4 = 0\nx2 +ax-3x  = 2a2 -6a+4\n\nx= (-a +- sqrt(a2 +12))/2\nare roots of x\n(-a+sqrt(a2 + 12))/2  =  2 * (-a-sqrt(a2 + 12))/2\nas sqrt of any no is +ve\ncancel den. 2\n-a + sqrt(a2 + 12) = 2 * (-a-sqrt(a2+12))\na = -3sqrt(a2+12)\na2 = 9(a2 +12)\n8a2 = -12 * 9\na = sqrt(-27/2) = 3i * sqrt(0.5)
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
896553
4x^2+8x-x-2\n(4x-1)(x+2)
784622
16x^2 - 4x - 2\n= (4x)^2 -(4x)- 2\n= {(4x)^2 - 1} - (4x+1)\n={(4x+1)(4x-1)} - (4x+1)\n=(4x+1)(4x-2)\n= 2*(4x+1)(2x-1)
78524
(1-a)+3(a-2)-(2a-3)=1