algebra problems. PLEASE HELP!!?
955983
The answer is 42.  Prove or disprove by doing your homework.
1106533
1) 6(x-4)(x-2)\n= 6(x^2 - 2x - 4x +8)\n= 6(x^2 - 6x + 8)\n= 6x^2 - 36x + 48\n\n2) (x+7)(x+4)-(x+3)(x+2)\n= (x^2 + 4x + 7x + 28) - (x^2 + 2x + 3x + 6)\n= (x^2 + 11x + 28) - (x^2 + 5x + 6)\n= x^2 + 11x + 28 - x^2 - 5x - 6\n= 6x + 22
859744
(x + 4y)(x - 4y)\nI answered 2 of your questions. Do the rest of the questions yourself.
896553
4x^2+8x-x-2\n(4x-1)(x+2)
58450
use the distributive property (poor u, u seem to be having a lot of trouble with your math homework tonight!! hope this helps:)\n\nthe property states that a(b+c)=ab+ac, basically you take the "a" and multiply it by the other 2, or distribute it.  say a=2, b=3, and c=1.  your equation would be 2(3+1) and, using the distributive property, make it become 6+2 which equals 8.  so its the same when your using x.  3(-4x+7)= -12x+21!!! yay!! hope this helped!
978722
You can't, algebraically.  You have two equations and three variables.  Sorry.  You need one more equation, such as a = b.
131300
-8x+2=26\nsubtract 2 from both sides.\n-8x=24\ndivide both sides by -8\nx=-3
1102373
i wont answer.... log onto www.mathguru.com and discover solutions with explanations.
708644
Use the quadratic formula.\nx1 = 1\nx2 = -1/2
821450
go to a math website off yahoo. it should help.
326220
You actually need to solve this a different way.\nYou can't multiply both sides by x+3 because this could be positive or it could be negative...you don't know.\nInstead, bring everything to one side by subtracting 2, then get a common denominator.\n(x - 2)/(x + 3) <= 2\n(x - 2)/(x + 3) - 2 <= 0\n(x - 2)/(x + 3) -2(x+3)/(x+3) <= 0\n[(x - 2) - 2(x+3)]/(x + 3) <= 0\n[(x - 2) - 2x - 6)]/(x + 3) <= 0\n(-x - 8)/(x + 3) <= 0\nYour important values are -8 and -3\nTest values less than -8, between -8 and -3, and greater than -3.\nYou will see that you get negative values for\nx <= -8 and x >= -3
123892
3(x – 2)^2 – 3(x – 2) – 6\n3((x-2)^2 - (x-2) -2)\n3(x^2-4x+4-x+2-2)\n3(x^2-5x+4)\n\n3(x-4)(x-1)\n\nYay!!
747340
-2x-x > -6 + 4\n-3x > -2 \nx < 2/3
283601
1) First we move all the factors with x to one side, and the ones without x to the other.  Result will be:\n\n4x/3 - 2x + x = 3/2 + 1/5\n\n2)  Then, we factor out an x from the left side and get:\n\nx(4/3 - 2 + 1) = 3/2 + 1/5\n\n3)  Calculate the actual numbers on both sides\n\nx(4/3 - 6/3 + 3/3) = 15/10 + 2/10\nx(1/3) = 17/10\n\n4)  Divide both sides by 1/3 and get a final result of...\n\nx = (17/10)/(1/3)\nx = (10/17)*(3/1)\nx = 30/17\n\n5)  Comform that I PWN n00bs.  And... I'm pretty sure I pwn n00bs.  Therefor, we are done!  ^-^
888367
You need to give some directions.  What is the problem asking you to do?\nNote: The person below is incorrect.  (4+3x)^5 does NOT equal\n4^5 + (3x)^5
831674
1. replace x with -3, calculate and see if you get 0\n2. divide by (x+3) and solve the remaining quadratic equation
1181037
You will find the worksheets at the website of http://www.algebra.com. This website is a free site. You can also pose Maths. questions there and will get answers at the cost of $0 (zero dollar).\n\nI am a free tutor there and you may see my lessons and solutions to problems at\nhttp://www.algebra.com/tutors/aboutme.mpl?userid=psbhowmick\n\nYou will find the worksheets at\nhttp://www.edhelper.com/algebra.htm