I need help with Algebra again!! I so hate Algebra!!?
915004
all u have to do before attempting algebra is stop panicking.\nBe cool and try to substitute unknown with some variable say x.\nSo the radiator has a capacity of 16 QT.\nAnd you need 40% antifreeze solution.\nSo u need (40*16)/100 quart of pure antifreeze\nwhich is equal to 6.4 pure antifreeze.\nSo suppose u need to drain x quart of antifreeze solution already in the radiator. You replace the drained solution with pure antifreeze solution.\nso form a equation:\n25% of (16-x)+ x= 6.4 \n25*(16-x)/100 + x = 6.4\n(16-x)/4 + x= 6.4\n16-x+4x=4*6.4\n3x = 25.6- 16\nx=9.6/3\nx=3.2 quart\nso u drain 3.2 quart of 25% antifreeze solution originally in the radiator and replace it with equal amount of pure antifreeze.\nHope this helps
771776
http://mathforum.org/dr.math/\n\nThis is Ask Dr. Math, a website that can offer help for general math through Algebra and beyond. There are other good websites for math out online, type in keywords,  homework  math help and see all that you come up with. Eventually you will find one that can walk you through most problems. Good luck.
1098902
Try these websites: http://www.algebrahelp.com/lessons/\nhttp://www.sparknotes.com/math/#algebra1\n\nI hope I helped. Good Luck!
1039099
This is the answer i got when i worked out the problem\n1. You have to distribute the (x + 2) & (x + 1)....you do the by using the FOIL method....Firsts Outsides Insides Last and you get: x + 2 = x^2 + 3x + 2\n\n2. Get rid of the 2 on the left side by subtracting it on the right side and you end up with : x = x^2 + 3x\n\n3. Now you can move the x on the left to the right side too so you have: 0 = x^2 + 2x\n\n4. Pull out and x leaving you with 0= x (x+2)\n\n5. x = 0 and x = -2\n\nMy algebra is rusty but I think this is how it's done.
362703
Free Video lessons!   \nForum/Help Center  Ask for help and help others \nand much more\nhttp://www.algebra.com/
826369
I have found sites like the following helpful:\n\nhttp://www.algebra.com/\nhttp://www.saab.org/mathdrills/factor.cgi\nhttp://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=algebra&s2=factor&s3=basic\n\nThe last one is a solution program, but it can be helpful to check your answers.
396545
http://www.math.com/ or http://www.gomath.com/algebra.html are good places to start.
966804
I completely understand! I have high schoolers, and have to nearly re-learn to help them!\n\nYahoo education has help with algebra;\nhttp://education.yahoo.com/homework_help/math_help/chapter_list?id=minialg1gt \n\nSo does math.com:\nhttp://www.math.com/homeworkhelp/Algebra.html \n\nPhysics help:\nhttp://www.physics247.com/ \n\nGood luck to you!
754643
im taking Trig. try finding a homework helper web page. or go to the book that ur learning from web page because the majority of the times they have practice help on there.
953801
Get a tutor, get your parents to help you, read through the chapter in your Algebra book that you're currently doing, if your school has extra help with teachers go to the help sessions, study with your friends, form a study group, and do the extra practice problems for the current chapter in the back of your book for learning how to do the equations and stuff. I am also in your same grade doing Algebra 1 Honors. Hehe, I should take my own advice!
572480
breaking it apart, we have:\n\n(x+1)^3=(x+1)(x+1)(x+1)=\n(x^2 + 2x + 1)(x+1) =\n(x^3 + 3x^2 + 3x + 1) SO\n6(x+1)^3 = 6(x^3 + 3x^2 + 3x + 1) = 6x^3 + 18x^2 + 18x + 6\n\n(x+1)^2 = (x+1)(x+1)=\nx^2 + 2x + 1   SO\n-3(x+1)^2 = -3(x^2 + 2x + 1)=\n-3x^2 - 6x - 3\n\n4(x + 1) = 4x + 4\nand -9 stays the same.\n\nSo putting everything together, you have\n(6x^3 + 18x^2 + 18x + 6) + (-3x^2 - 6x - 3) + (4x + 4) - 9 \ncombine like terms and you get:\n6x^3 + (18x^2 -3x^2) + (18x - 6x + 4x) + (6 - 3 + 4 - 9)\n6x^3 + 15x^2 +  16x - 2\n\nWHEW!  That took me back to college pre-calc!
571127
Try going to this website. The will explain it step by step and you can ask all the math questions you want. www.algebra.com
485040
Algebra can get difficult but i get used to it and i grasped it pretty quickly.  But when i first learn some of the concepts i need to do the homework and look at it for a while.  Then, it just sticks to me.  I can understand how people can find it really confusing and difficult.
913123
I think you can try this website \nhttp://www.algebra-test.com \nhttp://www.algebraavenue.com\nhttp://www.algebra-online.com/algebra-equation.html \nhttp://www.algebrahelp.com\nhttp://www.algebra-net.com\nhttp://www.algebra-online.com/solve-algebra-equation.html\nhttp://www.algebrahelp.com/lessons/equationbasics/pgw.htm\nhttp://www.algebrasolver.com\nhttp://www.edhelper.com/algebra.htm\nhttp://www.math.com/students/worksheet/algebra_sp.htm
202280
http://www.aaamath.com/ Try this. Hope it works!!
70409
Go back to math class and figure it out!
919291
I will show one example on how to do it and then you take that example and apply it to the other two problems:\n\nOn the first problem, start with these two equations:\n2x-y+z=15\nx-y+2z=12\n\nTake the second equation and multiply it by -2:\n2x-y+z=15\n-2x+2y-4z=-24\ny-3z=-9\n\nLet's focus on the other part:\nx-y+2z=12\nx+2y-z=-3\n\nMultiply the second equation by -1:\nx-y+2z=12\n-x-2y+z=3\n-3y+3z=15\n\nLet's put those two equations together:\ny-3z=-9\n-3y+3z=15\n-2y=6\ny=-3\n\nPlug it into a y-value:\n-3-3z=-9\n-3z=-6\nz=2\n\nPut the y and z value into this equation:\nx-(-3)+2(2)=12\nx+3+4=12\nx+7=12\nx=5\n\n(5,-3,2)\n\nCheck:\n5-(-3)+2(2)=12\n5+3+4=12\n8+4=12\n12=12\n\nI hope that helps!\n\nP.S. I checked with the original equations and they equaled out also. You can check the answers with a graphing calculator. Whoever is giving the people for thumbs down for their answers, whether it is right or wrong, is being very unkind and rude for doing so. I am sure that I am not only person who does not appreciate the thumbs down given too.