Need help with college algebra. I must show my work on the problems.?
826321
If a is positive and b is negative, or vice versa, then /a-b/>/a/-/b/. If a and b are both positive or both negative, then /a-b/=/a/-/b/. This covers all possible cases of a and b, thus /a-b/>=/a/-/b/.
616179
here are the general steps you want to take when approaching these math problems\n\n1.) check for the greatest common factor\n2.) write out your parenthesis\n3.) since it's a squared problem, fill in the x's\n4.) write out your + & - signs\n5.) list the multiplication factors for the last term\n6.) plug in factors that will lead to the correct product\n\nwe'll start with #1 to get a better idea\n\n1.) -x^2+5x+24\n\n1. there is no greatest common factor\n2. write out your parenthesis\n\n(    )(    )\n3. fill in the x's\n\n( x   )(x   )\n\n4. fill in + or - signs\n\n-since we're dealing with -x^2 we know that one of the x's need to be + and the other one -\n\n(x   )(-x   )\n\n-since our last term is positive we know that either both the #'s are + or both of them are -\n\n(x-  )(-x-  )\n\nor\n\n(x+  )(-x+  )\n\n\n\n5.  list the multiplication factors for the last term\n\n24\n\n1*24=24\n2*12=24\n3*8=24\n4*6=24\n\n6. plug in factors that will lead to the correct product\n\n(x-8)(-x-3)=-x^2+5x+24\n\n**now, with a little more practice, you'll be able to figure these out faster...here are the answers to the rest of the problems\n\n2. (x-12)(x-12)\n3. (x-9)(x-12)\n4. 2(x^2-8x+16)=2(x-4)(x-4)\n5. 3(x^2-6x+72)=3(x-12)(x+6)
443949
Use quadratic formula and do your own homework.
794018
you should read your textbook and do practice problems to help.  ask your teacher or classmate for help.
352026
I would like to be able to give you the answer but I don't think I would be doing you a favour.  If you are to get practice in division, it is better for you to work it out on your own.  Does your teacher know that you are getting homework answers here on Yahoo?
680155
i'd love to help but i'm having trouble with it myself\n\nuse ur math textbook it will have it unless ur math book say algebra and is red ( that my alg. book and it doesn't help) \n\nSRY good luck
775488
Ok, Victoria, if you're going to ask for help you have to write the questions correctly.  Problem 1 cannot be solved using algebra the way you wrote it, I doubt you can use complex integration or any other advanced tool to solve that one.  Unfortunately, I just spent the last half hour manipulating it before I remembered that little fact.\n\nPut parentheses in.  Had you put the parentheses in the problem would have been trivial.  Kudos to the helpers who were smarter than I in figuring out the constants were in the exponent.  It's not that hard Victoria, especially when you're asking for help.\n\nWrite the question correctly.\n\n\nPS. Irma, you forgot to multiply the 2 from 4=2^2 through the whole parenthetical statement.  The answer is x=4.\n.
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.
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.
154380
I posted these sources earlier to someone else who was looking for math help.\n\nGood Luck!!
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
498659
1) y=x+4\n   3x+y=16 (I guess you made a mistake by writing +16)\n--------------\nsubstitute x+4 for y in second equation and you'll get\n\n   3x+x+4=16\n    4x+4=16\n    4x=12\n     x=3\n\nnow y=x+4=3+4=7\n\nsolution (x,y)=(3,7)\n\n2) 4x+5y=48\n    y=3x+2\n--------------\nsubstitute 3x+2 for y in first eq.\n\n4x+5(3x+2)=48\n\n4x+15x+10=48\n\n19x=38\n\nx=2\n\nnow y=3x+2=3*2+2=6+2=8\n\nsolution (x,y)=\n(2,8)\n\n3)3x-y=0\ny=6\nin first equation instead of y write 6\n\n3x-6=0\n3x=6\nx=2\n\nso, the solution is (2,6)\n\n4) x-2y=-10\n3x-y=0\n\nfrom the first equation you have\nx=2y-10\nnow substitute 2y-10 for x in second one\n\n3(2y-10)-y=0\n6y-30-y=0\n\n5y=30\ny=6 \nnow x=2y-10=2*6-10=12-10=2\nso x=2 and y=6
272295
There's lots of help online. Try these web sites:\n\nhttp://www.algebrahelp.com/\nhttp://www.purplemath.com/\nhttp://www.sosmath.com/algebra/algebra.html\nhttp://www.helpalgebra.com/\n\nThere's even more. So start with these, and good luck!
1070900
5x - 2y + 4 = 0 (i)\n3x + 2y + 12 = 0 (ii)\n(i) + (ii)\n\n8x + 16 = 0\n8x = -16\nwherefore x = -2\n\nsub x = -2 into eq (i)\n-10 - 2y + 4 = 0\n-6 = 2y\nwherefore y = -3\n\nx = -2 y = -3 wherefore your homework is done
932261
hey there is a good website to check your work \nits www.webmath.com i use it all the time
939068
Is this one problem?  If so, you can solve it like this:\n\n(a/b) / (c/d) = (a/b) * (d/c)\n\n~ ~ ~ ~ \n\n( (x² - 4) / x² ) / ( (x²+5x+6) / x ) = \n\n( (x² - 4) / x² ) * ( x / (x²+5x+6) ) = \n\n( (x² - 4) * x) ) / (x² * (x²+5x+6) ) = \n\n( (x² - 4)  / (x * (x²+5x+6) ) = \n\n{{ now notice you can factor your quadratic equations, to get the following }}\n\n(x+2)(x-2)  / (x * (x+2)(x+3) ) = \n\n(x-2) / (x(x+3))\n\nHope that helped!\n\n~ ♥ ~
959479
all you need to do is go to www.webmath.com, it'll help u there!\n\nI always do my math problems to that site if ever i get stuck on something! It's helpful!