my first question was to help on 4x+1/2y=3\nbut nowi am on # 27 in my text book and i am really tired and i really need help.My teacher didnt explain it to us so a lot of ppl i know are asking for help on these questions so please help me.Thank u to everyone and i hope u guys can help me!\nlots of love \nKat
119856
cmon dont do this again, give the other equation
376421
The first simultaneous equation is confusing. If you rearrange them to give this:\n1.5x - 4 = 4y\nx - 2 = 2y\n(I just took the 2 across to the x side on the bottom equation then swapped the sides around so that it's easier to understand when compared with the top equation)\nthen you will see that if you were to multiply the bottom equation by 2 throughout and to keep the top equation the same, you would have:\n1.5x - 4 = 4y\n2x - 4 = 4y\nBoth of these have 4y on the right so it can be said:\n1.5x - 4 = 2x - 4\nTake 1.5x from 2x and -4 from -4:\n0 = 0.5x\nDivide by 0.5 throughout\n0 = x\n\nSo now that x = 0, you can put this into either equation to work out a value for y - here's the bottom one again:\n2y + 2 = x\nWe know x = 0 so:\n2y + 2 = \nTake 2 across\n2y = -2\nDivide by 2\ny = -1\nAnd so your answers are x = 0 and y = -1\n\n\nNow onto the next one, I presume you mean that the top one is "x + 8y = 24"\nx + 8y = 24\nx + y = 3\nWhat you can do here is take the bottom equation away from the bottom one (i.e take the x away from the x, the y away from the 8y and the 3 away from the 24) which gives you:\n7y = 21\nAnd so divide by 7 throughout\ny = 3\nNow you have y, you can look at the second equation again:\nx + y = 3\nWe know that y = 3 so:\nx + 3 = 3\nNow if we take 3 from both sides we get:\nx = 0\nAnd so y = 3 and x = 0\n\nI hope this is clear to you
472382
Very understandable....I just posted a question about my pre-calculus homework and i didnt ask for the answer, i just need help finding how to get it....i'd like to know how to do it myself, i dont just want someone to tell me the answer....does me no good @ all!
513486
hey well look i was not so good in math myself but i started to get some help and started 2 ask more??? and now am passing my Adv Alg with Trig with a  B .... Hope to hear from u soon good luck
441522
Do you mean (2x^2 + 3x + 1/ 2x^2 + x - 3) for your first fraction?  I'm going to assume you do.\n\nFirst, factor each numerator and denominator:\n\n2x^2 + 3x + 1 = (2x+1)(x+1)\n\n2x^2 + x - 3 = (2x-3)(x+1)\n\n4x - 4 = 2(2x-2)\n\nNow let's look out our multiplied fraction, rewritten using our factors we just found:\n\n[ (2x+1)(x+1)2(2x-2) ] / [ (2x-3)(x+1)(2x+1) ]\n\nNotice we can cancel out the (2x+1)'s and the (x+1)'s, to get:\n\n2(2x-2)/2x-3\n\nANSWER:   2(2x-2)/2x-3
413203
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.
349091
Okay, here we go:\n\n1) -2x-10-7 = 3x+2x-10\n    -2x-17 = 5x - 10\n    -2x -5x = -10+17\n    -7x = 7\n       x = -1\n\n2) 12x+6-6x = 5x-10-5x\n     6x+6 = -10\n     6x = -10 - 6\n     6x = -16\n     x = -8/3\n\n3) 30x-5-10x+5 = 4x+40-4x\n    20x = 40\n    x =2\n\n4) -24+6x+12 = 3x-x+4\n    6x-12 = 2x+4\n    6x-2x = 4+12\n    4x=16\n     x = 4\n\nSee? It's quite easy.
355067
For the first one, using the first equation, 3X = 8 + 2Y, or X = (8+2Y)/3.\n\nSubstituting this into the second equation gives us\n-6[(8-2Y)/3] + 4Y = -16, or -16 -4Y + 4Y = -16, which is true, but does not lead us to a solution.  The same is true if you substitute for X.\n\nLet's try adding them together and seeing if that helps.\nWe get -3X + 2Y = -8.  Using the same substitution we get \n-6[(8-2Y)/3] + 2Y = -8, or -16 -4Y + 2Y = -8, or -2Y = 8, or Y = -4 and X is 0.\nGoing the other way you get X = 8/3 and Y = 0.\n\nUse the same logic on #2, which is easier since you don't have zero clouding the picture.
435671
OK,as the first one is already done,I am doing the rest for you\n4(2x-6)=x+2 =>8x-24=x+2  =>8x-x=2+24  =>7x=26  =>x=26/3=8 2/3  (3) 2(x+5)+1=3(x-1)+4   =>2x+10+1=3x-3+4 =>2x-3x=-3+4-10-1  => -x=-4  =>x=1  (5)x+6=4(4x-1)   =>x+6=16x-4  =>x-16x=-4-6  => -15x=-10  =>x=-10/-15=2/3 [after reducing]
369872
y = 2 - 3x\ny = 2x + 4\ny = 4x/3 + 4\ny = 3x - 4\ny = 7x/2 + 5\n\nNot triying to bother the other guy, but i think you did #4 the wrong way. Check the "-" before the "y" in the original equation.
520944
About Me\nMember since: October 03, 2006\nTotal points: 100 (Level 1)\nPoints earned this week:\nTotal answers:\nBest answers:\n\nI need math help again... PLEASE HELP! I m desperately in need.?\n\nAnswer PARTB and PART C in the space provided. Evaluation of your solution to each problem will be based on the following.\n\n*A correct mathematical method for solving the problem is shown\n\n*The final answer is accurate and a check of the answer is shown where asked for by the question.\n\nDetermine if the equation is a linear equation, and if it is, write in standard form.\n\na. 5x+1=3y\n\nb. y=2x   x=y/2\n\nc. 4(2y-x)=1  8y-4x=1\n\nd. 25x-2y+2=0\n\ne. 2x-5y/2=7x\n\n2. a. f(x)= x (exponent 2)+2x+1 f(0),f(2)\n\nb. f(x)=2x-5 f(3),f(-3)\n\nc. f(x)=(-3x)exponent 2 f(0),f(-1)\n\nd. f(x)=2x(exponent 3)+4 f(0),f(-1)\n\ne. f={(1,5),(3,7),(5,9),(7,11)} f(3),F(7)
615323
Good job!  You're halfway there!\n\nAs you are writing it out on your homework, underline the (4x+1) in each of the two terms.  That helps you see what you are factoring.\n\nFactoring out the (4x+1) from each term, you should get your final answer.\n\n=(5x + 4y)(4x+1)
184372
goyalnarinder2 is correct.\n\nIf you also want to understand it, not only have the answer, then you should first know how to get rid of those brackets, for instance:\n\n(3x-1)(2x+3)= 3x(2x+3)-1(2x+3) \n\nGot the picture? It's called distributivity, the fact that you can distribute the factor 2x+3 to each of the terms of the first bracket - these would be 3x and -1.\n\nThen u continue distributing:\n3x(2x+3)=3x.2x+3x.3=6x^2+9x \n-1(2x+3)=-2x-3\n\nThat is how u get to:  \n6x^2+9x-2x-3=6x^2+7x-3 and this is just the result of the product of the first 2 brackets.\n\nAre u still w me? :-)\n\nYou now try to do the second product - which is (x-4)(x-1) - by yourself and see if you get this:\n(x-4)(x+1)=x^2-3x-4\n\nThen u add the 2 things (or rather substract...):\n(3x-1)(2x+3)-(x-4)(x+1)=(6x^2+7x-3)-(x^2-3x-4)\nand here u take care of the signs in the second bracket, which u should reverse (because say if I take -3 apples from u what I do is really give u +3 apples, got it?)\nSo u continue w:\nthe above = 6x^2+7x-3-x^2+3x+4=5x^2+10x+1\n\nHope this helped. Good luck!
482020
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
517117
(1)*4 +(2)\n-4x+16y=72\n4x+2y=0\n18y=72\ny=72/18=4\nsubstituting\n4x+2(4)=0\n4x=-8\nx=-2\nso x=-2 y=4
408944
Have you typed this correctly?\n\nEq 1: 2x² + 4xy - 3\nEq 2: x² - 2xy - 4\n\nSubtraction:\n2x² + 4xy - 3 - (x² - 2xy - 4)\n= 2x² + 4xy - 3 - x² + 2xy + 4\n= (2x² - x²) + (4xy + 2xy) + (-3 + 4)\n= x² + 6xy + 1\n\nI don't see that in your list.\nIf your first equation is 2x² + 4xy *plus* 3, then your answer is:\nD) x² + 6xy + 7
350751
I can't believe I'm actually doing your homework for you.\n\nPretty funny how bored I am.\n\n2x{3x-2(3+x)} - x2\n= 2x(3x - 6 - 2x) - x2 =2x2 - 12x - x2 = x2 - 12x = x(x - 12)\n\n3/2(4x-6)-x(3-x)=1/3x(3x+6)\n6x - 9 - 3x + x2 = x2 + 2x\n(3x-9+x2) - (x2 + 2x) = (x2 + 2x) - (x2 + 2x)\nx - 9 = 0, x = 9\n\n4 (y-7)-2y(1-3y)=6y2\n4y - 28 - 2y + 6y2 = 6y2\n2y - 28 = 0\ny = 28/2 = 14\n\nr2-{4-r(3-r)}=2(2r-3)\nr2 - 4 +3r - r2 = 4r - 6\n0 = r - 2\nr = 2\n\n1=2x(x-2)-{5-2x(1-x)}\n1 = 2x2 - 4x - 5 + 2x - 2x2\n2x = - 6\nx = -3