How do you solve the question - Algebra - pls help!!!?
142007
1/x+1/x+3=1/60\nfirst find the lcd for the left side\nthat's x(x+3)\n\nand u know how to get the numerator with lcd right\nthat's will be x+3-x all over x^2+3x\nwill give u 3 over x^2+3x=1/60 cross multiply\nwill give u 180=x^2+3x\nx^2+3x-180=0\nfactor this one out\n(x+15)(x-12)=0\nequate this to zero\nx=-15\nx=12
962594
you sure you have this problem right ???????\n\nhere's what i came up with\n\nm + ( m + 3 ) = ( m - 2 ) - ( m - 1 ) \nthis is wrong \n\nm + ( m + 3 ) = 2m + 3\n\n( m - 2 ) - ( m - 1 ) \n\n= ( m - 2 ) ( m + 12 ) \n\n= m^2  + 1m - 2m  + 2 \n\n= m^2 - 1m + 2
908426
Plug in each of the four answers.  The only one that works is (2), because you get\n\n | 3 (-2/3) - 3 | = 5\n | -2 - 3 | = 5\n | -5 | = 5\n 5 = 5
835153
.more kids homework... factoring?\n1. rewrite the middle term as the sum of two terms and then factor completely 15x^2 -21x +6\n\n3(5x^2-7x+2)\n3(5x^2-5x-2x+2)\n3(5x-2)(x-1)\n\n2. rewrite the middle term as the sum of two terms and then facto by grouping. x^2 - 14x +45 \n\nx^2-9x-5x+45\n(x-9)(x-5)\n\n3. rerewrite the middle term as the sum of two terms and then facto by grouping x^2 + 19x +90\n\nx^2-10x+9x+90\n(x+10)(x+9)\n\n4. rewrite the middle term as the sum of two terms and then facto by grouping x^2 -2x -63\n\nx^2-9x+7x-63\n(x-9)(x+7)\n\n5.solve x^2 = -6x\n\nx=-6
919864
1.  ax + xb\n2.  21x^3 + 24x*2y\n3.  24x^2 + 31x + 10\n4.  24x^2 - 8x - 6\n5.  40x^3 + 66x^2 + 38x + 6\n6.  x^2 - 1\n7.  x^3 - 8
842934
Ok, first you must get the -8 and the x/-2 out of parentheses, so u just multiply them both by one. So then is -8 * x / -2. First, you do -8 * x, which is -8x. The problem then becomes -8x / 2. In this step, just don't mind the negative or the x and just put them back in at the end. So you're really just doing 8 / 2, which is 4. You must put the - and the x back in, so the answer is -4x. Ta Da!
303479
ok. \n\nThe trick is this. SQUARE THE EQUATION.\nNote that : [{2 + (3^1/2)}^1/2]^x =  [{2 + (3^1/2)}^(x/2)] \n\nWe get\n\n[ {2 + (3^1/2)}^1/2]^x + [{2 - (3^1/2)}^1/2]^x = 4\n\n{2 + (3^1/2)}^x \n+ 2 [{2 + (3^1/2)}^(x/2)]*[{2 - (3^1/2)}^(x/2)]  \n+ {2 - (3^1/2)}^x = 16\n\nand [{2 + (3^1/2)}^(x/2)]*[{2 - (3^1/2)}^(x/2)]  equals to\n [ 4 - 3 ] ^(x/2) = 1.\n\nThus, we get,\n{2 + (3^1/2)}^x + 2 + {2 - (3^1/2)}^x = 16\n{2 + (3^1/2)}^x + {2 - (3^1/2)}^x = 14  ...(1)\n\nNow note that \n[{2 + (3^1/2)}^(x)]*[{2 - (3^1/2)}^(x)]  = [4 - 3]^x = 1\nmeans that,\n{2 - (3^1/2)}^x = 1 / [{2 + (3^1/2)}^(x)]\nSubstituiting this into (1), we get,\n{2 + (3^1/2)}^x + 1 / [{2 + (3^1/2)}^(x)]  = 14  \n\nLet y = [{2 + (3^1/2)}^(x)] \nThe equation above becomes,\n\n y + 1 / y = 14\n y^2 - 14y + 1 = 0 \n\nA quadratic equation in y.\nWhich gives y = [14 ± sqrt(192)]/2 \n              \nSince y = [{2 + (3^1/2)}^(x)], Thus,\n\n [{2 + (3^1/2)}^(x)] = [14 ± sqrt(192)]/2 \n                     = [14 ± 8sqrt(3)]/2\n                     =  7 ± 4*3(1/2)\nSo,\n    x = ln [ 7 ± 4*3^(1/2)] / ln [2 + (3^1/2)]\n      =   ± 2\n\nAlternately, \nOne can use the identity,\n{2 + (3^1/2)}^(2) = 7 + 4*3(1/2)\n\n{2 + (3^1/2)}^(-2) = 7 - 4*3(1/2)\n\nTo get x = ± 2 again.\n\nDone.\n\n \nThe below is ADDITIONAL information solving a slightly different problem though similar to the one you posted.\n\nSay we have a similar problem but x is a positive real.\nWe want to find solutions for \n [ 2 + {(3^1/2)}^1/2]^x + [2 - {(3^1/2)}^1/2]^x = 4\n\nBy observation, x = 1 is a solution since, \n\n[ {2 + (3^1/2)}^1/2] + [{2 - (3^1/2)}^1/2] = 4\n\nHow to show they arent any other solutions?\n\n( x > 0 )\n\nLet f(x) =[ {2 + (3^1/2)}^1/2]^x + [{2 - (3^1/2)}^1/2]^x - 4\nSo, our equation amounts for solutions for f(x) = 0.\nWe show that f(x) is increasing for x>0 and we are done.\nNote that when we show f(x) increasing, it means that \nf(x)=0 only at one value because for any x > y, increasing means f(x)> f(y).\nSince f(1) = 0, Thus,\nf(y) > f(1) = 0 for y > 1  and\nf(y) < f(1) = 0 for  0 < y < 1   \n\nAll that remains to be shown is that f(x) is increasing.\nSimplest way, is to use basic calculus. \nIts necessary and sufficent to show f'(x)>0 for x >0.\nUsing the fact for constants any constant m, if y = m^x,\nThen dy/dx = (m^x)(ln m) \n\nf'(x)=  [{2 + (3^1/2)}^1/2]^x *ln[{2 + (3^1/2)}^1/2]\n      + [{2 - (3^1/2)}^1/2]^x *ln[{2 - (3^1/2)}^1/2] \n\nNow, note that\n[{2 + (3^1/2)}^1/2] = 3.316074013\n[{2 - (3^1/2)}^1/2] = 0.683925987\nln[{2 + (3^1/2)}^1/2] = 1.198781557\nln[{2 - (3^1/2)}^1/2] = - 0.379905573\n\nSo,\nf'(x) = (3.316074013^x)*(1.198781557) \n        - (0.683925987^x)*(0.379905573)\n\nSo to show f'(x) >0, its equivalent to showing,\n\n(3.316074013^x)*(1.198781557) >(0.683925987^x)*(0.379905573)\n              (4.848586069)^x > (0.316909757)\n           x ln (4.848586069) > ln (0.316909757)\n                 1.57868713 x > -1.149138224\n                            x > - 0.727907513 \n\nwhich is true for all x > 0.\nThus, for x>0, the only solution to our modified\nproblem is x=1.\n\nAnd there is no "chance" in mathematics.\nIt must be reasoned out.\n\nRegards
1089643
http://www.algebra.com
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
982041
Well dear\n\n-If the first root is " α " and the second roots is " α + 3"  so have  \n▪x ² - (α + β) + α β = 0 \n▪( x - α )(x - β ) = 0 \n-b/a = α+ β  \n&\nc/a = α β\nso we have ; \nc/a = m/4  \n-b/a = - ( -16) / 4 = 4 \nα + α + 3 = 4 , 2α + 3 = 4 \nα = 1/2  ,  β = 3 1/2 \n2 /a = α β  =>  m/4 =  1/2 * 7/2  , m = 7\n\nGood Luck dear
817481
We can use faux-algebra to solve these. You probably haven't taken algebra yet. With algebra these would be extremely easy to solve. It's why algebra was invented.\n\na. 16 x ___ has to be -42, because if we add 51 to it we get 9. Thus, ____ has to be -42/16, or -21/8.\n\nb. 2 x ____ has to be 22 - 9 x _____, because if we subtract 8 from it we get 14 - 9 x ____. And 22 has to be 11 x ____, because if we subtract 9 x ____ from it we get 2 x ____. Thus, ____ has to be 22/11, or 2.\n\nc. 9 x ___ has to be 3 x ____ + 3, because if we add 2 to it we get 3 x ___ + 5.  And 3 has to be 6 x ____, becuase if we add 3 x ____ to it we get 9 x ____. Thus, ____ has to be 3/6, or 1/2.\n\nWith algebra, no such thinking would be required. You just plug in the formulas and grind away. You'll see when you study algebra.
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~ ♥ ~
559085
Let me check... I need to see the full equation by hovering... give me a sec.\n\nOkay you have:\n= (5a-12)/(a^2-8a+15) - (3a-2)/(a^2-8a+15)\n\nFortunately the denominators are the same, so you just have to subtract the numerators and keep the same denominator:\n= [ (5a-12) - (3a - 2) ] / (a^2-8a+15)\n\nTurn -(3a-2) into -3a + 2:\n= ( 5a - 12 - 3a + 2 ) / (a^2-8a+15)\n\nGroup the like terms:\n= ( 5a - 3a - 12 + 2 ) / (a^2-8a+15)\n\nSimplify:\n= ( 2a - 10 ) / (a^2 -8a + 15)\n\nFactor out a 2 in the front\n= 2 ( a - 5 ) / (a^2 -8a + 15)\n\nFactor the denominator:\n= 2 ( a - 5 ) / [ (a - 3)(a - 5) ]\n\nCancel the ( a -5 ) from numerator and denominator:\n= 2 / (a - 3)\n\nThis is different from your answer.  I'm not sure where you went off, but hopefully this explanation will help.  If you need more details, add a comment to your question and I'll check back.\n\nThe answer is: 2 / (a - 3)
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
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
854860
(3x - a)(x - 2)(x - 7) = \n(3x - a)(x^2 - 7x - 2x + 14) = \n(3x - a)(x^2 - 9x + 14) = \n3x^3 - 27x^2 + 42x - ax^2 + 18ax - 14a\n3x^3 + (-27 - a)x^2 + (42 + 18a)x - 14a\n\n3x^3 - 32x^2 + 81\n\nso\n\n(-27 - a)x^2 + (42 + 18a)x - 14a = -32x^2 + 81\n(-27 + 32 - a)x^2 + (42 + 18a)x + (-14a - 81) = 0\n(5 - a)x^2 + (42 + 18a)x - (14a + 81) = 0\n(-a + 5)x^2 + (18a + 42)x - (14a + 81) = 0\n-(a - 5)x^2 + (18a + 42)x - (14a + 81) = 0\n\nx = (-b ± sqrt(b^2 - 4ac))/(2a)\n\nx = (-(18a + 42) ± sqrt((18a + 42)^2 - 4(-(a - 5)(-(14a + 81)))/(2(-(a - 5))\n\nx = (-(18a + 42) ± sqrt(((18a + 42)(18a + 42)) - 4(a - 5)(14a + 81)))/(-2(a - 5))\n\nx = (-(18a + 42) ± sqrt((324a^2 + 756a + 756a + 1764) - 4(14a^2 + 81a - 70a - 405))) / (-2(a - 5))\n\nx = (-(18a + 42) ± sqrt(324a^2 + 1512a + 1764 - 4(14a^2 + 9a - 405))) / (-2(a - 5))\n\nx = (-(18a + 42) ± sqrt(324a^2 + 1512a + 1764 - 56a^2 - 36a + 1620))) / (-2(a - 5))\n\nx = (-(18a + 42) ± sqrt(268a^2 + 1476a + 3384)) / (-2(a - 5))\n\nx = (-6(3a + 14) ± sqrt(4(67a^2 + 369a + 846)))/(-2(a - 5))\n\nx = (-6(3a + 14) ± 2sqrt(67a^2 + 369a + 846)) / (-2(a - 5))\n\nx = (3(3a + 14) ± sqrt(67a^2 + 369a + 846))/(a - 5)\n\nUnless you mistyped something, 3x^3 - 32x^2 + 81 does not factor into (3x - a)(x - 2)(x - 7)\n\n--------------------------------------------------------------------------------\n\nx^3 - x^2 + 16x - 16\n(x^3 - x^2) + (16x - 16)\nx^2(x - 1) + 16(x - 1)\n(x^2 + 16)(x - 1)\n\nif you were to take this further, then\n\n(x - 4i)(x + 4i)(x - 1)
91753
factorize\nx^2 + x = x(x+1)\n\nfactorize\nx^2 - 1 = .... (you should know)\n\n2 = 2 (!!!)\nTake the lcm of the three denominators (each different factor (=parenthesys) once and with the highest power)\n\nIt is 2x(x+1)(...)   (fill the dots)\n\n.....\n\nnow if you can't go on, you need to open your book again and study, or pay more attention at school at your math classes, and ask immediately for further explanation to your teacher when you feel you've not understood.\n\nGood work and good study