Need help on these very confusing. We have to simpfly rational expressions.?
117018
the first one all you do is simplify. since both the top and bottom have 5x, both would cancel eachother out. \n5x+2y/5x+14y = 2y/14y\nto simplify that further, both have factors of 2 so divide both numbers by two and you end up with\n2y/14y = y/7y\nif you need to simplify further you could divide y out of the equation and end up with 1/7\n\nthe second question you have to find the factors of each complex equation. x^2-5x-14 has factors of (x-7)(x+2)\n2x^2-x-10 has factors of (2x-5)(x+2)\nsince both share a factor of (x+2) they cancel eachother out. so you are left with x-7/2x-5.
722836
I think you are asking to simplify the rational expression:\n[(x^3+27)/(9x+27)]/\n[(3x^2-9x+27)/(4x)]\n\nIf this is what you are asking, here's what you do.\n\nWe are dividing by a fraction, so we must flip the second fraction and multiply.  Our problem now becomes:\n[(x^3+27)/(9x+27)]*\n[(4x)/(3x^2-9x+27)]\n\nNow we must factor each expression individually.\n\nThe first expression--x^3+27--is a sum of two cubes and factors as:\n(x+3)(x^2-3x+9)\n\nThe second expression--9x+27--has 9 as a common factor and factors as:\n9(x+3)\n\nThe third expression--4x--cannot be factored, so we leave it alone.\n\nThe fourth expression--3x^2-9x+27--has 3 as a common factor and factors as:\n3(x^2-3x+9)\n\nUsing substitution, our expression:\n[(x^3+27)/(9x+27)]*\n[(4x)/(3x^2-9x+27)]\nbecomes:\n{[(x+3)(x^2-3x+9)]/[9(x+3)]}*\n{(4x)/[3(x^2-3x+9)]}\n\nMultiplying these expressions into one large expression, we get:\n[(x+3)*(x^2-3x+9)*(4x)]/\n[27*(x+3)*(x^2-3x+9)]\n\nCancelling the (x+3)'s and the (x^2-3x+9)'s that are common to the numerator and denominator leaves us with our final answer:\n\n(4x)/27\n\nGood luck.
1150392
No problem, i just finished a calc3 course in college.\n\nYour problem is the same as 125 * (6x^2 + 5x + 7)^(-1/3)\nYou have to use the chain rule so you get\n125 * (-1/3) * (12x + 5) * (6x^2 +5x + 7)^(-4/3)\n\nthen you can simplify how you like from there
144290
This is a better link that the previous one posted here!! try it! \n\nhttp://pubs.logicalexpressions.com/Pub0009/LPMArticle.asp?ID=305
741149
1.f(x)=x^2+4x-7\nx=[-4+/-rt(16+28)]/2\n=[-4+/-2rt(11)]/2\n=-2+/-rt11\n\n2.3x^2-11x-20=0\n=3x^2-15x+4x-20=0\n=(3x+4)(x-5)=0\nx=-4/3 or 5
1007589
1. b^2-4ac>0 : b>2sqrt(ac)\n2. b^2-4ac=0 : b=2sqrt(ac)\n3. b^2-4ac<0 : b<2sqrt(ac)\n\nThe rest should be a piece of cake.
900659
1. f(x) = 200x - 15/40 - 2x\n\nI assume you mean\nf(x) = (200x - 15)/(40 - 2x)\n\n= 5(40x - 3)/[2(20 - x)]\n= 5/2 * (40x - 3)/(20 - x)\n= 5/2 * (40x - 800 + 797)/(20 - x)\n= 5/2 * -40(20 - x)/(20 - x) + \n= 100 + 5/2 * 797/(20 - x)\n\nNow we can deal with it, as messy as it looks as before it was 'improper' since the degree of the numerator was the same as the degree of the denomonator and a rational function is 'improper' if degree numerator ≥ degree denominator\n\nNow lets look at 5/2 * 797/(20 - x)\n\nA. When x < 20, 20 - x > 0 so 5/2 * 797/(20 - x) > 0. \nTherefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x)) > 100\nAND when x > 20, 20 - x < 0 so 5/2 * 797/(20 - x) < 0. \nTherefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x)) < 100\n\nie for x > 20, f(x) < 100 and for x < 20, f(x) > 100\n\nB. When x = 20 20 - x = 0 so 5/2 * 797/(20 - x) is undefined.\nTherefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x)) is undefined.\nAs x → 20 from the left ,5/2 * 797/(20 - x) → +∞\nTherefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x))→ +∞\nAND as x → 20 from the right, 5/2 * 797/(20 - x) → -∞\nTherefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x))→ -∞\n\nie vertical asympotes at x = 20\n\nC As x → ±∞, 5/2 * 797/(20 - x) → 0\nTherefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x)) → 100\n\nie horitontal asymptotes at f(x) = 100\n\nD. You now have enough information to sketch this\n\n2. Work the function g(x)=x^2/x-2\n\nI assume you mean g(x)=x^2/(x - 2)\n\nAgain this is improper as deg numerator = 2 and deg denominator = 1 (< deg numerator)\n\nSo g(x) = x^2/(x - 2)\n= (x^2 - 4 + 4)/(x - 2)\n= ((x - 2)(x + 2) + 4)/(x - 2)\n= x + 2 + 4/(x - 2)\n\nDo steps A. to D as above\n\nA. Check out the signs of the denominator of the fraction\nWhen x < 2,  x - 2 < 0 so 4/(x - 2) < 0 \nTherefore g)x) = x + 2 + 4/(x - 2) {= x^2/(x - 2)} < x + 2\n\nWhen x > 2,  x - 2 > 0 so 4/(x - 2) > 0 \nTherefore g(x) = x + 2 + 4/(x - 2) {= x^2/(x - 2)} > x + 2\n\nie for x < 2, g(x) < x + 2 and for x > 2, f(x) > x + 2\n\nB. When x = 2,  x - 2 = 0 so 4/(x - 2) is undefined.\nTherefore g(x) = x^2/(x - 2) (= x + 2 + 4/(x - 2)) is undefined.\nAs x → 2 from the left, 4/(x - 2) → -∞\nTherefore g(x) = x^2/(x - 2) (= x + 2 + 4/(x - 2)) → -∞\nAND as x → 2 from the right, 4/(x - 2) → +∞\nTherefore g(x) = x^2/(x - 2) (= x + 2 + 4/(x - 2)) → +∞\n\nie vertical asymptotes at x = 2\n\nC. As x → ±∞, 4/(x - 2) → 0\nTherefore g(x) = x + 2 + 4/(x - 2) {= x^2/(x - 2)}  → x + 2\n\nTherefore g(x) is assymptotic to g(x) = x + 2\n\nD. You now have enough information to sketch this
890645
Wow, there is a lot here!  I'll start with some tips for solving algebra problems.  The first thing you should always remember is to play the equal game.  Whenever you add, subtract, multiply, or divide, you should do the same to the other side.  The other tip is to use the order of operations when simplifying.  Remember the sentence: Please excuse my dear Aunt Sally.  The initial letters of each word stands for the order of operations.  Do the Parenthesis first, followed by the Exponents, then Multiplication or Division, and Addition and/or subtraction next.  Finally, you should always simplify first using the order of operations before you start shifting numbers from one side of the = to the other.  \n\nI'm not going to work all of these problems, but I'll show you a few to get you started.\n\n1)\n3x-7 = 2   Add 7 to both sides\n3x = 9   Divide by 3\nx = 3\n\n4)\n6x - 3 = 2x - (4-3)/3   Simplify the stuff inside the parenthesis first\n6x - 3 = 2x - 1/3  Add 3 to both sides\n6x = 2x - 5/3  Subtract 2x from both sides\n4x = -5/3  Multiply both sides by 3\n12x = -5  Divide both sides by 12\nx = -12/5\n\nI hope this helps.  Good luck!
677039
(1) { -2x-y-z=0\n(2) { -x-2y-z=0\n(3) { -x-y-2z=0\n\nLet's look at (2):\n-x-2y-z=0 => x=-2y-z\nNow at (3):\n-x-y-2z=0 => x=-y-2z\nNow we can see that\n-2y-z=-y-2z => *** y=z ***\n\nNow let's put these values into (1):\n-2x-y-z=0 => -2x-y-y=0 => -2x-2y=0 => 2x=-2y => *** x=-y ***\n\nFor now, we know that x=-y=-z\nLet's put all of what we know into (2), and express everything using 'x':\n-x-2y-z=0 => -x-(-2x)-(-x)=0 => -x+2x+x=0 => 2x=0 => *** x=0 ***\n\nThus, x=0 => y=-0=0 => z=0\nSo the only solution is (0,0,0).\n\nGood luck!
859187
I'll do one for u.\nIn the first,\nMultiplying both sides by x*,we get\nx+3x+2x*=7x*\n4x+2x*=7x*\nMoving 4x and 2x* to the R.H.S.\n5x*-4x+0=0\nThus 5x*-4x=0\nAttempt the others in the same manner.
889816
1/rt3 *1/rt2-1/rt3\n[1/rt3](1/rt2-1)\n[1/rt3](1-rt2/rt2)\nrationalising\n[(1-rt2)*rt6]/6
970218
Do you require mathematical signs, or are standard operations OK (like square root)?\n\n1: (1 + 1 + 1)!\n2: 2 + 2 + 2\n3: 3 * 3 - 3\n4: sqrt(4) + sqrt(4) + sqrt(4)\n5: 5 + 5/5\n6: 6 + 6 - 6\n7: 7 - 7/7\n8: 8 - sqrt(sqrt(8 + 8))\n9: 9/sqrt(9) + sqrt(9)\n\nAnd for a bonus: (0! + 0! + 0!)! = 6
150854
i bet your thinking, my god, it has 3 values (x,y,z), how the heck does one attack this\n\nbelieve it or not, its simple\n\njust like you would in R^2\n\nsqrt[ (x-x1)^2 + (y-y1)^2 + (z-z1)^2]\n\nso \n\nsqrt [ (3-1)^2 + (-3- -4)^2 + (2-7)^2]\n\nso\n\nsqrt[ (2)^2 + 1^2 + (-5)^2]\n\nso\n\nsqrt [ 4 +1+25]\n\nso\n\nsqrt [30]\n\nhope that helps
753754
I presume you mean\n\n[(6x^2) / (4x^2y-12xy)] * [(x^2+x-12) / (3x^2+12x)]\n\nThere seems to be something wrong. Are you sure about those "y" terms?
201908
Ah, systems.\n\nthe first problem- there are many ways you can choose to do this, but I think we should multiply both sides of the first equation by two so that they positive and negative y's will cancel out.\n\nSo now we have:\n2(2x + y)=2(4)\n3x - 2y = -1\n\nSimplified:\n4x + 2y = 8\n3x - 2y = -1\n\nnow you add down, so you get:\n7x +0y = 7\n7x =7\nx=1\n\nsubstitute 1 for x in any equation:\n2(1)+y = 4   \n2 + y = 4\ny = 2\n\n\n\nfor the second one- lets cancel out the y's again.\n\n5x - 3y = 14\n-3(2x - y) = (6)-3\n\n5x - 3y = 14\n-6x +3y = -18\n\nAdd them together:\n-x = -4\nx = 4\n\nThen solve for y:\n5x - 3y = 14\n5(4) - 3y = 14\n20 -3y = 14\n-3y = -6\ny = 2
859715
1) 4-3x<_8+x\n4+8<_x+3x\n12<_4x\n3<_x or x in the interval [3,plus infinity)\n\n2) 3x-3>-x+1\n3x+x>1+3\n4x>4\nx>1 or x in the interval (1, plus infinity)\n\n3) 3(x-1)_<4x\n3x-3<_4x\n-3<_x or x in the interval [-3, plus infinity)\n\n4)1/2x<x+1\n-1<x/2\n-2<x or x in the interval (-2, plus infinity)\n\n5)-1/2x>1-x\nx/2>1\nx>2 or x in the interval (2, plus infinity)
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!!