Please help me with solving equations with rational expressions.?
819535
sqrt(x) - sqrt(3x + 3) = 1\nsqrt(x) = sqrt(3x + 3) + 1\n\nsquare both sides\n\nx = (sqrt(3x + 3) + 1)^2\nx = (sqrt(3x + 3) + 1)(sqrt(3x + 3) + 1)\nx = (3x + 3) + sqrt(3x + 3) + sqrt(3x + 3) + 1\nx = 3x + 3 + 2sqrt(3x + 3) + 1\nx = 3x + 4 + 2sqrt(3x + 3)\n\n-2x - 4 = 2sqrt(3x + 3)\n-x - 2 = sqrt(3x + 3)\n\nsquare both sides\n\n(-x - 2)^2 = 3x + 3\n(-x - 2)(-x - 2) = 3x + 3\nx^2 + 2x + 2x + 4 = 3x + 3\nx^2 + 4x + 4 = 3x + 3\nx^2 + x + 1 = 0\n\nx = (-b ± sqrt(b^2 - 4ac))/(2a)\n\nx = (-1 ± sqrt(1^2 - 4(1)(1)))/(2(1))\nx = (-1 ± sqrt(1 - 4))/2\nx = (-1 ± sqrt(-3))/2\n\nx = (1/2)(-1 ± isqrt(3))\n\nThere are no real solutions.
780298
first u distribute the - to the stuff in the parenthesis so you have\n7x^2-3+x^2\nthen combine like terms\n8x^2-3
998496
x^3-3x^2+10x-30 R 10\nit's x+3 / x^4+0x^3+x^2+0x-80\n              x^4+3x^3\n             -------------\n                    -3x^3+x^2\n                    -3x^3-9x^2\n                    -------------\n                             10x^2+0x\n                             10x^2+30x\n                             --------------\n                                      -30x-80\n                                      -30x-90\n                                     ------------\n                                              10
720929
Let me just work through the steps to solve the problem:\n\nsqrt(2x^2-1)=x   <---Original Equation\n2x^2-1=x         <---Square both sides\n2x^2-x-1=0       <---Subtract x\n\nNow, apply the quadradic equation:\nx=(-b+/-(sqrt(b^2-4ac))/(2a)\n\nFor this equation, the equation is:\nx=(-(-1)+/-(sqrt((-1^2)-4*2*(-1))/(2*(2))\n\nSimplified it is:\nx=(1+/-sqrt(1+8))/(4)\nx=(1+3)/4 or x=(1-3)/4\n\nSo, x=1 or x=-1/2
1053403
Write it as:\n2x^2-4x+3x-6\n=2x[x-2]+3[x-2]\n=[2x+3][x-2]\nSimple middle-term factorisation!!!
594770
Well we are not going to do your homework for you. But we will help you along. why not start out by writing down everything you know. they writing down what you need to find. Then look at your equations and find one that uses what you know to find one of the values you need to find.\n\nSo write it up and put it on here and we will help you along.
303592
Yes, I can solve them.\n\nLike any other fractions, find a common denominator.  Though with (a) you can just multiply everything by 7x and simplify the equation to:\n\n21 + 1 = -14x\n\nYou can do the same with (c).  With (b) it's a little more complicated, and you have to find the common denominator of x and (x + 6).  I'll leave that for you.
263286
the problem you presented above can be represented as,\n\n(8x^8/2x^2) + (16x^6/2x^2) - (12x^3/2x^2)\n\nthen you can simply divide each of the term\nyou'll get\n\n4x^6 + 8x^4 - 6x
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
311264
Well. Do you know how to work rational expressions in general? The first link below is a useful general resource and has a section on rational expressions and polynomial long division; the second is specifically on rational expressions at college intro-level.\n\nI used to tutor math and I found that most people's difficulty with word problems was not the actual math, it was extracting the appropriate equations from the words. Once they had the equations, they could solve them (assuming they'd learned the basics). \n\nI'd find it pretty difficult to give you general guidelines on "how to solve word problems with rational expressions" but the main trick with word problems is to figure out a) what variables are involved, and assign each of them a letter (e.g., speed of first train is x, time it has been travelling is y, speed of second train is z...), b) sort out the relationship between them (e.g. x = 2z/5), and c) make sure you know what *exactly* is being asked. It is not at all uncommon for word problems to include extraneous information (colour of train, age of driver) to muddle the issue.\n\nYour best bet is almost certainly to go ask your teacher, professor, TA, tutor, study group, or classmate-who-understands-it for help. They will be better able to figure out where your problem is if they can watch you work through an actual example.
368125
if this is the problem:\n((x+3)/2)*((x^2-x)/(x+3))=21\nthen you solve by\ncancelling out (x+3)\n(1/2)*(x^2-x)=21\nmultiplying by 2\nx^2-x=42\nmaking it quadratic\nx^2-x-42=0\nand using x=(-b(+-)srt(b^2-4ac))/2a\nx=(1(+-)srt(1-4*1*(-42)))/2\nx=(1(+-)srt(169))/2\nx=(1+13)/2  or x=(1-13)/2\nx=7   or x= -6
825795
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]
706946
In general it is difficult to solve a cubic equation; there is an algebraic formula for the solutions of a general cubic, but it is extremely complicated.\n\nTry to find a solution by inspection. First I will use the rational-roots test, which is that all rational roots of a polynomial (if they exist) can be written in the form a/b, where a divides the constant term and b divides the leading coefficient. This means that the only possible rational roots are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60, and their negatives. It is not hard to see that none of these is a root.\n\nNow I will try elements of the extension fields of the rationals by square roots of factors of 60.\n\nIf x=a+b√2, then a^3+3a^2*b√2+6ab^2+2b^3√2+2a+2b√2=60, so a^3+6ab^2+2a=60, and 3a^2*b+2b^3+2b=0, so 3a^2+2b^2=-2, which is impossible if a and b are rational; similar things happen with any other extension fields of degree 2.\n\nYou can also try extension fields of higher degree to see whether you get lucky.\n\nAnother idea is to find two values of x such that x^3+2x-60 is negative for one value and positive for another (for example, 3 and 4), and use procedures like Newton's method or regula falsi to approximate the real root (a graph indicates that there is only one real root).\nI cheated with my calculator and found no easy factorization.
255958
Substitute 2 for x and solve for c.\n2^2-3(2)+c=0\nRemember do powers, then multiplication, then subtraction.
953940
Hi again \n\nY = √2x – 3 \n2x – 3 = y ²\nX =( y ² + 3 )/ 2\nf –¹ (x)  =( y ² + 3 )/ 2  =>   f –¹ (x)  = (x ² + 3 )/ 2   &  x ≥ 0 \n\nGood Luck \n try to solve them yourslef again if you got wrong , am sure you can solve them correctly , if you just try your best
817835
x^3-x^2+3x-1\nx+4&#92;x^4+3x^3-x^2+11x-4\n      -(x^4+4x^3)\n              -x^3-x^2\n            -(-x^3-4x^2)\n                       3x^2+11x\n                     -(3x^2+12x)\n                                   -x-4\n                                 -(-x-4)\n                                         0\nx^4+3x^3-x^2+11x-4=(x+4)(x^3-x^2+3x-1)
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.