Can anyone help with this math problem, I can't seem to figure it out. x/5[SQRT(2x+10)]+1/[SQRT(2x+10)]=1/5
346405
You have to solve the equation so that you can eliminate the sqrt by raising it to the second power .for instance,\n\nsqrt(2x+10)=5     you square it and get\n\n2x+10=25   easy, isnt it?\n\nbe careful if you have something like this\n\nsqrt(x+3)+sqrt(x-2)= 5\n\nhere you cannot do that you have to distribute one sqrt on each side of the equation.\n\nsqrt(x+3)=5-sqrt(x-2)\n\nnow you can square them
787725
To solve a problem like that:\n\n2x-3+4x = 10 (i'm not sure if this is your answer but I'm using it as an example)\n\nFirst combine the x's on the left side to get 6x-3 = 10\n\nThen add 3 to both sides to cancel the 3 on the left side.  You will not have 6x = 10.\n\nDivide both sides by 6 to isolate the x.  You will now have x = 10/6 (or 5/3)
1157065
(x+5)(2x^2+4x+2)\nx(2x^2+4x+2)+5(2x^2+4x+2)\n2x^3+4x^2+2x+10x^2+20x+10\n2x^3+14x^2+22x+10
58026
a + 1/a = 5\na = 5 - 1/a\na = 5a - 1\n4a = 1\na = 1/4
824496
The numerator is 2x - 5 and the denominator is x.\n\nWe want x > 2x - 5, subject to the condition that both x and 2x - 5 are positive integers.\n\nFirst, solve for x.\n\nx > 2x - 5\n-x > -5\nx < 5\n\nx must be less than 5. Because of the requirements of the problem, x must be greater than 0. x must also be an integer.\n\nIf x = 4: (2*4 - 5)/ 4 = 3/4\nIf x = 3: (2*3 - 5) / 3 = 1/3\nIf x = 2: (2*2 - 5) / 2 = -1/2.\n\nx = 2 doesn't work, because the numerator is negative.\n\nPut another way, we have an additional constraint that:\n\n2x - 5 > 0\n2x > 5\nx > 5/2.\n\nTo be an integer that satisfies all of these condtions, x can be equal to 3 or 4.\n\nTo recap, the fractions are 3/4 (if x = 4) and 1/3 (if x = 3).
342762
x2+5x-2x-10\n\n*answer*\n\nx2+3x-10
677314
To do this, you do (2/3) / (1/5). You must flip the denomiator fraction over and change the operation to multiplication. So... that would be (2/3) * (5/1) which would be 10/3.
790653
In order to solve, you need a common denominator:\n2(sqrt(3/5)) - 5(sqrt (5/3))\n2(sqrt 3/sqrt 5) - 5 (sqrt 5/sqrt 3)\n2(sqrt3/sqrt3)(sqrt 3/sqrt 5) - 5(sqrt 5/sqrt 5)(sqrt 5/sqrt 3)\n2(3/sqrt 15) - 5(5/sqrt 15)\n6/sqrt 15 - 25/sqrt 15\n-19/sqrt 15
46798
if (1/x) + 2= (c/2)\n--------------------------------\n\n1/x = (c/2) - 2\n----------------------\nx = 1/((c/2) - 2)\n       (OR)\nx = ((c/2)-2)^-1\n-----------------------
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
712992
I will try to explain step by step....\n\nA) x-2/3 + x+3/4=5/6\nFirst you would place like terms together;\nx+x-2/3+3/4=5/6\nThen try to make all the denominators equal.  Do this by thinking of what the denominators can all go into.  In this case, they can all go into 12 using whole numbers.  Since multiplying by 1 doesn't change the value, you can do that by multiplying the top and bottom of each fraction by the same number.  \n  So the -2/3 would all be multiplied times 4/4 giving you -8/12.  The 3/4 would all be multiplied times 3/3 giving you 9/12.  The 5/6 would all be multiplied times 2/2, giving you 10/12.\n\nThe equation would now look like;\nx+x-8/12+9/12=10/12\n\nYou then combine like terms;\nx+x=2x\n-8/12+9/12= 1/12\nSo your equation is now;\n2x+1/12=10/12\n\nSubtract 1/12 from both sides;\n2x=9/12\n\nAnd divide by 2;\nx=4.5/12 or 9/24\n\n9/24=3/8\n\nSo, the answer to A would be x=3/8\n\nAs for B...I'm not to good at word problems, sorry :)\n\n\nAnd NS in line 3 of your work, it should be -8/12 because -2*4= -8 not 8
17351
The equation to solve looks like:\n\n(x+10)/(y+5)=3x/y\n\nwhere x is the numerator and y is the denominator.\n\nIf you cross multiply the expression, it yields:\n\n-15x+10y=2xy\n\nThere is no single answer to that problem. Any (x,y) pair that satisfies the equation above works. (y cannot be 0 or -5). x=2 indeed yields y = 5. x=1 yields y=15/8 etc...
968475
(2x sqrt(2x)) (3 sqrt(5x))\n= (2x*3) (sqrt(2x*5x))\n= (6x) (sqrt(10x^2))\n= (6x) sqrt(x^2) sqrt(10)\n= (6x)(x) sqrt(10)\n= (6x^2) sqrt(10)
148612
(A-B)^2=A^2-2AB+B^2\nThen\n(3/5)x= 2*x*N\nThen N=3/10\n\n(x-3/10)^2\nx^2-3/5x+9/100
965349
I believe you meant (2x^2 - 3x - 2 )/ (10 + x - 3x^2) right?\n\nWell we can rearrange the problem to look like (2x^2 - 3x - 2 )/ (- 3x^2+ x +10)\n\nWe now factor out the top and factor out the bottom.\n[(2x +1) (x-2)] / [(-3x-5)(x-2)]\n\nSince this is a fraction, we can cancel out the (x-2) because it is on top and the bottom.\n\nWe are left with:\n(2x +1) / (-3x-5)\n\nI think this is what you wanted, good luck and be careful of how you write your notation!
259375
X - 2/3 X =  2X - 5\n\n2X - X + 2/3 X =5\n\n5/3 X = 5\n\nX = 3\n\n\n3 exceeds two thirds of 3 (=2) by 1 which is equal to 5 less twice of three (6-5=1)
991546
you divide both sides by 25, you should have either a fraction or decimal, and then you should have x^2= 13/25, and then take the square root of both sides. you will have x=__.\n2(x-5)^2=3.............25 divide into 2 minus 3.........correct to solve for x\n\nx^2+4x+4=7\n.. You should get all numbers on one side to where it would be.. x^2+4x-3=0. then factor\n\ncompleting the square.\nx^2-6x-3=0\n\n2x^2-8x+5=0\n\nI never did understand completing the square. If you don't have to show your work on completing the square, then do it a way you do know. That's how I always did it. And then have your teacher sit down and show you.