how would you write this as an algebraic equation?
920013
Let x be the amount of gallons you used:\n\nThere is a flat charge of $55 for the first 6,000 gallons.  So start with that.\n\nThen you need to add the excess charge.  The charge is 0.20 per gallon over 6000.  \n\nThe hard part is figuring out how to write the excess of 6000 gallons.  If I had used 7000, it would be 1000.  If I had used 8000, it would be 2000.  So in general, you would write (x - 6000).\n\nPutting it all together:\nWater bill = 55 + (x - 6000) * 0.20\n\nNow technically, you have two formulas.  You have one for amounts under 6000, and one for amounts over 6000.\n\nFor x <= 6000,\nf(x) = 55\n\nFor x > 6000\nf(x) = 55 + (x - 6000) * 0.20
982146
Just add and substract terms in both sides.\n\nSince \ny-x^2+8x\nis equal to 7, \nyou can add x^2 and substract 8x to both sides.\n(y-x^2+8x) +x^2-8x = y\n7 +x^2-8x = x^2-8x+7\n\nso y=x^2-8x+7\n\nby the way, if you factorize in terms of x then you'd get\ny=(x-1)(x-7)
807939
= (1 + 8x)*(1 - 8x) = 1 - 64*x^2\n\nAnd what do you say?
460242
ur answer is [sqrt(3) - sqrt(7)].it can written as sqrt[x +/- 2*sqrt[y]] when x=a^2 + b^2 and y=a*b is equal to sqrt(a) +/- sqrt(b)
121228
here is a simple program algo\n\nprogram quad\nbegin\nvar\n     declare variables;\nget variable values of a,b,c for equation ax2+bx+c=0\nd=)b*b)-4*a*c;\nif(d<0)\nbegin\n\n   root are imaginary;\nend\nelse\n\n    root1=b(sqrt(d))/2a;\n    root2=-b(sqrt(d))/2a;\n ( note : formula i am not sure make changes as per required)\nprint the value of root1 and root2;\nend
288929
-> 2x^2 - 1 = 4x + 15\n-> 2x^2 - 1 - 4x - 15 = 0\n-> 2x^2 -4x -16 = 0\n-> 2( x^2 -2x -8 )= 0\n-> x^2 - 2x - 8 = 0\n-> factoring the equation you will come up with\n-> (x - 4)(x + 2) = 0\n-> equate both factors to zero\n-> x - 4 =0 , x + 2 = 0\n-> x = 4, x = -2
1053464
if you want to simplify it \n\nit is 3x - 4x + 4 = -x + 4
508029
The equation is:\n(x+a+b+c)^2 = 2x^2+(x+a)^2-(x+b)^2-(x+c)^2\nA simple rearrangement yields\n(x+a+b+c)^2 - x^2 = (x+a)^2-(x+b)^2 + x^2-(x+c)^2\nFactorize using the identity a^2-b^2=(a+b)(a-b)\n(x+a+b+c+x)(x+a+b+c-x) = (x+a+x+b)(x+a-x-b) + (x+x+c)(x-x-c)\nwhich can be written as\n(2x+a+b+c)(a+b+c) = (2x+a+b)(a-b)-c(2x+c)\n==> 2x(a+b+c)+(a+b+c)^2 = 2x(a-b)+(a-b)(a+b)-2xc-c^2\n==> 2x(a+b+c-a+b+c) = (a+b+c)^2 + a^2-b^2-c^2\n==> 4x(b+c) = a^2+b^2+c^2+2ab+2bc+2ca+a^2-b^2-c^2\n            = 2(a^2+ab+bc+ca) = 2(a+b)(c+a)\n==> x = (a+b)(c+a)/[2*(b+c)]
739098
Move the constant to the other side of the equation:\n2x^2 -6x = 11\nDivide out the x^2 coefficient:\nx^2 - 3x = 11/2\nTake half of the coefficient of x, square it, and add it to both sides of the equation:\n(-3/2)^2 = 9/4\nx^2 -3x + 9/4 = 31/4\nThe left-hand side of the equation can now be expressed as a square where the constant added to x is the number you just squared and added to both sides:\n(x - 3/2)^2 = 31/4\nTake the square root of both sides, remembering that the right side must be expressed in plus-minus form:\nx - 3/2 = +/- sqrt 31/4\nMove the constant on the left side of the equation to the right in order to isolate the variable:\nx = 3/2 +/- sqrt 31/4\nSimplify.  Since sqrt (31/4) is the same thing as (sqrt 31)/(sqrt 4), you have:\nx = 3/2 +/- (sqrt 31)/2\nx = (3 +/- sqrt 31)/2
725157
assuming the numbers after the parenthesis and variables are exponents...\n(x-1)^(3/2) [6x(x-1)^2-x^2(x-1)^3]\njust add the exponents together from the outside to the inside- (3/2) + (2) = 7/2\n\n(x-1)^(3/2) [6x(x^2-2x+1)-x^2(x^3-3x^2+3x-1)]\n(x-1)^(3/2) [6x^3-12x^2+6x-x^5-3x^4+3x^3-x^2]\n(x-1)^(3/2) [-x^5-3x^4+9x^3-13x^2+6x]\nx(x-1)^(3/2)(-x^4-3x^3+9x^2-13x+6)\nokay, that took me for friggin ever to type on a regular keyboard. Please, seriously consider me for the best answer!
947283
You can do this using "synthetic division"\n....... ______________________\nx - 5 ) x^4 - 0x^3 - 3x^2 + 0x + 6\n\nI'm going to simplify this leaving out the x-terms and only showing the coefficients, but just imagine they are there:\n\n.............. 1 . 5 .22 110\n1 -5 ) 1 . 0 .-3 . 0 . 6\n......... 1. -5\n......... ------\n.............. 5 .-3\n.............. 5 -25\n.............. -------\n................... 22. 0\n....... ........... 22 -110\n....... ............----------\n....... ........ ........ 110  6\n....... ........ ........ 110 -550\n....... ........ ........ ------------\n....... ........ ........ ....... 556\n\nSo if I did it correctly the answer should be:\n1 5 22 110 remainder 556\n\nThis equates to:\n(x^3 + 5x^2 + 22x + 110) + 556/(x-5)
1137023
(2x-9)(2x+9)\n\na guess... its been a long time since I did that stuff
11476
algebraic simplification\n\nx over 4+3 = 2x+3 over 5\nx / (4+3) = (2x+3) / 5\nx/7 = (2x+3)/5\n5x/7 = 2x+3\n9x/7 = -3\n\nx = -7/3
197473
This is not an equation unless it is equal to something.  It is an expression.\n\nFirst, you can simplify the middle: a-5a = -4a\n\nThen, multiply the first 2 together: 3a(-4a)=-12a^2\n\nNow multiply this by the last part: -12a^2(a-35)= -12a^3 + 420a^2.\n\nIf you were supposed to set this equal to 0 and solve for a, you would not need to multiply it out.  You would just set each part = 0 and solve\n\n3a = 0\n-4a = 0\na-35 = 0
580733
First equation:\n(x+a+b)^2 = x^2 + (x+a)^2 - (x+b)^2\nor x^2 + 2x(a+b) + (a+b)^2 = x^2 + x^2 + 2ax + a^2 - x^2 - 2bx - b^2\nor x^2 + 2ax + 2bx + a^2 + 2ab + b^2 - [x^2 + x^2 + 2ax + a^2 - x^2 - 2bx - b^2] = 0\nor 4bx + 2ab + 2b^2 = 0\nor 2b(2x+a+b) = 0\nor  2x + (a + b) = 0\nor 2x = -(a+b)\nTherefore x = -(a+b)/2\n\nSecond equation:\n(x+a)^2=(x+b)^2\nor (x+a)^2 - (x+b)^2 = 0\nor (x+a+x+b)(x+a-x-b) = 0\nor (2x+a+b)(a-b) = 0\nIf a = b then the equation becomes identity. So a is not equal to b.\nHence 2x + a + b = 0 \nor x = -(a+b)/2\n\nThus from both equations we have x = -(a+b)/2
134065
Your statement translates to the following algebraic equation.\na+b=a-b=a*b\nSolving a+b=a-b, we get b = 0; \nSubstituting b=0 and solving a+b=a*b, we get a = 0;\nSo the two numbers are zeroes.
202779
This is quadratic. A quadratic equation looks like:\nax^2+bx+c = 0\n\nSo a=4, b=-1, and c=3\n\nSo by using the quadratic equation: \n\nx= (-b+sqrt(b^2-4ac))/2a\nand\nx= (-b-sqrt(b^2-4ac))/2a\n\nTry to get the answer from that.... note that imaginary numbers may be involved!