solvoing a problem: Multiply by a reciprocal first?
816596
x-11=2(-7)\nx=11-14\nx=-3\n\nCheck:\n  -1/7 (-3-11)=(-1/7)(-14)=2 OK\n\nOr did you mean\n\n-1/(7(x-11))=2    Then\n-1=14(x-11)\nx=11 - 1/14\nx=10 13/14\n\nCheck:\n  -1/(7/-1/14))=14/7=2   OK
766444
time to the minus one.\n\nDo a dimensional analaysis on a rate equation to confirm this.
660791
You just switch the top number and bottom number\n\nhttp://www.mathleague.com/help/fractions/fractions.htm#reciprocal
101802
I don't think there is a unique solution for it. You have more unknown variables than the equations.
888367
You need to give some directions.  What is the problem asking you to do?\nNote: The person below is incorrect.  (4+3x)^5 does NOT equal\n4^5 + (3x)^5
714493
This *is* the way to write it so that x is the input and w is the output. If you want to solve for x in terms of w (i.e. have w the input and x the output), you are better off doing numerical approximations than using the cubic formula (it is really, really nasty).
954059
there is a logical way to indicate solutions\n\nfind smallest A such that 1/ A < 3/7\nA= 4\nNow if A = 4\n1/B+1/C = 3/7 -1/4 = 5/28\n\n5/28 = (1+4)/28 = 1/28+1/7 \n            (2+3)/28 no\nif we take A = 5 we may get more solutions\n\n1/B+1/C= 3/7-1/5= 8/35\n= (1+7)/35 = 1/35+1/5 allow if duplicates permitted else no\nno other\n\ntake A =6 and we get\n\n1/B+1/C = 3/7-1/6 = 11/42\nno other solution as 11/42 is not sum of reciprocal
1166133
Multiply both sides by R+r\nI(R+r) = E\nDistribute the I\nIR + Rr = E\nSubtract IR form each side\nRr = E - IR\nDivide both sides by R\nr = (E-IR)/R
435388
damnt i should have paid attention calculus...\n\nlet pam...pam...breat....dimension......carry the one.....as x approaches zero......carry  the two....hold on...\n\nthis could take a while.
867789
Try starting with a proportion. Let K be the cost we are looking for.\n\nb / x = K / y\n\nNow solve for K.\n\nK = by / x
878962
first you get one variable on one side by its self (lets say x), solve for that variable (x), then you put the answer in for "x" and solve for the second variable, Just like you would if you only had one variable to begin with.
1160760
what you have is something like\na(b+c)\n\nthe solution is \nab+ac\n\n(x^3)(2x^-2)+(x^3)(4x)\nnow to multiply x^a * x^b you get x^(a+b)\n\n2x^((3+(-2))+4x^(3+1)\n\n2x+4x^4
1082141
divide. multiply. subtract. bring down
921908
1\n\njust do a little bit of trial you will see that the smaller number you use it nears 1.
991053
Substitute or plug in the value for x an solve the problem using PEMDAS\n\nParentheses\nExponents\nMultiplication&\nDivision\nAddition&\nSubtraction
415757
I hate to tell you, but I don't think there is an algebraic method for arriving at a value for x. Sometimes, this can be the case, in which case all you can do is find a numeric approximation, here around x=1.0587...\n\nI'll see if another user can prove me wrong, but I'm going to go with there's no way to solve this problem using analytic methods.
244823
The amount (A) will be doubling, so substitute "2P" in for A. Then you can drop the "P" from each side by division and solve for t.\n\nA = Pe^(rt)\n2P = Pe^(rt)\n2 = e^(rt)\nln 2 = (rt) ln e\nln 2 = .054t (ln e = 1)\n(ln 2)/.054 = t\n(approx.) 12.836 = t\n\nDoes that sound correct? (It's been awhile since I've done these kinds of problems.)