an equation to solve?
371319
The answer is d), i.e. "c"\n\n\na/b = b/c = c/d = d/e = x\n\na = x.b\nb = x.c\nc = x.d\nd = x.e\n\na = x.x.x.x.e = x^4.e\nb = x.x.x.e = x^3.e\nc = x.x.e = x^2.e\nd = x.e = x^1.e\ne = x^0.e\n\nabcde = (x^4.e).(x^3.e).(x^2.e).(x^1.e).(x^0.e)\n\nabcde = x^10.e^5\n\n(abcde)^(1/5) = (x^10.e^5)^(1/5)\n\n(abcde)^(1/5) = x^2.e\n\n(abcde)^(1/5) = c
928663
Are you asking for help or an answer?\n\nTake any 2 equations and multiply all of the terms or one or both by numbers that will make one of the terms in each equation have the same coefficient. Now subtract one from the other and the term will disappear. Now take the resulting equation and do the same with it against the third. Keep doing this until you are down to one term and solve for the variable. Now substitute the variable into the equation and solve for the other variables.
288520
cmon dont do this again, give the other equation
57195
is that your homework?\nthat's a very easy equation...
576570
algebra...usually (well i do) set up a equation for everything, and solve for the variable. If u need help u can always get help in www.algebra.com .....
636051
x=4\n\nplease solve...\n\n8-(1) :-) ;-(
511948
There are an infinite number of ordered pairs that would satisfy this equation. If you want to find a few of them, assign a value to x and then solve for y.
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
663777
SOrry, I can't help you. If I had the formulas I could do it, but there is no way I remember that off the top of my head! Good luck
869887
If you have 2 equations, just solve for one variable for one of the equations. Take that equation and plug it into the same variable for the other equation. You will solve for that variable, and plug in that number to any of the 2 equations to solve for the other variable.
195201
i think u typed the question wrong, can u please correct it and if u have 2 unknowns, please give 2 equations, else u cant get a definite solution
1084156
The answer to that is a line.  If you want a unique solution to a two-variable problem, you need two equations.
224079
try this site... it shows step by step how to go about sloving these equations, good luck:
720847
Use logarithms. If you haven't learned them yet, it just helps to solve equations with powers.\na = b^x\nlna = ln(b^x) = xlnb\nx = lna/lnb\nx = ln(10)/ln(32.6087)\nx = 0.6607928852\nYou can use your calculator to check. I guarentee that it works.
956102
Here is the equation\n\n(246,92)U ---> (4,2)He + (242,90)Th + 2Gamma
106013
y = -2x + 8\n\nsolve it yourself
363348
Are we solving for x?