## Need help, Solving equation with multi variables

http://tutorial.math.lamar.edu/AllBr...es/eq0031M.gif

i tried solving for C first, but i dont get the same answer

if someone could just show me how to work it out i'd greatly appreciate it
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 can u please just work out the problem. if i see how you do it i would get it. i'm this is actually for my Calc II class where i'm trying to do integration by partial fraction by i get the algebra wrong

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## Need help, Solving equation with multi variables

Algebra is one of the most important mathematical skills! You shouldn't think any less of it just because you're in calc II. (In fact, some mathematicians research algebra for their entire career!)

Are you sure you don't know how to solve this sort of arithmetic problem? From your edit, you appear to have a good idea of what to do. The most common mistake I see when people (myself included) do these kinds of problems is a silly arithmetic error. You could try checking each of your individual calculations, but my approach is usually to try solving the problem again from scratch.
 undertoes, as Hurkyl mentioned, you can try checking each of your individual calculations. A good way to do so on these types of problems (especially since you have the advantage of knowing the answer) is to simply plug the answer into each step along the way to see which equation is wrong. Start somewhere in the middle of your calculations and work forward or backward, until you narrow it down to where you went wrong.
 man, im getting some weired fraction as an answer for any variable i try to solve for... and i didnt even think algebra research existed, damn, that sucks i've wasted like 2 hours on this!
 b u m p!!!
 Recognitions: Homework Help Hint: always look for an equation containing only two unknowns, and plug any of these unknowns into the rest of the equations. After repeating this for two times, you'll get two equations with two unknowns, which you'll solve easily.
 C=-A sub "C" for "A": -4a + b + 8a + D=1----> D=-4a-b samething again:3a + 16c - 8d=29---->-8d=-29+13a samething again:-12a +3b +16b= 5 ===> im stuck someone help with all the steps, thankssss
 Recognitions: Homework Help Ok, let's do it again. A + C = 0 ...(1) -4A + B -8C + D = 1 ...(2) 3A + 16C -8D = -29 ...(3) -12A + 3B + 16D = 5 ...(4) -------------------------------- First: (1) => A = -C. Put this into equations (2), (3) and (4). After that, you'll get three equations (2'), (3') and (4') with the variables B, C and D. Equation (3') is the equation which will contain only C and D, so from that equation you'll obtain C in terms of D (*) (i.e. C = something*D + something else). Plug this into equations (2') and (4') and you'll get two equations with only two variables, B and D. Now, solve these two equations in order to obtain B and D, and simply plug D into equation (*) to get C, and, finally, plug C into the first equation A = -C to obtain A.
 guys, i know the method, but like i said i'm getting some weird number here. plz plz do this for me. plz just show every method mathematically. plz i beg you. i need to get place this here. its for my calc II class. i have to move on to the next subject but i cant cuz this is holding me me up. plz plz plz
 bummmmppppp
 Recognitions: Gold Member Science Advisor Staff Emeritus First, the lecture: (1) If you can't do the algebra then you need to learn. I said it once, and I'll say it again: algebra is one of the most important mathematical skills. You seem to have a deficiency in your algebra skills -- although you seem to know the method to use, you do not appear to have enough experience to identify and/or recover from your mistakes, so you need to develop that skill. Being in calc II is not an excuse for slouching on algebra. (2) If you wanted a solutions manual, then you should have bought one. Simply giving a solution is against our site policy. (3) It's unethical to take others' work and pass it off as your own. I somehow doubt you are intending, in your homework, to give credit to those who actually did the work. --------------------------------------------------------------------- Now, you've told (by drpizza) a very effective way of finding your error when you know what the answer is supposed to be -- simply go through each step of your work, and plug in the actual values to make sure they work. Have you tried this? (Incidentally, that's exactly what I did to find your mistake)
 ok, i understand the site policy of not just giving the answer. but nothing else has worked. and i would acutally give credit to whoever gave me the answer. but i'm not, because this isn't homework. If you look at the link i gave you all, its from some calc tutorial website. so come on, just help a bro out. i thought that was the policy of this site. you guys are my only resoure for the weekend
 Have you tried putting these equations in a matrix? That is what I would do. I whould either use kramers rule or take the inverse of the matrix and multiply it by the values.

 Quote by marika Please, have a look on this weird system of equations: a*(sinh(x)-sin(x))+b*(cosh(x)-cos(x))=c*(sinh(y)+sin(y))+d*(cosh(y)+cos(y)) b*(sinh(x)+sin(x))+a*(cosh(x)-cos(x))=-d*(sinh(y)-sin(y))-c*(cosh(y)+cos(y)) a*(sinh(x)+sin(x))+b*(cosh(x)+cos(x))=c*(sinh(y)-sin(y))+d*(cosh(y)-cos(y)) b*(sinh(x)-sin(x))+a*(cosh(x)+cos(x))+d*(sinh(y)+sin(y))+c*(cosh(y)-cos(y))+r*t*[ a*(sinh(x)-sin(x))+b*(cosh(x)-cos(x))]=0 Does anyone knows how to find a,b,c,d? (x,y,r,t are given) I will be very grateful for any hints!

Well, since you are given x,y,r,t this means that those things in parenthesis that include sin, cos, sinh, cosh, (and their sum and difference) are simply constants.

So, what you could do is take the coefficient matrix, where your entries of the matrix would actually be these things including the difference or sum of sin, cos, sinh, cosh.

THen, you could use el. operations to find a,b,c,d.