smart_worker said:Homework Equations
If a(y+z)=x, b(z + x)=y, c(x+y)=z, prove that x2/a(1-bc) = y2/b(1-ca) = z2/c(1-ab)
The Attempt at a Solution
ay+az=x,bz+bx=y,cx+cy=z.
i solved these equations and found x,y and z values,pluged them onto the proof equation.but it took me about half an hour for that.is there a quicker or easier way to approach these kind of questions?
epenguin said:I don't know in what level of course this question came - group theory or abstract algebra?
Whatever, I wonder if you are supposed to say something like: you have an equality* that is unaltered by moving its terms one step forward alphabetically in cyclic fashion:
Does that work? If the given equations were a=x, b=y, c=z, does it follow that a/y = b/z = c/x?epenguin said:you have an equality* that is unaltered by moving its terms one step forward alphabetically in cyclic fashion:
a → b → c → a, x → y → z → x.
This must be true still of any equality you derive from the given ones.
Therefore there is no way x2/a(1 - bc) could not be equal to the last two.
epenguin said:Yes what I said is wrong. You do have to show one of the target equalities, the other is then automatically true, it is saying the same thing as the first with names changed.
I couldn't resist the thought that because we have a form f(a, b, c, x, y, z) = f(b, c, a, y, z, x) ( = f(c, a, b, z, x, y) ) then any other function g(a, b, c, x) = g(b, c, a, y) (I mean when the terms are connected by the f relation) . But it can't be generally true or else I could invent various different g of which some would give results contradicting others. It would make things like this so easy I wondered if there is a condition for it to be true that one could easily see which could save a lot of work, but this is probably a rubbish thought.
The solution expected from you is these are just three simultaneous linear equations from which you can eliminate z, say in three different ways to get
x/y = a(1 + c)/(1 - ac) = (1 - bc)/b(1 + c) = a(1 + b)/b(1 + a) .
You then get the result for x2/y2 multiplying the second expression above by the third.
You are meant to realize that the three apparently different expressions for x/y are the same thing as each other in virtue of a relation that must hold between x, y, z (independently of a, b, c)
ab + ac + bc + 2ab -1 = 0.
for the equations to have solutions (other than x = y = z = 0). You can get this condition from any pair of the above equations.
So you can only fix at will two of a, b, c then the third is dependent. After which the ratios x: y: z are fixed and you can decide at will only one of these three.
An algebra proof question is a type of mathematical problem that requires you to use algebraic principles and equations to prove or demonstrate a given statement or theorem.
The key to solving an algebra proof question is to start with what you know and work towards what you need to prove. This involves breaking down the given statement into smaller, more manageable equations and using algebraic manipulation to transform them into the desired result.
Some common strategies for solving algebra proof questions include substitution, simplification, and using known identities or properties of algebra. It is also helpful to keep track of your steps and justify each one with a logical explanation.
A good way to check if your algebra proof is correct is to work backwards from the desired result and see if each step can be reversed or undone. You can also plug in different values for the variables to see if the equation still holds true.
Some common mistakes to avoid when solving algebra proof questions include forgetting to apply the order of operations, making careless errors in simplification, and assuming properties that may not necessarily be true. It is important to carefully check each step and have a clear understanding of the algebraic principles being used.