I have been given the following equation and believe that it is not true. If I am correct, how can i go about proving it in a sophisticated way?
a*x+b+c=k*x*(b*c)
I don't think that this can work, am I correct?
What, exactly, do you mean by "prove an equation"? What do you mean by "it is not true"? Equations are typically true for some values of the variables and false for other. Often one either solves an equation (determines values of the unknowns so that the equation is true) or proves it is an identity (true for all possible values of the unknown). But then what are the "unknowns" and what are the constants?
If you are trying to solve the equation for x, that's easy. From ax+ b+ c= kbcx, subtract ax from both sides: kbcx- ax= (kbc- a)x= b+ c. Now, if kbc- a is not 0, that is if $\displaystyle kbc\ne a$, $\displaystyle x= \frac{b+ c}{kbc- a}$. If kbc= a, there is no such x.
Or are you trying to prove that there exist numbers a and k so that this is true, no matter what x, b, and c? One thing you could do certainly is solve for a: $\displaystyle a= [kbcx- b- c]/x$. As long as x is not 0, given any values for b, c, x, and k, that gives a value of a that makes the equation true.
If you are trying to prove that is an "identity", true for all values of x, a, b, c, and k, that is obviously not true.
Thank you for your reply. I don't think I was very clear in my original question, that's my mistake. I basically intend both the LHS and RHS of the equation to be functions of an independant variable 'x' with the other values remaining constant. Therefore, there will be a value of 'x' which would satisfsy this equation (x=((−(b+c))/(a-b*c*k)) I think) but for all other values of 'x' the equation would be false. I just need some help proving this algebraically in a formal mathematical proof format, something I haven't really done before. I suppose I am hoping for some type of general mathematical rule to exist that states that this type of equality will not exist if you have an equation like that. I would appreciate any further help.