Simultaneous equation/expression re-arrangement I'm having a little difficulty (being a physicist). I have the following two equations (the constants have been changed, but it is identical to the original expression which relates to the virial theorem):
$$
Y_0=A+B
$$
$$
Y=xA+x^2B
$$
I want to show:
$$
Y=x(2x-1)Y_0
$$
Can anyone just completely humiliate me and show me how this is done?
 A: I assume that you want your equations to hold for all $x$. Let's take this backwards, and assume all three of your equations are true, and find what must be true about $A$ and $B$.
Substituting your first equation into your third,
$$\begin{align}
Y &= x(2x-1)Y_0 \\
  &= (2x^2-x)(A+B) \\
  &= x(-A-B)+x^2(2A+2B)
\end{align}$$
Comparing that with your second equation we see that for both equations to be true for all $x$ the coefficients of $x$ and $x^2$ must be equal, and so
$$-A-B=A \qquad\text{and}\qquad 2A+2B=B$$
Solving those two linear simultaneous equations with the two variables $A$ and $B$ we find that the equations are linearly dependent and have the solution

$$B=-2A$$

and so from your first equation

$$Y_0=-A$$

If we try that solution in all three of your equations we see that it all checks.
If those conditions are satisfied, it is easy to show that your third equation is true for all $x$ if your first two equations are also true for all $x$. If those conditions are not satisfied, your third equation is not true.
Does this satisfy your needs?
