When substituting, how do I know if I'll get a new equation or just an identity?

Question

My question is more general, but it is illustrated in this example:

I was trying to answer this question, which asks to solve for $xy$ in the equation $x^2 + xy + y^2 + \sqrt3 y + 1 = 0$. One idea that I had was to set $xy = c$, substitute $x = \dfrac cy$, and then solve for $y$ to get $y = ($some expression in terms of $c)$. Then I was going to substitute $y = \dfrac cx$ to get $x = ($ another expression in terms of $c)$. Lastly, I was going to set up the equation

$($some expression in terms of $c) \cdot ($another expression in terms of $c) = c$

How do I know if my last equation will just reduce to $c = c$ and all my work will have been for nothing?

More generally, how do I know when making substitutions will give me a new equation with new information, or just an identity?

Answer 1

From your first substitution you got $y=f(c)$. Then, when you plug in $y=\frac{c}{x}$ you only get $x=\frac{c}{f(c)}$, i.e. $c=\frac{c}{f(c)}f(c)=c$ so you're going in circles.

The problem is you're not utilizing any new information, but recycling what you already used. Basically, the only substitution made here was $xy=c$, rewritten the first time as $x=\frac{c}{y}$. The second substitution was again just a rewrite of this same equation, no additional information was given and this will typically get you going in circles.

It's a lot like when, for example, you have a linear system \begin{align} x-y&=2\\ x+y &=4 \end{align} You would get from the first equation that $x=2+y$, but you wouldn't plug that back into the same equation, but into the second one, which gives you one more piece of information.

Answer 2

The question at first sight seems to be a single equation in two unknowns, so without further information seems unsolvable

The additional information here is that there is apparently a solution

There are two possibilities:

• The equation is a convoluted way of saying $xy$ takes a particular value: an example of this could be $2x-5 = \frac{4-15y}{3y}$. You would have discovered this with the first step of your approach as using $x=\frac{c}{y}$ would lead to $\frac{2c}{y}-5= \frac{4-15y}{3y}$ then $6c-15y=4-15y$ and finally $c=\frac{2}{3}$. But you did not in the example in your question; if $c$ varies with $y$ then additionally using $y=\frac{c}{x}$ will not help

• Some special other feature of the question restricts the number of real solutions. Since your example is a degree-two equation in two variables, attempting to complete the square seems natural, so look at $\left(x^2+2x\frac{y}{2} + \frac{y^2}{2^2}\right) + \left(\frac{\sqrt{3}^2}{2^2}y^2+2\frac{\sqrt{3}}{2}y + 1^2\right)=0$, i.e. $\left(x+\frac{y}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}y+1\right)^2=0$ to get $x+\frac{y}{2}=0$ and $\frac{\sqrt{3}}{2}y+1=0$, two degree-one equations in two unknowns and easily solvable, as in the answers to the linked question