$y=f(x)\Leftrightarrow x^2 y+y=7$

How do I go about even reading this, let alone solving this?

I also have to find the domain of "$f$"

  • 2
    $\begingroup$ It is just saying that $y=f(x)$ is the same as saying $x^2y+y=7$. So solve the second for $y$ to get the definition of $f$. $\endgroup$ – Terra Hyde Sep 15 '15 at 1:27
  • $\begingroup$ Perhaps the symbol $\equiv$ would be more appropriate than $\iff$? $\endgroup$ – John Joy Sep 15 '15 at 12:37

"$y$ equals $f(x)$ if and only if $x^2y+y=7$"

That doesn't really tell you anything though. Are you supposed to isolate $y$ in the right-hand side above? If so, then you would have $y(x^2+1)=7$ which you could then write as $y=\frac{7}{x^2+1}$, and since $x^2$ is never negative (and particularly never $-1$), you can say that the domain is all of the real numbers. But this assumes whatever instructional source you are using intends you to read $y$ and $f(x)$ as the same symbolism. Regardless, it's sloppy wording/phraseology, and I don't fault you for your confusion. What I outlined above is the only thing I can think of that makes marginal sense.

  • $\begingroup$ As far as I can tell with how we've been (poorly) instructed, this goes with the ideology. Thank you for making it coherent! I believe I should be able to solve similar problems like this. $\endgroup$ – Diego27865 Sep 15 '15 at 1:30
  • $\begingroup$ @Diego27865 Yes, it's a stupid problem because saying "if $y=f(x)$..." is completely useless. Nonetheless, I'm glad I could help! $\endgroup$ – Daniel W. Farlow Sep 15 '15 at 1:32

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