Confused by a proof from Solow book

Question

I am working on the Solow book about how to do proofs.

Consider the problem of proving that, “If x and y are real numbers that $x^2 +6y^2 = 25$ and $y^2 +x = 3$, then $y = 2$.” In working forward the hypothesis, which of the following is not valid? Explain.

a) $y^2 = 3−x$

=> a) is considered valid. But why?

In my opinion a) is not valid because the hypothesis doesn't imply a)
Hypothesis: $x^2 +6y^2 = 25$ and $y^2 +x = 3$
Counter example:
$x^2 +6y^2 = 25 \implies x = \sqrt{(25-6y^2)}$
$y^2 +x = 3 \implies x = 3-y^2$
$\sqrt{(25-6y^2)} \neq 3-y^2$

Answer 1

The hypotesis is

• $x^2 +6y^2 = 25$ and $y^2 +x = 3$

then $y^2 +x = 3$ is true.

From the hypotesis we obtain

• $x^2 +6(3-x) = 25 \iff x^2-6x-7=0 \implies x=-1,7 \implies y=\pm 2,y=\pm 2i$

and then the statement “If x and y are real numbers that $x^2 +6y^2 = 25$ and $y^2 +x = 3$, then $y = 2$” is not true, indeed

• $P\implies Q$

is equivalent to

• $\neg Q \implies \neg P$

which is false.