From Basic Abstract Algebra (Robert Ash):
...the element $t \in E$ depends algebraically on $T$ over $F$ if $t$ is algebraic over $F(T)$, the field generated by $T$ over $F$...We say that $T$ spans E algebraically over $F$ if each $t$ in $E$ depends algebraically on $T$ over $F$, that is $E$ is an algebraic extension of $F(T)$.
I was trying to show that if $\alpha$ is algebraic over $F(x)$ and if x is algebraic over $F(y_1, y_2)$ (and if $\alpha$ depends algebraically on $x$ over F), then $\alpha$ must be algebraic over $F(y_1, y_2)$, but I was a bit confused.
Let's suppose that (I'm just making these up) $\alpha x + a = 0$ for $a \in F$ and $y_1x + y_2x^2 = 0$. Now I cannot see how it is possible for $\alpha$ to be algebraic over $F(y_1, y_2)$, because we know that $x$ is not included in $F(y_1, y_2)$. Then how can $\alpha$ be algebraic over it?
Thanks in advance
One way to see it: a field extension $F\subseteq F(x)$ is algebraic iff $F(x)$ has finite dimension as a vector field over $F$.
In your case, the dimension of $F(x,y_1,y_2)$ over $F(y_1,y_2)$ is finite, and dimension of $F(x,\alpha)$ over $F(x)$ is finite. Use those two to show that the dimension of $F(x,\alpha,y_1,y_2)$ over $F(y_1,y_2)$ is finite.
Edit: In hindsight, I think it's easier to use another equivalent condition, namely that $F\subseteq F(x)$ is algebraic iff $F[x]$ has finite dimension as a vector field over $F$, so we have to show that $F(y_1,y_2)[x,\alpha]$ has finite dimension over $F(y_1,y_2)$.