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Yes I'm pretty sure you did this correctly but it's not really a question of setting $z$ equal to zero.

$f(x,y)=x^{3}y^{2}$ is a function whose graph lives in three dimensions. This should make sense because it defines a third variable, $z$, in terms of the other two, $x,y$.

So in other words it's like saying $z=x^{3}y^{2}$ and you find a level curve by setting $z$ to a constant.

You're just "undoing" this at the end of part b to make the equations from a and b equal to each other by solving for $z$ at the end of part (b). Not setting it equal to zero.

Yes I'm pretty sure you did this correctly but it's not really a question of setting $z$ equal to zero.

$f(x,y)=x^{3}y^{2}$ is a function whose graph lives in three dimensions. This should make sense because it defines a third variable, $z$, in terms of the other two, $x,y$.

So in other words it's like saying $z=x^{3}y^{2}$ and you find a level curve by setting $z$ to a constant.

You're just "undoing" this at the end of part b to make the equations from a and b equal to each other by solving for $z$ at the end of part (b). Not setting it equal to zero.

Yes I'm pretty sure you did this correctly but it's not really a question of setting $z$ equal to zero.

$f(x,y)=x^{3}y^{2}$ is a function whose graph lives in three dimensions. This should make sense because it defines a third variable, $z$, in terms of the other two, $x,y$.

So in other words it's like saying $z=x^{3}y^{2}$ and $z=x^{3}y^{2}$. And you find a level curve by setting $z$ bring $x^{3}y^{2}$ over to a constant.the other side to create $F(x,y,z)$.

You're just "undoing" this at the end of part b to make the equations from a and b equal to each other by solving for $z$ at the end of part (b). Not setting it equal to zero.