91Ó°ÊÓ

The given equations are:

$$\begin{gathered} y-3x^3=2 \\ x+y^2=3 \\ 3x-2y=6 \end{gathered}$$

For a given value of $x$, we cube it and multiply it by $3$and then add $2$to the product to find the value of $y$.

$y=3x^3+2$

For example, if $x=2,$

$$\begin{gathered} y=3{\left(2\right)}^3+2 \\ y=3\left(8\right)+2 \\ y=24+2 \\ y=26 \end{gathered}$$

We have that when $x=2,$then$y=26$. It would work similarly for any value of $x$.Since each value of $x$corresponds to only one value of$y$the equation defines a function.

Isolating $y,$we get,

$y^2=3-x$

When $x=-1$, we have, $y^2=4$.

This gives us two values of $y$: $y=2$and $y=-2$

We have shown that when $x=-1,$then $y=2$and $y=-2.$It would work similarly for any value of $x.$Since each value of $x,$does not correspond to only one value of$y,$the equation does not define a function.

For a given value of $x,$we multiply it by $3$and it to $6$. We then divide the sum by $-2.$