91Ó°ÊÓ

The given equations are:

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

For each value of $x,$we cube it and multiply it by 3. We then add $2$to this product.

$y=3x^3+2$

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 if $x=2,$then$y=26.$It would work similarly for all values 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 $x=-1$,

$$\begin{gathered} y^2=3-(-1) \\ {y}^2=4 \end{gathered}$$

This gives us two values of $y.$

That is, $y=2$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 each value of $x,$we multiply it by $3$and it to $6.$We then divide this sum by $-2.$

For example, if $x=1$

$$\begin{gathered} y=\frac{3\left(1\right)+6}{-2} \\ y=\frac{3+6}{-2} \\ y=\frac{9}{-2} \\ y=-4.5 \end{gathered}$$

We have that when $x=1,$then $y=-4.5$. It works similarly for any value of $x.$Since each value of $x,$corresponds to only one value of$y,$the given equation is a function.

Therefore, the equations$y-3x^3=2$and$3x-2y=6$are functions.