## Precalculus (10th Edition)

Solve for $y$: $2x^2+3y^2=1\\3y^2=1-2x^2\\y^2=\frac{1-2x^2}{3}\\y=\pm\frac{1-2x^2}{3}.$ This relation is s not a function, because a distinct $x$-value corresponds to two different $y$-values. For example, $(0, \frac{1}{3})$ and $(0, –\frac{1}{3}).$