## Precalculus (6th Edition)

(a) $f(x) = \frac{2}{5}x+\frac{9}{5}$ (b) $f(3) = 3$
(a) function notation. Solve for $y$: \begin{array}{ccc} &-2x+5y&=&9 \\&-2x+5y+2x&= &9+2x \\&5y&=&2x+9 \\&\frac{5y}{5}&=&\frac{2x+9}{5} \\&y&=&\frac{2}{5}x+\frac{9}{5} \end{array} Let $y=f(x)$. The equation becomes: $$f(x)=\frac{2}{5}x+\frac{9}{5}$$ (b) To find $f(3)$, substitute $3$ to $x$ in $f(x)$ to obtain: $f(x)= \frac{2}{5}x+\frac{9}{5} \\f(3) = \frac{2}{5}(3)+\frac{9}{5} \\f(3)=\frac{6}{5}+\frac{9}{5} \\f(3)=\frac{15}{5} \\f(3)=3$