<s>[INST] showing $f$ to be a bijection but stuck trying to show that $f$ is surjective Recall that the closed interval $[a,b]$={$x \in \mathbb{R} \vert a\le x \le b$}
prove that $\vert [2,5] \vert = \vert [-2,3] \vert$ 
(cardinality of [a,b] not absolute value)
proof: define $f:[2,5] \rightarrow [-2,3]$ by $f(x)=c+ \frac{d-c}{b-a} (x-a)=-2+ \frac{5}{3}(x-2)$
show that $f$ is injective:
let $a,b \in [2,5]$ such that $f(a)=f(b)$
then
$-2+\frac{5}{3}(a-2)=-2+\frac{5}{3}(b-2)$
$\frac{5a-10}{3}=\frac{5b-10}{3}$
$\frac{5a}{3}=\frac{5b}{3}$ $\rightarrow$ $a=b$
so f is injective.
now when i try to show that f is surjective, i let $y\in [-2,3]$
then if $x\in [2,5]$ such that $x=\frac{3(y+2)+10)}{15}$ 
we see that $f(x)=-2+\frac{5}{3}((\frac{3(y+2)+10}{15})-2)$ 
basically trying to show that this simplifies to $f(x)=y$ and hence is surjective, was gettng way too tedious and wasnt looking very promising. Can someone please show me where I went wrong here? 
 [/INST] You have 
$$f(x) = \frac{5}{3}(x - 2) - 2$$
Let $y = f(x)$, then solve for $x$.
\begin{align*}
y & = \frac{5}{3}(x - 2) - 2\\
y + 2 & = \frac{5}{3}(x - 2)\\
\frac{3}{5}(y + 2) & = x - 2\\
\frac{3}{5}(y + 2) + 2 & = x
\end{align*}
Substituting this value for $x$ in the function $f$ yields
\begin{align*}
f\left(\frac{3}{5}(y + 2) + 2\right) & = \frac{5}{3}\left[\frac{3}{5}(y + 2) + 2 - 2\right] - 2\\
& = \frac{5}{3}\left[\frac{3}{5}(y + 2)\right] - 2\\
& = y + 2 - 2\\
& = y
\end{align*}
so $f$ is surjective.
Since you did not show your work, I am not sure how you obtained a $15$ in your denominator.  However, that was your error.
Note:  Another approach to showing that a bijection exists would be to solve for 
$$g(x) = \frac{3}{5}(x + 2) + 2$$
then showing that $g = f^{-1}$ by demonstrating that 
\begin{align*}
(g \circ f)(x) & = x && \text{for each $x \in \text{Dom}_f = [2, 5]$}\\
(f \circ g)(x) & = x && \text{for each $x \in \text{Dom}_g = [-2, 3]$}
\end{align*}
</s>