If f : A → A and A = mathbbR - {8/5}, show that the function f(x) = (8x + 3)/(5x - 8) is one – one…

$f(x) = \frac{8x+3}{5x-8}$ is one-one and onto, and $f^{-1}(x) = \frac{8x+3}{5x-8}$. One-one: let $f(x_1) = f(x_2)$, $x_1,x_2\in A$. $(8x_1+3)(5x_2-8) = (8x_2+3)(5x_1-8)$ $40x_1x_2 - 64x_1 + 15x_2 - 24 = 40x_1x_2 - 64x_2 + 15x_1 - 24$ $-79x_1 = -79x_2$, so $x_1 = x_2$. Hence $f$ is one-one. Onto: let $y = \frac{8x+3}{5x-8}$, $y\in A$. Then $5xy - 8y = 8x + 3$, so $x(5y-8) = 8y+3$ and $x = \frac{8y+3}{5y-8}$. Since $y \neq \frac{8}{5}$, $x$ is defined; also $x \neq \frac{8}{5}$, because $x=\frac{8}{5}$ would give $8y+3 = 8y - \frac{64}{5}$, which is impossible. So $x\in A$ and $f(x) = y$: every $y\in A$ has a pre-image. Hence $f$ is onto. Since $f$ is one-one and onto, $f^{-1}$ exists and $f^{-1}(y) = \frac{8y+3}{5y-8}$. Hence $f^{-1}(x) = \frac{8x+3}{5x-8}$, that is, $f^{-1} = f$.