![\bf \begin{array}{llll} &[(-6,2),(2,3),(1,1),(-7,2),(4,2)]\\\\ inverse& [(2,-6),(3,2),(1,1),(2,-7),(2,4)] \end{array} \\\\\\ \textit{is the original a one-to-one?}\qquad \stackrel{rep eated~y-values}{(-6,\stackrel{\downarrow }{2}),(2,3),(1,1),(-7,\stackrel{\downarrow }{2}),(4,\stackrel{\downarrow }{2})}](https://img.qammunity.org/2018/formulas/mathematics/college/5najm2frqrj4oc3yagfo2mkvkalnu0eqy3.png)
notice, the inverse set is just, the same set with the x,y turned to y,x, backwards.
is it a one-to-one? well, for a set to be a one-to-one, it must not have any x-repeats, that is, the value of the first in the pairs must not repeat, and it also must not have any y-repeats, namely the value of the second in the pairs must not repeat.