Question 383537
Starting with:
{{{h(x)=(6x-5)/2}}}
replace h(x) with a y:
{{{y=(6x-5)/2}}}
switch the variables x for y and y for x:
{{{x=(6y-5)/2}}}
solve for y:
{{{x=(6y-5)/2}}}
{{{2x=6y-5}}}
{{{2x+5=6y}}}
{{{(2x+5)/6=y}}}
.
So, your inverse would be:
{{{h^-1(x)=(2x+5)/6}}}