Question 892333
{{{B=54+A}}}
.
.
.
{{{2B/A=6+16/A}}}
{{{2B=6A+16}}}
Substituting from above,
{{{2(54+A)=6A+16}}}
{{{54+A=3A+8}}}
{{{2A=46}}}
{{{A=23}}}
.
.
Then,
{{{B=54+23}}}
{{{B=77}}}
Their sum is 
{{{A+B=23+77=100}}}