Question 988820
For this kind of problem, I would start with
{{{x}}}= the tens digit
{{{y}}}= the ones digit
{{{10x+y}}}= the value of the original number
{{{10y+x}}}= the value of the number with the digits interchanged


1. The sum of the digits of a two-digit number is 8 translates as
{{{x+y=8}}} .
The number with the digits interchanged is 16 less than thrice the original number translates as
{{{10y+x=3*(10x+y)-16}}} .
Now you have two equations. You can simplify the second one a bit:
{{{10y+x=3*(10x+y)-16}}}--->{{{10y+x=30x+3y-16}}}--->{{{10y+x-30x-3y=-16}}}--->{{{7y-29x=-16}}}--->{{{29x-7y=16}}} .
{{{system(x+y=8,29x-7y=16)}}}--->{{{system(y=8-x,29x-7y=16)}}}--->{{{system(y=8-x,29x-7(8-x)=16)}}}
--->{{{system(y=8-x,29x-56+7x=16)}}}--->{{{system(y=8-x,36x-56=16)}}}--->{{{system(y=8-x,36x=16+56)}}}
--->{{{system(y=8-x,36x=72)}}}--->{{{system(y=8-x,x=72/36)}}}--->{{{system(y=8-x,x=2)}}}--->{{{system(y=8-2,x=2)}}}--->{{{highlight(system(y=6,x=2))}}}


2. The units digit of a two-digit number is one less than one-half the tens digit translates as
{{{y=(1/2)x-1}}}--->{{{2y=2((1/2)x-1)}}}--->{{{2y=x-2}}}--->{{{2y+2=x}}} .
The number with the digits interchanged is 5 more than thrice the sum of the digits translates as
{{{10y+x=3(x+y)+5}}}--->{{{10y+x=3x+3y+5}}}--->{{{10y+x-3x-3y=5}}}--->{{{7y-2x=5}}} .
{{{system(x=2y+2,7y-2x=5)}}}--->{{{system(x=2y+2,7y-2(2y+2)=5)}}}--->{{{system(x=2y+2,7y-4y-4=5)}}}
--->{{{system(x=2y+2,3y-4=5)}}}--->{{{system(x=2y+2,3y=5+4)}}}--->{{{system(x=2y+2,3y=9)}}}--->{{{system(x=2y+2,y=9/3)}}}
--->{{{system(x=2y+2,y=3)}}}--->{{{system(x=2*3+2,y=3)}}}--->{{{highlight(system(x=8,y=3))}}}