Question 890113
Let "x" be the 1st number
"y" be the 2nd number

{{{x = 4 + 5y}}}
{{{(x+y)-6 = 16}}}
{{{4 + 5y + y - 6 = 16}}}
{{{6y - 2 = 16}}}
{{{6y = 18}}}
{{{y = 3}}}

{{{x = 4 + 5(3)}}}
{{{x = 19}}}

Therefore, the 1st number is 19 and the other is 3