Question 258138
We have two equations
(i) {{{u = 2t + 1}}}
(ii) {{{u + t = 10}}}
by substitution, we get
(iii) {{{2t + 1 + t = 10}}}
Solving for t, we get
(iv) {{{3t + 1 = 10}}}
(v) {{{3t = 9}}}
{{{t = 3}}}
and 
{{{u = 7}}}
The number is 37