One number is 3 less than two times the other. If their sum is increased by 7

then the result is 37 the numbers are

Let the $1^{st}$ number be $x$

The second number be $2x-3$

As per the problem

Their sum increased by $7$ the result $=37$

$x+2x-3+7=37$

$3x-3+7=37$

$3x+4=37$

$3x=37-4$

$3x=33$

$x=\frac{33}{3}$

$x=11$

$\therefore$ the other number $=2x-3$

$=2\left(11\right)-3$

$=22-3$

$=19$

$\therefore$  the required number are $11,19$