Find out the number

Call $F$ is the first one and $S$ is the second one, we have this system of equation: $\begin{aligned}\ 2F+3S & = & 39 \\3F+2S & = & 36 \end{aligned}$ So $S-F=(2F+3S)-(3F+2S)=39-36=3$ And $S+F=\frac{(3+2) \times S+(2+3) \times F}{2+3}=\frac{(3S+2F)+(2S+3F)}{5}=\frac{39+36}{5}=\frac{75}{5}=15$ Finally, we have this $\begin{aligned}\ S+F & = & 15\\ S-F & = & 3 \end{aligned}$ Solve it $\cdot\cdot\cdot$ $F=\frac{F+F}{2}=\frac{S+F-S+F}{2}=\frac{(S+F)-(S-F)}{2}=\frac{15-3}{2}=\frac{12}{2}=6$ $\implies S=15-F=15-6=9$ Because $9>6$ so the answer is $\Large \boxed{9}$

Let the two positive numbers be $a$ and $b$ . The conditions given were:

$2a+3b=39\quad\quad\quad\implies$ Eq. (1)

$3a+2b=36\quad\quad\quad\implies$ Eq. (2)

Eq. (1) $\times 2 \quad\implies 4a+6b=78 \quad\quad\quad\implies$ Eq. (3)

Eq. (2) $\times 3 \quad\implies 9a+6b=108 \quad\quad\quad\implies$ Eq. (4)

Eq. (4) $-$ Eq. (3):

$(9a+6b)-(4a+6b) = 108 - 78\\ 5a=30 \implies a=6$

Substitute to find $b$ :

$2(6)+3b = 39\\ 3b=27 \implies b=9$

Therefore, the larger number is $\boxed{9}$