The smallest denominator is 3.
$${\frac{{\mathtt{7}}}{{\mathtt{12}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{9}}}}$$
$$\mathrm{ERROR} = {\frac{{\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{4}}}{{\mathtt{12}}}}{\mathtt{\,\times\,}}{\mathtt{4}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{3}}}{{\mathtt{9}}}}{\mathtt{\,\times\,}}{\mathtt{4}}$$
Umm...I had done what you done with your example on Math Formula and this came up.
And also when you done 3 & 5, 15 was bigger than both so you could times it by 5 & 3 but to times 12 and 9 by something to get 3. Would be a decimal.