Not hard to imagine

$ABCD$ is a square with area $k$ . $E$ is a point on $\overline{BC}$ such that $\overline{BE}:\overline{EC}=2:3$ and $F$ a point on $\overline{AD}$ such that $\overline{DF}:\overline{FA}=5:7$ . If G is the intersection point between $\overline{AE}$ and $\overline{BF}$ . The area of triangle $AGB$ can be written as $\frac{a}{b}k$ . Find $a+b$