Step 1:

The equation of a hyperbola and the combined equation of the asymptotes differ only in the constant term.

Therefore the combined equation of the asymptotes is of the form $8x^2+10xy-3y^2-2x+4y+k=0$-----(1)

Step 2:

Factorising the 2nd degree terms,

$8x^2+10xy-3y^2=8x^2+12xy-2xy-3y^2$

$\qquad\qquad\;\;\;\;\;\;\;\;\;\;\;=4x(2x+3y)-y(2x+3y)$

$\qquad\qquad\;\;\;\;\;\;\;\;\;\;\;=(4x-y)(2x+3y)$

Therefore the asymptotes have equations $4x-y+l=0,2x+3y+m=0$ and their combined equation is $(4x-y+l)(2x+3y+m)=0$------(2)

Step 3:

Equations (1) & (2) represent the same pair of lines.

Comparing the coefficient of $x$ and $y$

$4m+2l=-2$

$-m+3l=4$

$\Delta=\begin{vmatrix}4 & 2\\-1 & 3\end{vmatrix}=12+2=14$

$\Delta m=\begin{vmatrix}-2 & 2\\4 & 3\end{vmatrix}=-6-18=-14$

$\Delta l=\begin{vmatrix}4& -2\\-1 & 4\end{vmatrix}=16-2=14$

$l=\large\frac{\Delta l}{\Delta}$$=1$

$m=\large\frac{\Delta m}{\Delta}$$=-1$

Therefore the asymptotes are $2x+3y-1=0$ and $4x-y+1=0$ and their combined equation is $8x^2+16xy-3y^2-2x+4y-1=0$