Equation of the line passing through the point of intersection of  $x+y-4=0$and $2x+3y-10=0$ , parallel to the line  $5x-7y+5=0$

$\text{ The equation of any line passing through the point of intersection of two lines }x+y-4=0$

$\text{ and }2x+3y-10=0\text{ can be taken as }$

$\begin{array}{c}x+y-4+\lambda \left(2x+3y-10\right)=0\\ x\left(1+2\lambda \right)+y\left(1+3\lambda \right)-4-10\lambda =0\end{array}$

$\text{ Since the line }x(1+2\lambda )+y(1+3\lambda )-4-10\lambda =0\text{ is parallel to the line }5x-7y+5=0\text{, }$

$\frac{\left(1+2\lambda \right)}{5}=\frac{\left(1+3\lambda \right)}{-7}$

Cross multiply and then simplify 

$\begin{array}{c}-7\left(1+2\lambda \right)=5\left(1+3\lambda \right)\\ -7-14\lambda =5+15\lambda \\ 29\lambda =-12\\ \lambda =-\frac{12}{29}\end{array}$

$\text{ Substitute }\lambda =-\frac{12}{29}\text{ in the equation }x+y-4+\lambda (2x+3y-10)=0$

It implies that 

$\begin{array}{l}x+y-4-\frac{12}{29}(2x+3y-10)=0\\ 29x+29y-116-24x-36y+120=0\\ 5x-7y+4=0\end{array}$

$\text{ Therefore, the equation of the required line is }5x-7y+4=0$