The equation of the straight line passing through the point of intersection of the lines $x-y=1$ and $2x-3y+1=0$ and parallel to the line $3x+4y=14$  is 

Given lines $x-y-1=0$  and $2x-3y+1=0$

$\begin{array}{l}\begin{array}{ccc}x& y& 1\\ -1& -1& \begin{array}{cc}1& -1\end{array}\\ -3& 1& \begin{array}{cc}2& -3\end{array}\end{array}\\ \Rightarrow \frac{x}{-1-3}=\frac{y}{-2-1}=\frac{1}{-3+2}\\ \Rightarrow \frac{x}{-4}=\frac{y}{-3}=\frac{1}{-1}\\ \Rightarrow x=4,y=3\\ \Rightarrow (x,y)=(4,3)\end{array}$

The equation of the line parallel to $3x+4y=14$ and through $(4,3)$ is

$\begin{array}{l}\Rightarrow 3(x-4)+4(y-3)=0\\ \Rightarrow 3x-12+4y-12=0\\ \Rightarrow 3x+4y-24=0\end{array}$