We have 2 equations : 

\(\left\{{}\begin{matrix}3x-2y=5\\2y+3z=1\end{matrix}\right.\Leftrightarrow\left(3x-2y\right)+\left(2y+3z\right)=5+1\)

\(\Leftrightarrow3\left(x+z\right)=6\Leftrightarrow x+z=2\)

We also have \(x-z=4\)

\(\Leftrightarrow x=\dfrac{2+4}{2}=3;z=-1\)

Therefore; x = 3 ; z = -1 and y\(\in R\)

Let $ x be the price of a shirt; so a blouse costs : $ \(\dfrac{3}{5}x\)

We have the equation :

\(2x+4.\left(\dfrac{3}{5}x\right)=110\)

\(\Leftrightarrow2x+\dfrac{12}{5}x=110\)

\(\Leftrightarrow\dfrac{22}{5}x=110\)

\(\Leftrightarrow x=\dfrac{110.5}{22}=25\) 

The price of 6 such shirts is : \(25.6=150\)

Answer : $ 150.