[tex]\displaystyle \left \{ {{\frac{x+2y}{5}+\frac{x-3y}{10}=\frac{3}{5} } \atop {\frac{9x-2y}{12}+\frac{4x+8y}{2}=1,5 }} \right. \\\\\left \{ {{3x+y=6} \atop {33x+46y=18}} \right.[/tex]

решение:

[tex]\displaystyle \frac{x+2y}{5}+\frac{x-3y}{10}=\frac{3}{5}\\ \\2(x+2y)+x-3y=6\\ 2x+4y+x-3y=6\\3x+4y-3y=6\\3x+y=6[/tex]