solve the system using substitution method 3x+y=62x-4y=10

[tex]\left\{\begin{array}{ccc}3x+y=6&|-3x\\2x-4y=10\end{array}\right\\\\\left\{\begin{array}{ccc}y=6-3x\\2x-4y=10\end{array}\right\\\\substitute\ y=6-3x\ to\ the\ second\ equation\\\\2x-4(6-3x)=10\\\\2x-4\cdot6-4\cdot(-3x)=10\\\\2x-24+12x=10\\\\14x-24=10\ \ \ |+24\\\\14x=34\ \ \ \ |:14\\\\x=\dfrac{34}{14}\to x=\dfrac{17}{7}\\\\substitute\ the\ value\ of\ x\ to\ first\ equation\\\\y=6-3\cdot\dfrac{17}{7}=6-\dfrac{51}{7}=\dfrac{42}{7}-\dfrac{51}{7}=-\dfrac{9}{7}[/tex]

[tex]\boxed{\left\{\begin{array}{ccc}x=\dfrac{17}{7}\\\\y=-\dfrac{9}{7}\end{array}\right}[/tex]