Помогите,пожалуйста Решите систему уравнений 2х-1/10=2-9(y-3)/20 y-4/4=2x-9/3-3/4

$\displaystyle \left \{ {{\frac{2x-1}{10}=2-\frac{9(y-3)}{20}\;|\cdot 20} \atop {\frac{y-4}{4}=\frac{2x-9}{3}-\frac{3}{4}\;|\cdot 12\quad }} \right.\\\\\left \{ {{4x-2=40-9y+27} \atop {3y-12=8x-36-9}} \right.\\\\\left \{ {{4x+9y=69\qquad \qquad } \atop {3y-8x=-33\;|\cdot (-3)}} \right.\\\\\left \{ {{4x+9y=69\; \;} \atop {24x-9y=99}} \right.+\\\\4x+24x+9y-9y=69+99\\28x=168;\;4x=24;\;x=6\\4x+9y=69;\;4\cdot 6+9y=69\\9y=69-24=45\\y=45:9=5$

Ответ: (6;5).