Question

помогитi плизз......

Answer 1

$\left \{ {{x/2-y/3=1} \atop {x/4+2y/3=8}} \right. \\ \\ \left \{ {{}\frac{3x-2y}{6}=1  \atop {}\frac{3x+8y}{12}=8 } \right. \\ \\ \left \{ {{3x-2y=6} \atop {3x+8y=96}} \right. \\ -----------------\\ 10y=90;y=9\\ \\ x/2-9/3=1\\x/2=4;x=8$

$otvet:(8;9)$

$\left \{ {{x/4+y/4=2} \atop {x/6+y/3=2}} \right. \\ \\ \left \{ {{x+y=8} \atop {x+2y=12}} \right. \\ -------------\\ y=4\\ \\ x=8-4=4\\ \\ otvet:(4;4)$

Answer 2

$1)\; \; \left \{ {{\frac{x}{2}-\frac{y}{3}=1\; |\cdot 6} \atop {\frac{x}{4}+\frac{2y}{3}=8\; |\cdot 12}} \right. \; \left \{ {{3x-2y=6} \atop {3x+8y=96}} \right. \ominus \left \{ {{x=\frac{1}{3}(2y+6)} \atop {-10y=-90}} \right. \; \left \{ {{x=8} \atop {y=9}} \right. \; \Rightarrow \; \; (8,9)$

$2)\; \; \left \{ {{\frac{x}{4}+\frac{y}{4}=2\; |\cdot 4} \atop {\frac{x}{6}+\frac{y}{3}=2\; |\cdot 6}} \right. \; \left \{ {{x+y=8} \atop {x+2y=12}} \right.\ominus \left \{ {{x=8-y} \atop {-y=-4}} \right. \; \left \{ {{x=4} \atop {y=4}} \right. \; \Rightarrow \; \; (4,4)$