Question

Решить систему методом постановки ​

Answer 1

$(2; \quad -2); \quad (3; \quad 1);$

Объяснение:

$\left \{ {{x-3y=8} \atop {5x-y=12}} \right. \Leftrightarrow \left \{ {{x=3y+8} \atop {5x-y=12}} \right. \Leftrightarrow \left \{ {{x=3y+8} \atop {5(3y+8)-y=12}} \right. \Leftrightarrow \left \{ {{x=3y+8} \atop {15y+40-y=12}} \right. \Leftrightarrow \left \{ {{x=3y+8} \atop {(15-1)y=12-40}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{x=3y+8} \atop {14y=-28}} \right. \Leftrightarrow \left \{ {{x=3y+8} \atop {y=-2}} \right. \Leftrightarrow \left \{ {{x=3 \cdot (-2)+8} \atop {y=-2}} \right. \Leftrightarrow \left \{ {{x=2} \atop {y=-2}} \right. ;$

$(2; \quad -2);$

$\left \{ {{2x-y=5} \atop {2y-x=-1}} \right. \Leftrightarrow \left \{ {{y=2x-5} \atop {2y-x=-1}} \right. \Leftrightarrow \left \{ {{y=2x-5} \atop {2(2x-5)-x=-1}} \right. \Leftrightarrow \left \{ {{y=2x-5} \atop {4x-10-x=-1}} \right. \Leftrightarrow \left \{ {{y=2x-5} \atop {(4-1)x=-1+10}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{y=2x-5} \atop {3x=9}} \right. \Leftrightarrow \left \{ {{y=2x-5} \atop {x=3}} \right. \Leftrightarrow \left \{ {{y=2 \cdot 3-5} \atop {x=3}} \right. \Leftrightarrow \left \{ {{x=3} \atop {y=1}} \right. ;$

$(3; \quad 1);$