Question

Решить систему методом постановки ​

Answer 1

Ответ:

$(1; \quad 1); \quad (-2; \quad -6);$

Объяснение:

$\left \{ {{x+3y=4} \atop {3x-4y=-1}} \right. \Leftrightarrow \left \{ {{x=4-3y} \atop {3x-4y=-1}} \right. \Leftrightarrow \left \{ {{x=4-3y} \atop {3(4-3y)-4y=-1}} \right. \Leftrightarrow \left \{ {{x=4-3y} \atop {12-9y-4y=-1}} \right. \Leftrightarrow \left \{ {{x=4-3y} \atop {(-9-4)y=-1-12}} \right. \Leftrightarrow$$\Leftrightarrow \left \{ {{x=4-3y} \atop {-13y=-13}} \right. \Leftrightarrow \left \{ {{x=4-3y} \atop {y=1}} \right. \Leftrightarrow \left \{ {{x=4-3 \cdot 1} \atop {y=1}} \right. \Leftrightarrow \left \{ {{x=1} \atop {y=1}} \right. ;$

$(1; \quad 1);$

$\left \{ {{5x-y=-4} \atop {13x-4y=-2}} \right. \Leftrightarrow \left \{ {{y=5x+4} \atop {13x-4y=-2}} \right. \Leftrightarrow \left \{ {{y=5x+4} \atop {13x-4(5x+4)=-2}} \right. \Leftrightarrow \left \{ {{y=5x+4} \atop {13x-20x-16=-2}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{y=5x+4} \atop {(13-20)x=-2+16}} \right. \Leftrightarrow \left \{ {{y=5x+4} \atop {-7x=14}} \right. \Leftrightarrow \left \{ {{y=5x+4} \atop {x=-2}} \right. \Leftrightarrow \left \{ {{y=5\cdot (-2)+4} \atop {x=-2}} \right. \Leftrightarrow \left \{ {{x=-2} \atop {y=-6}} \right. ;$

$(-2; \quad -6);$