Question

Реши систему уравнений методом подстановки.


{2−5(0,2v−2k)=3(3k+2)+2v,

4(k−2v)−(2k+v)=2−2(2k+v).


Ответ:

k=

;v=

.

Answer 1

Ответ:

$(k;v)=(-2;-2)$

Объяснение:

$\displaystyle \left \{ {2-5(0,2v-2k)=3(3k+2)+2v,\atop {4(k-2v)-(2k+v)=2-2(2k+v);}} \right. \Leftrightarrow\left \{ {{2-v+10k=9k+6+2v,} \atop {4k-8v-2k-v=2-4k-2v;}} \right. \Leftrightarrow \left \{ {{2-v+k=2v+6,} \atop {6k-7v=2;}} \right. \Leftrightarrow \left \{ {{k=2v+6+v-2,} \atop {6k-7v=2;}} \right. \Leftrightarrow\left \{ {{k=3v+4,} \atop {6k-7v=2;}} \right. \Leftrightarrow \left \{ {{k=3v+4} \atop {6*(3v+4)-7v=2;}} \right. \Leftrightarrow$

$\displaystyle\Leftrightarrow\left \{ {{k=3v+4,} \atop {18v+24-7v=2;}} \right.\Leftrightarrow\left \{ {{k=3v+4,} \atop {11v=2-24;}} \right. \Leftrightarrow\left \{ {{k=3v+4,} \atop {v=-\frac{22}{11}=-2; }} \right. \Leftrightarrow\left \{ {{k=3*(-2)+4,} \atop {v=-2;}} \right. \Leftrightarrow\left \{ {{k=-6+4,} \atop {v=-2;}} \right. \Leftrightarrow\left \{ {{k=-2,} \atop {v=-2}} \right.$