Question

-24.6.-16 STA Sa+3+ 360424b-oxing zlata "H12 -82-30 102 -Holte-20 Blad 276*14 15) 3a 3b + c -23 a+26-30 - 25 40-25 3(2)-2-30 -1993 16) 6x2 sx+

Answer 1

15.

Let:

$\begin{gathered} 3a-3b+4c=-23_{\text{ }}(1) \\ a+2b-3c=25_{\text{ }}(2) \\ 4a-b+c=25_{\text{ }}(3) \end{gathered}$

Using elimination method:

$\begin{gathered} (1)-3(2) \\ 3a-3a-3b-6b+4c-9c=-23-75 \\ -9b+13c=-98_{\text{ }}(3) \end{gathered}$$\begin{gathered} 4(2)-(3) \\ 4a-4a+8b-b-12c-c=100-25 \\ 9b-13c=75_{\text{ }}(4) \end{gathered}$

so:

$\begin{gathered} (3)+(4) \\ -9b+9b+13c-13c=-98+75 \\ 0=-23 \\ False \\ 0\ne-23 \end{gathered}$

Therefore, the system has no solution