Question

suma a 4 numere este 43. daca din fiecare se scade acelasi numar se obtin 7,2,9,13. Care sunt numerele?

Answer 1

[tex]\begin{bmatrix}a+b+c+d=43\\ a-x=7\\ b-x=2\\ c-x=9\\ d-x=13\end{bmatrix} \\ \\ \begin{bmatrix}a+b+c+d=43\\ -b-c-d-x=-36\\ b-x=2\\ c-x=9\\ d-x=13\end{bmatrix} \\ \\ \begin{bmatrix}a+b+c+d=43\\ -b-c-d-x=-36\\ -c-d-2x=-34\\ c-x=9\\ d-x=13\end{bmatrix} \\ \\ \begin{bmatrix}a+b+c+d=43\\ -b-c-d-x=-36\\ -c-d-2x=-34\\ c-x=9\\ d-x=13\end{bmatrix} \\ \\ \begin{bmatrix}a+b+c+d=43\\ -b-c-d-x=-36\\ -c-d-2x=-34\\ -d-3x=-25\\ -4x=-12\end{bmatrix} \\ \\ -4x=-12 =\ \textgreater \ x= 3 \\ \\ -d-3\cdot \:3=-25=\ \textgreater \ d= 16 \\ \\ [/tex][tex]-b-c-d-x=-36 =\ \textgreater \ c=12 \\ \\ a+b+c+d=43 =\ \textgreater \ d=16\quad \:c=12\quad \:b=5=\ \textgreater \ a=10 \\ \\ \\ \\ d=16,\:c=12,\:b=5,\:a=10 [/tex]

Answer 2

I nr   /------/ + 7
II nr  /------/ + 2              }43
III nr /------/ + 9
IV nr /------/ + 13 

43 - (7 + 2 + 9 + 13) = 43 - 31 = 12 

1 + 1 + 1 +1 + 1 = 4 s.e. 

12 : 4 = 3  un s.e. 

I nr = 3 + 7 = 10 

II nr = 3 + 2 = 5 

III nr = 3 + 9 = 12 

IV nr = 3 + 13 = 16