01) Seja o sistema de equações lineares dadas por:
[tex]\begin{cases}\sf{6y_{1} + y_{2} + y_{3} + y_{4} +y_{5 } = 10} \\ \sf{y_{1} + 6y_{2} +y_{3} + y_{4} +y_{5} = 20} \\ \sf{y_{1} + y_{2} +6y_{3} + y_{4} +y_{5} =40} \\ \sf{y_{1} + y_{2} +y_{3} + 6 y_{4} +y_{5} =80} \\ \sf{ y_{1} + y_{2} +y_{3} + y_{4} + 6y_{5} =160} \end{cases}[/tex]
O valor de [tex]\sf{7y_{1} + 3 y_{5}}[/tex] é:
(A) 12
(B) 24
(C) 36
(D) 48
(E) 60
[tex]\begin{cases}\sf{6y_{1} + y_{2} + y_{3} + y_{4} +y_{5 } = 10} \\ \sf{y_{1} + 6y_{2} +y_{3} + y_{4} +y_{5} = 20} \\ \sf{y_{1} + y_{2} +6y_{3} + y_{4} +y_{5} =40} \\ \sf{y_{1} + y_{2} +y_{3} + 6 y_{4} +y_{5} =80} \\ \sf{ y_{1} + y_{2} +y_{3} + y_{4} + 6y_{5} =160} \end{cases}[/tex]
O valor de [tex]\sf{7y_{1} + 3 y_{5}}[/tex] é:
(A) 12
(B) 24
(C) 36
(D) 48
(E) 60
Lista de comentários
Verified answer
Resposta:
[tex]\textsf{Leia abaixo}[/tex]
Explicação passo a passo:
[tex]\begin{cases}\sf{6y_{1} + y_{2} + y_{3} + y_{4} +y_{5 } = 10} \\ \sf{y_{1} + 6y_{2} +y_{3} + y_{4} +y_{5} = 20} \\ \sf{y_{1} + y_{2} +6y_{3} + y_{4} +y_{5} =40} \\ \sf{y_{1} + y_{2} +y_{3} + 6 y_{4} +y_{5} =80} \\ \sf{ y_{1} + y_{2} +y_{3} + y_{4} + 6y_{5} =160} \end{cases}[/tex]
[tex]\sf 10y_1 + 10y_2 + 10y_3 + 10y_4 + 10y_5 = 310[/tex]
[tex]\sf y_1 + y_2 + y_3 + y_4 + y_5 = 31[/tex]
[tex]\sf 5y_1 = -21[/tex]
[tex]\sf y_1 = -\dfrac{21}{5} \Leftrightarrow 7y_1 = -\dfrac{147}{5}[/tex]
[tex]\sf 5y_5 = 129[/tex]
[tex]\sf y_5 = \dfrac{129}{5} \Leftrightarrow 3y_5 = \dfrac{387}{5}[/tex]
[tex]\sf 7y_1 + 3y_5 = -\dfrac{147}{5} + \dfrac{387}{5}[/tex]
[tex]\sf 7y_1 + 3y_5 = \dfrac{240}{5}[/tex]
[tex]\boxed{\boxed{\sf 7y_1 + 3y_5 = 48}}\leftarrow\textsf{letra D}[/tex]