• Matéria: Matemática
  • Autor: CeselianaKawaii
  • Perguntado 5 anos atrás

Resolva o sistema:
 \frac{2m + 3n}{5}  = 10 -  \frac{n}{3}
 \frac{4n - 3m}{6}  =  \frac{3m}{4}  + 1

Respostas

respondido por: Armandobrainly
3

Explicação passo-a-passo:

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\Large \bf{\begin{cases} { \frac{2m + 3n}{5} = 10 -  \frac{n}{3}  }\\{ \frac{4n - 3m}{6} =  \frac{3m}{4}   + 1} \end{cases}} \\ \Large \bf{3(2m + 3n) = 150 - 5n} \\ \Large \bf{6m + 9n = 150 - 5n} \\ \Large \bf{6m + 9n + 5n = 150} \\ \Large \bf{6m + 14n = 150} \\ \Large \bf{\begin{cases} {6m + 14n = 150}\\{ \frac{4n - 3m}{6}  =  \frac{3m}{4}  + 1} \end{cases}} \\ \Large \bf{ \frac{4n - 3m}{6} =  \frac{3m}{4}  + 1 } \\ \Large \bf{2(4n - 3m) = 9m + 12} \\ \Large \bf{8n - 6m = 9m + 12} \\ \Large \bf{8n - 6m - 9m = 12} \\ \Large \bf{8n - 15m = 12} \\ \Large \bf{ - 15m + 8n = 12} \\ \Large \bf{\begin{cases} {6m + 14n = 150}\\{ - 15m + 8n = 12} \end{cases}} \\ \Large \bf{5(6m + 14n) = 5 \times 150} \\ \Large \bf{5 \times 6m + 5 \times 14n = 750} \\ \Large \bf{30m + 5 \times 14n = 750} \\ \Large \bf{30m + 70n = 750} \\ \Large \bf{\begin{cases} {30m + 70n = 750}\\{ - 15m + 8n = 12} \end{cases}} \\ \Large \bf{ - 15m + 8n = 12} \\ \Large \bf{2( - 15m + 8n) = 2 \times 12} \\ \Large \bf{ - 2 \times 15m + 2 \times 8n = 24} \\ \Large \bf{ - 30m + 2 \times 8n = 24} \\ \Large \bf{ - 30m + 16n = 24} \\ \Large \bf{\begin{cases} {30m + 70n = 750}\\{ - 30m + 16n = 24} \end{cases}} \\ \Large \bf{30m + 70n - 30m + 16n = 750 + 24} \\ \Large \bf{\cancel{\red {30m}} + 70n\cancel{\red { -  30m}} + 16n = 750 + 24} \\ \Large \bf{70n + 16n = 750 + 24} \\ \Large \bf{86n = 750 + 24}  \\ \Large \bf{86n = 774} \\ \Large \bf{n =  \frac{774}{86} } \\ \Large \bf{n = 9} \\ \Large \bf{ - 15m + 8 \times 9 = 12} \\ \Large \bf{ - 15m + 72 = 12} \\ \Large \bf{ - 15m = 12 - 72} \\ \Large \bf{ - 15m =  - 60} \\ \Large \bf{m =  \frac{ - 60}{ - 15} } \\ \Large \bf{m = 4} \\ \Large \bf{\begin{cases} { \frac{2 \times 4 + 3 \times 9}{5}  = 10 -  \frac{9}{3} }\\{ \frac{4 \times 9 - 3 \times 4}{6}  =  \frac{3 \times 4}{4} + 1 } \end{cases}} \\ \Large \bf{\begin{cases} {7 = 7}\\{4 = 4} \end{cases}} \\ \Large \bf{\red{ \boxed{ \green{ \boxed{ \pink{ \boxed{(m,n) = (4,9)} }} } } }}

\mathcal{ATT : ARMANDO}

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