sistemas b-c/4=4 3c+4b/6-3c+b/9=11
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Bom dia Manuela!
Solução!
Vamos organizar as equações e em seguida substitui-la na outra.
![Equacao~~i\\\\\\
\dfrac{b-c}{4}=4\\\\\
b-c=16\\\\\
b=16+c](https://tex.z-dn.net/?f=Equacao%7E%7Ei%5C%5C%5C%5C%5C%5C%0A+%5Cdfrac%7Bb-c%7D%7B4%7D%3D4%5C%5C%5C%5C%5C%0Ab-c%3D16%5C%5C%5C%5C%5C%0Ab%3D16%2Bc "Equacao~~i\\\\\\
\dfrac{b-c}{4}=4\\\\\
b-c=16\\\\\
b=16+c")
![Equacao~~ii\\\\\
\dfrac{3c+4b}{ \dfrac{6}{ \dfrac{6-3c+b}{9} } }=11\\\\\\\\\
\dfrac{3c+4b}{6} \times \dfrac{9}{6-3c+b}=11\\\\\\\\\
\dfrac{3c+4b}{2} \times \dfrac{3}{6-3c+b}=11\\\\\\\\
\dfrac{9c+12b}{12-6c+2b}=11\\\\\\\
9c+12b=132-66c+22b\\\\\\\
9c+66c+12b-22b=132\\\\\\
75c-10b=132\\\\\\](https://tex.z-dn.net/?f=Equacao%7E%7Eii%5C%5C%5C%5C%5C%0A+%5Cdfrac%7B3c%2B4b%7D%7B+%5Cdfrac%7B6%7D%7B+%5Cdfrac%7B6-3c%2Bb%7D%7B9%7D+%7D+%7D%3D11%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A+%5Cdfrac%7B3c%2B4b%7D%7B6%7D+%5Ctimes+%5Cdfrac%7B9%7D%7B6-3c%2Bb%7D%3D11%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A+++%5Cdfrac%7B3c%2B4b%7D%7B2%7D+%5Ctimes+%5Cdfrac%7B3%7D%7B6-3c%2Bb%7D%3D11%5C%5C%5C%5C%5C%5C%5C%5C%0A+%5Cdfrac%7B9c%2B12b%7D%7B12-6c%2B2b%7D%3D11%5C%5C%5C%5C%5C%5C%5C%0A9c%2B12b%3D132-66c%2B22b%5C%5C%5C%5C%5C%5C%5C%0A9c%2B66c%2B12b-22b%3D132%5C%5C%5C%5C%5C%5C%0A75c-10b%3D132%5C%5C%5C%5C%5C%5C%0A%0A%0A+%0A "Equacao~~ii\\\\\
\dfrac{3c+4b}{ \dfrac{6}{ \dfrac{6-3c+b}{9} } }=11\\\\\\\\\
\dfrac{3c+4b}{6} \times \dfrac{9}{6-3c+b}=11\\\\\\\\\
\dfrac{3c+4b}{2} \times \dfrac{3}{6-3c+b}=11\\\\\\\\
\dfrac{9c+12b}{12-6c+2b}=11\\\\\\\
9c+12b=132-66c+22b\\\\\\\
9c+66c+12b-22b=132\\\\\\
75c-10b=132\\\\\\")
Vamos agora substituir a equação 1 na equação 2 encontrar o valor de c.
![Equacao~~i\\\\\\\
b=16+c\\\\\\
75c-10(16+c)=132\\\\\
75c-160-10c=132\\\\\\
75c-10c=132+160\\\\\\
65c=292\\\\\\\
c= \dfrac{292}{65}](https://tex.z-dn.net/?f=Equacao%7E%7Ei%5C%5C%5C%5C%5C%5C%5C%0Ab%3D16%2Bc%5C%5C%5C%5C%5C%5C%0A%0A%0A%0A75c-10%2816%2Bc%29%3D132%5C%5C%5C%5C%5C%0A75c-160-10c%3D132%5C%5C%5C%5C%5C%5C%0A75c-10c%3D132%2B160%5C%5C%5C%5C%5C%5C%0A65c%3D292%5C%5C%5C%5C%5C%5C%5C%0Ac%3D+%5Cdfrac%7B292%7D%7B65%7D+ "Equacao~~i\\\\\\\
b=16+c\\\\\\
75c-10(16+c)=132\\\\\
75c-160-10c=132\\\\\\
75c-10c=132+160\\\\\\
65c=292\\\\\\\
c= \dfrac{292}{65}")
Agora o valor de b
![b=16+c\\\\\\\
b=16+ \frac{292}{65}\\\\\\
65b=1040+292\\\\\\
65b=1332\\\\\
b= \dfrac{1332}{65}](https://tex.z-dn.net/?f=b%3D16%2Bc%5C%5C%5C%5C%5C%5C%5C%0Ab%3D16%2B+%5Cfrac%7B292%7D%7B65%7D%5C%5C%5C%5C%5C%5C%0A65b%3D1040%2B292%5C%5C%5C%5C%5C%5C%0A65b%3D1332%5C%5C%5C%5C%5C%0Ab%3D+%5Cdfrac%7B1332%7D%7B65%7D++ "b=16+c\\\\\\\
b=16+ \frac{292}{65}\\\\\\
65b=1040+292\\\\\\
65b=1332\\\\\
b= \dfrac{1332}{65}")
Bom dia!
Bons estudos!
Solução!
Vamos organizar as equações e em seguida substitui-la na outra.
Vamos agora substituir a equação 1 na equação 2 encontrar o valor de c.
Agora o valor de b
Bom dia!
Bons estudos!
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