Me ajudem na resolução desta equação de sistema:
2p - 3q -1 = 0
3p -2q - 34 =0
Obrigada :)
Soluções para a tarefa
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Boa tarde!
Solução!
![\begin{cases}
2p-3q-1=0\\
3p-2q-34=0\end{cases}\\\\\\\\\
\begin{cases}
2p-3q=1\\
3p-2q=34
\end{cases}\\\\\\\
\begin{cases}
2p-3q=1.(-2)\\
3p-2q=34.(3)
\end{cases}\\\\\\\
\begin{cases}
-4p+6q=-2\\
~~9p-6q=102
\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%0A%0A2p-3q-1%3D0%5C%5C%0A%0A3p-2q-34%3D0%5Cend%7Bcases%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%5Cbegin%7Bcases%7D%0A%0A2p-3q%3D1%5C%5C%0A%0A3p-2q%3D34%0A%0A%5Cend%7Bcases%7D%5C%5C%5C%5C%5C%5C%5C%0A%5Cbegin%7Bcases%7D%0A%0A2p-3q%3D1.%28-2%29%5C%5C%0A%0A3p-2q%3D34.%283%29%0A%0A%5Cend%7Bcases%7D%5C%5C%5C%5C%5C%5C%5C%0A%5Cbegin%7Bcases%7D%0A%0A-4p%2B6q%3D-2%5C%5C%0A%0A%7E%7E9p-6q%3D102%0A%0A%5Cend%7Bcases%7D "\begin{cases}
2p-3q-1=0\\
3p-2q-34=0\end{cases}\\\\\\\\\
\begin{cases}
2p-3q=1\\
3p-2q=34
\end{cases}\\\\\\\
\begin{cases}
2p-3q=1.(-2)\\
3p-2q=34.(3)
\end{cases}\\\\\\\
\begin{cases}
-4p+6q=-2\\
~~9p-6q=102
\end{cases}")
![-4p+9p=102-2\\\\
5p=100\\\\\
p= \dfrac{100}{5}\\\\\
p=20\\\\\\\\\
2p-3q=1\\\\
2(20)-3q=1\\\\\
40-3q=1\\\\\
-3q=1-40\\\\\
-3q=-39\\\\\
q= \dfrac{-39}{-3}\\\\\
q=13](https://tex.z-dn.net/?f=-4p%2B9p%3D102-2%5C%5C%5C%5C%0A5p%3D100%5C%5C%5C%5C%5C%0Ap%3D+%5Cdfrac%7B100%7D%7B5%7D%5C%5C%5C%5C%5C%0Ap%3D20%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A2p-3q%3D1%5C%5C%5C%5C%0A2%2820%29-3q%3D1%5C%5C%5C%5C%5C%0A40-3q%3D1%5C%5C%5C%5C%5C%0A-3q%3D1-40%5C%5C%5C%5C%5C%0A-3q%3D-39%5C%5C%5C%5C%5C%0Aq%3D+%5Cdfrac%7B-39%7D%7B-3%7D%5C%5C%5C%5C%5C%0Aq%3D13 "-4p+9p=102-2\\\\
5p=100\\\\\
p= \dfrac{100}{5}\\\\\
p=20\\\\\\\\\
2p-3q=1\\\\
2(20)-3q=1\\\\\
40-3q=1\\\\\
-3q=1-40\\\\\
-3q=-39\\\\\
q= \dfrac{-39}{-3}\\\\\
q=13")
Boa tarde!
Bons estudos!
Solução!
Vou resolver pelo método da adição.
Boa tarde!
Bons estudos!
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