Question

Resolva o sistema de equações com o regra de Cramer:

$\begin{cases}\mathsf{5x-8y=64}\\\mathsf{5x-5y=40}\end{cases}$​

Answer 1

Resposta:

S= {0; -8}

Explicação passo a passo:

D= 5.(-5) - (-8).5

D= -25 -(-40)

D= -25 + 40

D= 15

Dx= 64.(-5) - (-8).40

Dx= -320 +320

Dx= 0

Dy= 5.40 - 64.5

Dy = 200 - 320

Dy = -120

x= dx/d = 0/15 = 0

y= dy/d = -120/15= -8

Answer 2

Olá

$\begin{cases}\mathsf{5x-8y=64}\\\mathsf{5x-5y=40}\end{cases}$

✍️ Para resolver o sistema utilizando a regra Cramer, primeiro liste todos os determinantes necessários.

$~~~~~~ \sf D = \sf \left[\begin{array}{cc} \sf5& \sf- 8\\ \sf5& \sf- 5\end{array}\right] \\ \\ \sf D_{1} = \sf \left[\begin{array}{cc} \sf64& \sf- 8\\ \sf40& \sf- 5\end{array}\right] \iff\sf D_{2} = \sf \left[\begin{array}{cc} \sf5& \sf64\\ \sf5& \sf40\end{array}\right]$

✍️ Fómula para encontrar o determinante:

$~~~~~~~~~~~\sf \left[\begin{array}{cc} \sf a & \sf b\\ \sf c& \sf d\end{array}\right] = ad - bc$

✍️ Agora, vamos encontrar os determinantes de cada um:

$~~~~~~~~~~~ \sf D = \sf \left[\begin{array}{cc} \sf5& \sf- 8\\ \sf5& \sf- 5\end{array}\right] \\ \\~~~~~~~~~~~ \sf{5 \times ( - 5) - ( - 8) \times 5} \\ ~~~~~~~~~~~ \sf{ - 5 \times 5 + 8 \times 5} \\ ~~~~~~~~~~~ \sf{5 \times ( - 5 + 8)} \\ ~~~~~~~~~~~ \sf{5 \times 3} \\~~~~~~~~~~~ \sf\large\boxed{{\sf D = 15{}}} \\ \\ \\ ~~~~~~~~~~~ \sf D_{1}=\sf \left[\begin{array}{cc} \sf64& \sf- 8\\ \sf40& \sf- 5\end{array}\right] \\ \\~~~~~~~~~~~ \sf{64\times ( - 5) - ( - 8) \times 40} \\ ~~~~~~~~~~~ \sf{ - 320+ 8 \times40} \\ ~~~~~~~~~~~ \sf{ - 320 + 320} \\ ~~~~~~~~~~~ \sf\large\boxed{{\sf D_{1}=0{}}} \\ \\ \\ ~~~~~~~~~~~ \sf D_{2}=\sf \left[\begin{array}{cc} \sf5& \sf64\\ \sf5& \sf40\end{array}\right] \\ \\~~~~~~~~~~~ \sf{5 \times 40 - 64 \times 5} \\ ~~~~~~~~~~~ \sf{ 200 - 320} \\ ~~~~~~~~~~~ \sf\large\boxed{{\sf D_{2}= - 120{}}}$

✍️ Dado D ≠ 0, a regra de Cramer pode ser aplicada, então encontre x, y, usando a fórmula $\sf x = \sf\frac{D_{1}}{ D } \: , \: y = \sf\frac{D_{2}}{ D }$

Logo,

$~~~~~~ \sf\sf x = \frac{0}{15} \iff \sf y = - \frac{120}{15} \\ \\ ~~~~~~ \sf\large\boxed{{\sf \large\boxed{{\sf \sf x = 0 \iff \sf y = - 8{}}}{}}}$

✍️ Sendo assim, a solução do sistema é o par ordenado (x, y):

$~~~~~~~~~ \sf\large\boxed{{\sf \large\boxed{{\sf (x,y) = (0, - 8) {}}}{}}}$