petewe Group Title Solve the system. 2y = 4x + 2 2y = -x + 7 one year ago one year ago

1. petewe Group Title

with matrices

2. richardlong369 Group Title

well

3. KKJ Group Title

2y = 4x + 2 2y = -x + 7 Thus, 4x + 2 = -x + 7 4x + x = 7 - 2 5x = 5 x = 1 Put x = 1 into any of the two equations 2y = 4(1) + 2 2y = 4 + 2 2y = 6 y = 3 Therefore (x, y) = (1, 3)

4. petewe Group Title

WITH matrices.

5. KKJ Group Title

$\left[\begin{matrix}-4 & 2 \\ 1 & 2\end{matrix}\right]\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}2 \\ 7\end{matrix}\right)$ $\left(\begin{matrix}x \\ y\end{matrix}\right) = \left[\begin{matrix}-4 & 2 \\ 1 & 2\end{matrix}\right]^{-1} \left(\begin{matrix}2 \\ 7\end{matrix}\right)$

6. KKJ Group Title

$\left(\begin{matrix}x \\ y\end{matrix}\right) = \frac{ 1 }{ -4(2) -2(1) }\left[\begin{matrix}2 & -2 \\ -1 & -4\end{matrix}\right]\left(\begin{matrix}2 \\ 7\end{matrix}\right)$

7. KKJ Group Title

$\[\left(\begin{matrix}x \\ y\end{matrix}\right) = \frac{ 1 }{ -10}\left[\begin{matrix}2 & -2 \\ -1 & -4\end{matrix}\right]\left(\begin{matrix}2 \\ 7\end{matrix}\right)$\]

8. KKJ Group Title

$\[\[\left(\begin{matrix}x \\ y\end{matrix}\right) = \frac{ 1 }{ -10}]\left(\begin{matrix}4 - 14 \\ -2 - 28\end{matrix}\right)$\]\]

9. KKJ Group Title

$\left(\begin{matrix}x \\ y\end{matrix}\right) = \frac{ 1 }{ -10}]\left(\begin{matrix}- 10 \\ -30\end{matrix}\right) = \left(\begin{matrix}1 \\ 3\end{matrix}\right)$