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If$\begin{bmatrix}x 3\\ 0 2y-x\end{bmatrix} +\begin{bmatrix}1 2\\ 0 -2\end{bmatrix} =\begin{bmatrix}4 5\\ 0 3\end{bmatrix}$.
Find the value of x and y.
Given:
$\begin{bmatrix} x & 3\\ 0 & 2y-x \end{bmatrix} +\begin{bmatrix} 1 & 2\\ 0 & -2 \end{bmatrix} =\begin{bmatrix} 4 & 5\\ 0 & 3 \end{bmatrix}$
To do:
We have to find the values of x and y.
Solution:
$\begin{bmatrix} x & 3\\ 0 & 2y-x \end{bmatrix} +\begin{bmatrix} 1 & 2\\ 0 & -2 \end{bmatrix} =\begin{bmatrix} 4 & 5\\ 0 & 3 \end{bmatrix}$
This implies,
$\begin{bmatrix} x+1 & 3+2\\ 0+0 & 2y-x+( -2) \end{bmatrix} =\begin{bmatrix} 4 & 5\\ 0 & 3 \end{bmatrix}$
$\begin{bmatrix} x+1 & 5\\ 0 & 2y-x-2 \end{bmatrix} =\begin{bmatrix} 4 & 5\\ 0 & 3 \end{bmatrix}$
Therefore,
$x+1=4$
$x=4-1$
$x=3$
And,
$2y-x-2=3$
$2y-( 3) -2=3$
$2y=3+5$
$2y=8$
$y=\frac{8}{2}$
$y=4$.
The values of x and y are 3 and 4 respectively.
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