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(D) Given the matrix equation: $x\begin{bmatrix} 3 \ 2 \end{bmatrix} + y\begin{bmatrix} 1 \ -1 \end{bmatrix} = \begin{bmatrix} 15 \ 5 \end{bmatrix}

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$x=4, y=3$

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(D) Given the matrix equation: $x\begin{bmatrix} 3 \ 2 \end{bmatrix} + y\begin{bmatrix} 1 \ -1 \end{bmatrix} = \begin{bmatrix} 15 \ 5 \end{bmatrix}$ Multiplying the scalars $x$ and $y$ into the matrices: $\begin{bmatrix} 3x \ 2x \end{bmatrix} + \begin{bmatrix} y \ -y \end{bmatrix} = \begin{bmatrix} 15 \ 5 \end{bmatrix}$ Adding the matrices on the left side: $\begin{bmatrix} 3x + y \ 2x - y \end{bmatrix} = \begin{bmatrix} 15 \ 5 \end{bmatrix}$ By equating the corresponding elements,we get a system of linear equations: $3x + y = 15$  --- $(i)$ $2x - y = 5$  --- $(ii)$ Adding equation $(i)$ and $(ii)$: $(3x + y) + (2x - y) = 15 + 5$ $5x = 20 \Rightarrow x = 4$ Substituting $x = 4$ into equation $(i)$: $3(4) + y = 15$ $12 + y = 15 \Rightarrow y = 3$ Thus,the values are $x = 4$ and $y = 3$.

If $x\begin{bmatrix} 3 \\ 2 \end{bmatrix} + y\begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 15 \\ 5 \end{bmatrix}$,then the values of $x$ and $y$ are:

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