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$$\quad \begin{bmatrix} 3 & 7.5 \\ -6 & 4.5 \end{bmatrix} \quad \begin{bmatrix} x \\y \end{bmatrix} \quad = \begin{bmatrix} 6 \\ -90 \end{bmatrix} \quad$$

which of the following represent the solution of the system of the equation.

1. 12, -4
2. -12, 4
3. 1, 2
4. 2, 3

Please enumerate the steps to solve this problem as I am looking for a procedure and want to learn this topic from this question possibly.

@Lakshman Patel RJIT Corrected.

Sorry for that mistake.

$x=12,y=-4$

Given that $\begin{bmatrix} 3&7.5 \\-6 &4.5 \end{bmatrix}_{2\times 2}.\begin{bmatrix}x \\y \end{bmatrix}_{2\times 1}=\begin{bmatrix}6 \\-90\end{bmatrix}_{2\times 1}$

We can simply multiply and get

$\begin{bmatrix} 3x+7.5y\\-6x+4.5y \end{bmatrix}_{2\times 1}=\begin{bmatrix}6 \\-90\end{bmatrix}_{2\times 1}$

compare and write     $3x+7.5y=6$

multiply by $10$ on both sides, we get

$30x+75y=60$

divide by $15$ into both sides, we get

$2x+5y=4$----------$>(1)$

and                    $-6x+4.5y=-90$

multiply by $10$ on both sides, we get

$-60x+45y-900$

divide by $15$ into both sides, we get

$-4x+3y=-60$-------$>(2)$

from equation $(1)$ and $(2)$,we can solve and get $x=12$ and $y=-4$