Solve for x and y :

$\begin{bmatrix$}[r] 2 & 5 \ 5 & 2 \end{bmatrix}\begin{bmatrix}[r] x \ y \end{bmatrix} = \begin{bmatrix}[r] -7 \ 14 \end{bmatrix}

Given,

$\Rightarrow \begin{bmatrix$}[r] 2 & 5 \ 5 & 2 \end{bmatrix}\begin{bmatrix}[r] x \ y \end{bmatrix} = \begin{bmatrix}[r] -7 \ 14 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 2 \times x + 5 \times y \ 5 \times x + 2 \times y \end{bmatrix} = \begin{bmatrix}[r] -7 \ 14 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 2x + 5y \ 5x + 2y \end{bmatrix} = \begin{bmatrix}[r] -7 \ 14 \end{bmatrix}

By definition of equality of matrices we get,

2x + 5y = -7

⇒ 2x = -(7 + 5y)

⇒ x = $\dfrac{-(7 + 5y)}{2}$ ……(i)

5x + 2y = 14

Substituting value of x from (i) in above equation we get,

$\Rightarrow 5 \times \dfrac{-(7 + 5y)}{2} + 2y = 14 \\[1em] \Rightarrow \dfrac{-35 - 25y}{2} + 2y = 14 \\[1em] \Rightarrow \dfrac{-35 - 25y + 4y}{2} = 14 \\[1em] \Rightarrow -35 - 21y = 28 \\[1em] \Rightarrow -21y = 28 + 35 \\[1em] \Rightarrow -21y = 63 \\[1em] \Rightarrow y = -3.$

Substituting y = -3 in (i) we get,

$x = \dfrac{-(7 + 5(-3))}{2} = \dfrac{-(7 - 15)}{2} = \dfrac{8}{2} = 4$.

Hence, x = 4 and y = -3.