We are given that,

\(\begin{bmatrix} 3x+y& -y \\[0.3em] 2y-x &3\\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 1& 2\\[0.3em] -5 & 3 \\[0.3em] \end{bmatrix}\)

We need to find the values of x and y. 

We know by the property of matrices,

 \(\begin{bmatrix} a_{11}& a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} b_{11}& b_{12} \\[0.3em] b_{21} & b_{22} \\[0.3em] \end{bmatrix}\)

This implies, 

a₁₁ = b₁₁, 

a₁₂ = b₁₂, 

a₂₁ = b₂₁ and 

a₂₂ = b₂₂ 

So, if we have 

 \(\begin{bmatrix} 3x+y& -y \\[0.3em] 2y-x &3\\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 1& 2\\[0.3em] -5 & 3 \\[0.3em] \end{bmatrix}\)

Corresponding elements of two matrices are equal.

That is,

3x + y = 1 …(i) 

- y = 2 …(ii) 

2y – x = - 5 …(iii) 

3 = 3

To solve for x and y, 

We have equations (i), (ii) and (iii).

From equation (ii),

-y = 2 

Multiplying both sides by -1, 

-1 × -y = - 1 × 2 

⇒ y = - 2 

Substituting y = - 2 in either of the equations (i) or (iii), say (i) 

3x + y = 1 

⇒ 3x + (-2) = 1 

⇒ 3x – 2 = 1 

⇒ 3x = 1 + 2 

⇒ 3x = 3 

⇒ x = \(\frac{3}{3}\) 

⇒ x = 1 

Thus, We get 

x = 1 and y = - 2.