We are given that,

\(\begin{bmatrix}2x+y &3y \\[0.3em]0 & 4 \\[0.3em]\end{bmatrix}\)= \(\begin{bmatrix}6&0 \\[0.3em]6 & 4 \\[0.3em]\end{bmatrix}\)

We need to find the value of x.

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 

\(\begin{bmatrix}2x+y &3y \\[0.3em]0 & 4 \\[0.3em]\end{bmatrix}\)= \(\begin{bmatrix}6&0 \\[0.3em]6 & 4 \\[0.3em]\end{bmatrix}\)

Corresponding elements of two elements are equal.

That is,

2x + y = 6 …(i) 

3y = 0 …(ii) 

To solve for x, 

We have equations (i) and (ii).

We can’t solve for x using only equation (i) as equation (i) contains x as well as y. 

We need to find the value of y from equation (ii) first.

From equation (ii),

3y = 0 

⇒ y = \(\frac{0}{3}\) 

⇒ y = 0 

Substituting y = 0 in equation (i),

2x + y = 6 

⇒ 2x + (0) = 6 

⇒ 2x = 6 – 0

⇒ 2x = 6 

⇒ x = \(\frac{6}{2}\) 

⇒ x = 3 

Thus, we get 

x = 3.