\(
2\begin{pmatrix}
3 & 4 \\
5 & x \\
\end{pmatrix}
\) + \(
\begin{pmatrix}
1 & y \\
0 & 1 \\
\end{pmatrix}
\) = \(
\begin{pmatrix}
7 & 0 \\
10 & 5 \\
\end{pmatrix}
\)
To solve this problem we will use the comparison that is we will use that all the elements of L.H.S are equal to R.H.S .
= \(
\begin{pmatrix}
6 & 8 \\
10 & 2x \\
\end{pmatrix}
\) + \(
\begin{pmatrix}
1 & y \\
0 & 1 \\
\end{pmatrix}
\)
= \(
\begin{pmatrix}
7 &8+ y \\
10 & 2x+1 \\
\end{pmatrix}
\)
Comparing with R.H.S
8 + y = 0
y = - 8
2x + 1 = 5
2x = 4
x = 2