Q.Using the equality of matrices $\begin{pmatrix} x+y & 2 \\ 5 & xy \end{pmatrix} = \begin{pmatrix} 6 & 2 \\ 5 & 8 \end{pmatrix}$, find the values of $x$ and $y$.

Since the two matrices are equal, every corresponding pair of entries must match, which turns the matrix equation into two ordinary equations in $x$ and $y$.

Both matrices have the same order ($2\times2$), so equality holds if and only if corresponding entries match:

(the entries $2$ and $5$ already match automatically and give no new information).

From $x+y=6$, write $y=6-x$ and substitute into $xy=8$:

Factorising: $(x-2)(x-4)=0$, so $x=2$ or $x=4$. Correspondingly $y=4$ or $y=2$.

Independent check: substitute $x=2,y=4$ back into both original conditions: $x+y=2+4=6$ ✓ and $xy=2\times4=8$ ✓. Substitute the other solution $x=4,y=2$: $x+y=6$ ✓ and $xy=8$ ✓. Both pairs genuinely satisfy the equality, confirming the quadratic was solved correctly.

Alternative: since $x$ and $y$ are the two roots of the same quadratic $t^2-(x+y)t+xy=0$, i.e. $t^2-6t+8=0$, solving this quadratic directly for $t$ gives the same pair $\{2,4\}$ without substitution.

Reporting only one of the two valid $(x,y)$ pairs — since $x+y$ and $xy$ are symmetric in $x,y$, the two unknowns are interchangeable and both orderings are genuinely correct answers unless extra information (e.g. $x<y$) is given.