Question

Find the value x, y, z. $\left[\begin{array}{cc}x+y& 2\\ 5+z& xy\end{array}\right]=\left[\begin{array}{cc}6& 2\\ 5& 8\end{array}\right]$.

Answer 1

It is clear that

$5+z=5⟹z=0$

$x+y=6⟹y=6-x\dots \left(1\right)$

$xy=8\dots \left(2\right)$

From (1) and (2),

$x(6-x)=8$

$x^2-6x+8=0$

$\therefore x=\frac{6\pm \sqrt{36-32}}{2}⟹x=4\text{or}2$

Now, if $x=2$, $y=4$ and if $x=4$, $y=2$.

Hence, there are 2 possible solution sets for $(x,y,z)$ one is $(4,2,0)$ and the other is $(2,4,0)$.