Find the values of x and y if
$\left[\begin{array}{cc}x+10 & y^{2}+2 y \\ 0 & -4\end{array}\right]=\left[\begin{array}{cc}3 x+4 & 3 \\ 0 & y^{2}-5 y\end{array}\right]$
Here,
$x+10=3 x+4 \quad[\because$ All the corresponding elements of the matrix are equal $]$
$\Rightarrow x-3 x=4-10$
$\Rightarrow-2 x=-6$
$\therefore x=3$
Also,
$y^{2}+2 y=3$
$\Rightarrow y^{2}+2 y-3=0$
$\Rightarrow y^{2}+3 y-y-3=0$
$\Rightarrow y(y+3)-1(y+3)=0$
$\Rightarrow(y+3)(y-1)=0$
$\Rightarrow y+3=0$ or $y-1=0$
$\Rightarrow y=-3$ or $y=1$
Now,
$-4=y^{2}-5 y$
$\Rightarrow y^{2}-5 y+4=0$
$\Rightarrow y^{2}-4 y-y+4=0$
$\Rightarrow y(y-4)-1(y-4)=0$
$\Rightarrow(y-4)(y-1)=0$
$\Rightarrow y-4=0$ or $y-1=0$
$\Rightarrow y=4$ or $y=1$
Since $y^{2}+2 y=3$ and $y^{2}-5 y=-4$ must hold good simultaneously, we take the common solution of these two equations.
Thus,
$y=1, x=3$ and $y=1$
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