# Find the value of x and y, when

Question:

Find the value of x and y, when

i. $\left[\begin{array}{l}x+y \\ x-y\end{array}\right]=\left[\begin{array}{l}8 \\ 4\end{array}\right]$

Solution:

If $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{ll}e & f \\ g & h\end{array}\right]$

Then a=e, b=f, c=g, d=h

Given $\left[\begin{array}{l}x+y \\ x-y\end{array}\right]=\left[\begin{array}{l}8 \\ 4\end{array}\right]$

So, $x+y=8$ and $x-y=4$

Adding these two gives $2 x=12$

$\Rightarrow x=6$

$y=2$

Conclusion : $x=6$ and $y=2$

ii. $\left[\begin{array}{cc}2 \mathrm{x}+5 & 7 \\ 0 & 3 \mathrm{y}-7\end{array}\right]=\left[\begin{array}{cc}\mathrm{x}-3 & 7 \\ 0 & -5\end{array}\right]$

Given, $\left[\begin{array}{cc}2 x+5 & 7 \\ 0 & 3 y-7\end{array}\right]=\left[\begin{array}{cc}x-3 & 7 \\ 0 & -5\end{array}\right]$

So, $2 x+5=x-3$ and $3 y-7=-5$

$\Rightarrow 3 y=2 \Rightarrow y=\frac{2}{3}$

$\Rightarrow 2 x+5=x-3 \Rightarrow x=-8$

Conclusion : $x=-8$ and $y=\frac{2}{3}$

iii. $2\left[\begin{array}{cc}\mathrm{x} & 5 \\ 7 & \mathrm{y}-3\end{array}\right]+\left[\begin{array}{cc}3 & -4 \\ 1 & 2\end{array}\right]=\left[\begin{array}{cc}7 & 6 \\ 15 & 14\end{array}\right]$

$2\left[\begin{array}{cc}x & 5 \\ 7 & y-3\end{array}\right]+\left[\begin{array}{cc}3 & -4 \\ 1 & 2\end{array}\right]=\left[\begin{array}{cc}7 & 6 \\ 15 & 14\end{array}\right]$

$\left[\begin{array}{cc}2 x+3 & 6 \\ 15 & 2 y-4\end{array}\right]=\left[\begin{array}{cc}7 & 6 \\ 15 & 14\end{array}\right]$

$2 x+3=7 \Rightarrow x=2$

$2 y-4=14 \Rightarrow y=9$

Conclusion : $x=2$ and $y=9$