Question

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

• A. $(2,6)$
• B. $(1,6)$
• C. $(2,9)$
• D. $(3,6)$

Answer 1

The correct option is C $(2,9)$

We have $2\left[\begin{array}{cc}x& 5\\ 7& y-3\end{array}\right]+\left[\begin{array}{cc}3& -4\\ 1& 2\end{array}\right]=\left[\begin{array}{cc}7& 6\\ 15& 14\end{array}\right]\Rightarrow \left[\begin{array}{cc}2x& 10\\ 14& 2y-6\end{array}\right]+\left[\begin{array}{cc}3& -4\\ 1& 2\end{array}\right]=\left[\begin{array}{cc}7& 6\\ 15& 14\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}2x+3& 10-4\\ 14+1& 2y-6+2\end{array}\right]=\left[\begin{array}{cc}7& 6\\ 15& 14\end{array}\right]\Rightarrow \left[\begin{array}{cc}2x+3& 6\\ 15& 2y-4\end{array}\right]=\left[\begin{array}{cc}7& 6\\ 15& 14\end{array}\right]$
$\Rightarrow 2x+3=7$ and $2y-4=14$
$\Rightarrow 2x=7-3$ and $2y=18$
$\Rightarrow x=\frac{4}{2}$ and $y=\frac{18}{2}$
$\therefore x=2;y=9.$