Question
Find the value of $x$ if $\left[\begin{array}{cc}3 x+y & -y \\ 2 y-x & 3\end{array}\right]=\left[\begin{array}{cc}1 & 2 \\ -5 & 3\end{array}\right]$
✓
Answer
$
\left[\begin{array}{cc}
3 x+y & -y \\
2 y-x & 3
\end{array}\right]=\left[\begin{array}{cc}
1 & 2 \\
-5 & 3
\end{array}\right]
$
Comparing the corresponding terms, we get.
$
\begin{aligned}
& -y=2 \\
& \Rightarrow y=-2 \\
& 3 x+y=1 \\
& \Rightarrow 3 x=1-y \\
& \Rightarrow 3 x=1-(-2) \\
& =1+2 \\
& =3 \\
& \Rightarrow x=\frac{3}{3} \\
& =1
\end{aligned}
$
Hence $x=1, y=-2$.
\left[\begin{array}{cc}
3 x+y & -y \\
2 y-x & 3
\end{array}\right]=\left[\begin{array}{cc}
1 & 2 \\
-5 & 3
\end{array}\right]
$
Comparing the corresponding terms, we get.
$
\begin{aligned}
& -y=2 \\
& \Rightarrow y=-2 \\
& 3 x+y=1 \\
& \Rightarrow 3 x=1-y \\
& \Rightarrow 3 x=1-(-2) \\
& =1+2 \\
& =3 \\
& \Rightarrow x=\frac{3}{3} \\
& =1
\end{aligned}
$
Hence $x=1, y=-2$.
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