Answer: y = 3
Step-by-step explanation:
Multiply the matrices on the left and set each row equal to the right side. This will result in a system of equations.
Use the following format for multiplying:
[tex]\left[\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22}\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] =\left[\begin{array}{c}c\\d\end{array}\right]\\ \\\\\rightarrow \left\{\begin{array}{cc}a_{11}(x) + a_{12}(y)&=c\\a_{21}(x)+a_{22}(y)&=d\end{array}\right\}[/tex]
[tex]\left[\begin{array}{cc}3&4\\6&1\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] =\left[\begin{array}{c}\ \ 6\\-9\end{array}\right]\\ \\\\\rightarrow \left\{\begin{array}{cl}3x+4y&=\ \ 6\\6x+1y&=-9\end{array}\right\}[/tex]
Now solve the system. Since you are looking for "y", eliminate "x" by multiplying Row 1 by -2:
[tex]\begin{array}{clll}3x+4y&=\ \ 6&\rightarrow -2(3x + 4y = 6)&\rightarrow -6x-8y=-12\\6x+1y&=-9&\rightarrow 1(6x + 1y = -9)&\rightarrow \underline{\ \ 6x + 1y =\ -9}\\& & &\ \quad \qquad -7y=-21\\& & &\ \ \qquad \qquad y=3\\\end{array}[/tex]
