Answer
$x=5, y=3, z=4$
Work Step by Step
We denote the number of 6 bladed packages by $x$, the number of 12 bladed packages by $y$ and the number of 24 bladed packages by $z$.
We are given the system:
$\begin{cases}
x+y+z=12\\
6x+12y+24z=162\\
2x+3y+4z=35
\end{cases}$
We will use the addition method. Multiply Equation 1 by -6 and add it to Equation 2 to eliminate $x$. Multiply Equation 1 by -2and add it to Equation 3 to eliminate $x$:
$\begin{cases}
6x+12y+24z+-6(x+y+z)=162+-6(12)\\
2x+3y+4z+-2(x+y+z)=35+-2(12)
\end{cases}$
$\begin{cases}
6y+18z=90\\
y+2z=11
\end{cases}$
Multiply Equation 2 by -6 and add it to Equation 1 to eliminate $y$ and determine $z$:
$\begin{cases}
6y+18z=90\\
-6y-12z=-66
\end{cases}$
$6z=24$
$z=4$
Substitute the value of $z$ in the Equation $y+2z=11$ to determine $y$:
$y+8=11$
$y=3$
Substitute the values of $y, z$ is Equation 1 of the given system to find $x$:
$x+3+4=12$
$x+7=12$
$x=5$
The system's solution is:
$(x=5, y=3, z=4)$
