Chapter 12.4, Problem 27PPSE

Solution

To find: The number of equations needed to represent the situation. To write the system using matrix equation.

The situation is represented by three equations:

  $\left\{\begin{array}{c}x+y+z=9\\ 2.45x+1.85y+0.80z=15\\ x+y-2z=0\end{array}\right.$

The matrix equation is given by:

  $\left[\begin{array}{ccc}1& 1& 1\\ 2.45& 1.85& 0.80\\ 1& 1& -2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}9\\ 15\\ 0\end{array}\right]$

Given information:

Given, to fill nine 1-lb tins with a snack mix. Almonds cost $2.45/lb, peanuts cost $1.85/lb, and raisins cost $0.80/lb. The mix contains twice as much nuts as raisins by weight. Exactly $15 is spent on the ingredients.

Calculation:

Let the weight of almonds be x lb, peanuts be y lb and raisins be z lb.

Nine 1-lb tins total weight is 9 lb i.e., $x+y+z=9$

The cost equation is given by $2.45x+1.85y+0.80z=15$

Given that the weight of nuts is twice as much as raisins. So, $x+y=2z\Rightarrow x+y-2z=0$

Hence the system of equations is:

  $\left\{\begin{array}{c}x+y+z=9\\ 2.45x+1.85y+0.80z=15\\ x+y-2z=0\end{array}\right.$

Hence the situation is represented by three equations.

The matrix equation is given by:

  $\left[\begin{array}{ccc}1& 1& 1\\ 2.45& 1.85& 0.80\\ 1& 1& -2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}9\\ 15\\ 0\end{array}\right]$

Conclusion:

The situation is represented by three equations:

  $\left\{\begin{array}{c}x+y+z=9\\ 2.45x+1.85y+0.80z=15\\ x+y-2z=0\end{array}\right.$

The matrix equation is given by:

  $\left[\begin{array}{ccc}1& 1& 1\\ 2.45& 1.85& 0.80\\ 1& 1& -2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}9\\ 15\\ 0\end{array}\right]$