Q11. - Algebra 2

Question

COOKING For exercise 10 and 11, use the following information. Jambalaya is a Cajun dish made from chicken, sausage and rice. Simone is making large pot of jambalaya for a party. Chicken costs \( $6$per pound, sausage costs \) $3$  per pound, and rice costs \($1$ per pound. She spends \)$42$ on $13\frac{1}{2}$ pounds of food. She buys twice as much as rice as sausage.

 11. How much chicken sausage, and rice will she use in her dish?

Step-by-Step Solution

Answer

 She uses$4\frac{1}{2}$pounds of chicken, $3$ pounds of rice and$6$ pounds of sausage in her diet.

Step-1 –Using elimination we obtain system of two equations with two variables

Given system of equations are 

$\begin{array}{l}6x+3y+z=42\\ x+y+z=13\frac{1}{2}\\ 2y-z=0\end{array}$

Multiplying the second equation by $6$and subtracting from the first equation, we get

$\begin{array}{l}3y-6y+z-z6=42-81\\ -3y-5z=-39\\ 3y+5z=39\end{array}$

Step-2 –Solving the system of two equations

$\begin{array}{l}2y-z=0\\ 3y+5z=39\end{array}$

From the above first equation we get 

$z=2y$

Putting the value of $z$in second equation, we get$\begin{array}{l}3y+5\left(2y\right)=39\\ 3y+10y=39\\ 13y=39\\ y=3\end{array}$

$\begin{array}{l}z=2y\\ z=2\left(3\right)\\ z=6\end{array}$

Step-3 –Substituting the value of y and z in any equation with three variables

$\begin{array}{l}6x+3y+z=42\\ 6x+3\left(3\right)+6=42\\ 6x+9+6=42\\ 6x=42-15\\ 6x=27\\ x=\frac{27}{6}\\ x=\frac{9}{2}\\ x=4\frac{1}{2}\end{array}$