Q10. - 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.

 10. Write a system of three equations that represents how much food Simone purchased.

Step-by-Step Solution

Answer

 Thesystem of equations are

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

Step-1 –Assumption of variables

 Let us assume that $x$ pound of chickens, $y$ pound of sausage and$z$ pound of rice have been brought.

Step-2 –Evaluation of first equation

Cost of chicken per pound$=\$6.$

Cost of$x$pound chicken$=\$6x.$

Cost of sausage per pound$=\$3.$

Cost of$y$pound sausage$=\$3y.$

Cost of rice per pound$=\$1.$

Cost of$z$ pound rice$=\$z.$

Total amount spend$=\$42.$

$=6x+3y+z=42.$

Step-3 –Evaluation of second equation

Amount of chicken bought$=x$pounds.

Amount of sausage bought$=y$pounds.

Amount of rice bought$=z$pounds.

Total amount of food$=13\frac{1}{2}$pounds.

$x+y+z=13\frac{1}{2}$ Pounds.

Step-4 –Evaluation of third equation

She buys twice as much rice as sausage.

So, the solution is,

$\begin{array}{l}2y=z\\ 2y-z=0\end{array}$