Question

Jambalaya is a Cajun dish made from chicken, sausage, and rice. Simone is making a 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 13.5 pounds of food. She buys twice as much rice as sausage. How much chicken, sausage, and rice will she use in her dish?
Can someone please show me how to make this into 3 algebraic equations. I should be good from there. I would really appreciate it.

Answer 1

$x- amount\ of\ chicken-\ cost\ 6\ \ per\ pound\\ y-amount\ of sausage\ - cost\ 3\ \ per\ pound\\ z-amount\ of rice-\ cost\ 1\ \ per\ pound \\\\ x+y+z=13,5\ and\\ 6*x+3*y+1z=42 \ and\\ z=2y\\\\ First\ substitude\ z=2y\ to \ both\ equations:\\ \left \{ {{x+y+2y=13,5} \atop {6x+3y+2y=42}} \right. \\\\ \left \{ {{x+3y=13,5} \atop {6x+5y=42}} \right. |First\ equation\ multiply\ by\ -6\\\\ \left \{ {{-6x-18y=-81 } \atop {6x+5y=42}} \right.|Adding\ equations\\ \+\ -\ -\ -\ -\ -\ -\- \\ -13y=-39|:-13$$y=3\\ x+3y=13,5\\ x+3*3=13,5\\ x=13,5-9=4,5\\ x=4,5\\ z=2y\\ z=2*3=6\\ She\ will \ use \ 4,5\ pounds\ of\ chicken, \ 3\ pounds\ of\ sausage\\ and\ 6\ pounds\ of \ rice$