Chapter 7.5, Problem 82STP

To determine

To Find:

The prices for a hot dog, a soda, and a bag of potato chips

Answer to Problem 82STP

Price of one hot dog = $1.95

Price of one soda = $1.50

Price of a bag of chips = $0.90

Explanation of Solution

Given Information:

The Hot Dog Grille offers the lunch combinations shown. Assume that the price of a combo meal is the same price as purchasing each item separately.

  

Formula Used:

CRAMER’S RULE FOR THE SYSTEM OF LINEAR EQUATIONS WITH 3 VARIABLES:

Given a linear system:

  $\begin{array}{l}{a}_{1}x+b_{1}y+c_{1}z=d_1\\ a_{2}x+b_{2}y+c_{2}z=d_2\\ a_{3}x+b_{3}y+c_{3}z=d_3\end{array}$

Coefficient matrix :

  $D=\left[\begin{array}{l}{a}_1{b}_1{c}_1\\ a_2{b}_2{c}_2\\ a_3{b}_3{c}_3\end{array}\right]$

X-matrix:

  $D_x=\left[\begin{array}{l}{d}_1{b}_1{c}_1\\ d_2{b}_2{c}_2\\ d_3{b}_3{c}_3\end{array}\right]$

Y-matrix

  $D_y=\left[\begin{array}{l}{a}_1{\text{ d}}_1{c}_1\\ a_2{\text{ d}}_2{c}_2\\ a_3{\text{ d}}_3{c}_3\end{array}\right]$

Z-matrix

  $D_z=\left[\begin{array}{l}{a}_1{b}_1{\text{ d}}_1\\ a_2{b}_2{\text{ d}}_2\\ a_3{b}_3{\text{ d}}_3\end{array}\right]$

To solve for x :

  $x=\frac{\left|D_x\right|}{\left|D\right|}$

To solve for y :

  $y=\frac{\left|D_y\right|}{\left|D\right|}$

To solve for z :

  $z=\frac{\left|D_z\right|}{\left|D\right|}$

Determinent:

  $\begin{array}{l}\left|A\right|=\left|\begin{array}{l}abc\\ \text{d e f}\\ \text{g h i}\end{array}\right|=a\cdot \left|\begin{array}{l}\text{e f}\\ \text{h i}\end{array}\right|-b\cdot \left|\begin{array}{l}\text{d f}\\ \text{g i}\end{array}\right|+c\cdot \left|\begin{array}{l}\text{d e}\\ \text{g h}\end{array}\right|\\ =a(ei-fh)-b(di-fg)+c(dh-eg)\end{array}$

Calculation:

Let

price of a hot dog = x

price of a soda = y

price of a bag of potato chips = z

From the lunch combo, we have :

  $\begin{array}{l}2x+y=5.40\\ x+z+y=4.35\\ 2x+2z=5.70\end{array}$

Solve the given set of linear equations using Cramer’s Rule:

  $\begin{array}{l}D=\left[\begin{array}{l}\text{2 1 0}\\ 1\text{ 1 1}\\ 2\text{ 0 2}\end{array}\right]\\ D_x=\left[\begin{array}{l}\text{5}\text{.40 1 0}\\ \text{4}\text{.35 1 1}\\ \text{5}\text{.70 0 2}\end{array}\right]\\ D_y=\left[\begin{array}{l}\text{2 5}\text{.40 0}\\ 1\text{ 4}\text{.35 1}\\ 2\text{ 5}\text{.70 2}\end{array}\right]\\ D_z=\left[\begin{array}{l}\text{2 1 5}\text{.40}\\ 1\text{ 1 4}\text{.35}\\ 2\text{ 0 5}\text{.70}\end{array}\right]\\ \left|D\right|=2(2)-1(2-2)+0(-2)=4\\ \left|D_x\right|=5.40(2)-1(2(4.35)-1(5.70))+0(-5.70)=7.8\\ \left|D_y\right|=2(2(4.35)-5.70)-5.40(0)+0=6\\ \left|D_z\right|=2(5.70)-1(5.70-2(4.35))+5.40(-2)=3.6\\ x=\frac{\left|D_x\right|}{\left|D\right|}=\frac{7.8}{4}=1.95\\ y=\frac{\left|D_y\right|}{\left|D\right|}=\frac{6}{4}=1.50\\ z=\frac{\left|D_z\right|}{\left|D\right|}=\frac{3.6}{4}=0.90\end{array}$

So,

Price of one hot dog = $1.95

Price of one soda = $1.50

Price of a bag of chips = $0.90