Chapter 6.7, Problem 30PPS

(a)

To determine

To write: a system of linear equations.

Answer to Problem 30PPS

System of equations is $\begin{array}{l}8n+6m=378\\ 6n+9m=333\end{array}$

Explanation of Solution

Given:

1. Cost of replace $8uniformsand6pomsis\$378$

2. Cost of replace $\text{6 }uniformsand\text{ 9 }pomsis\$333$

Calculation:

Suppose that $n$ represents the cost of one uniform and $m$ represents the cost of poms.

Price of $8$ uniforms is $8n$ and the price of $6$ poms is $6m$ .

Therefore, $8n+6m=378$

Price of $6$ uniforms is $6n$ and the price of $9$ poms is $9m$ .

Therefore, $6n+9m=333$

Hence, system of equations is $\begin{array}{l}8n+6m=378\\ 6n+9m=333\end{array}$ .

Conclusion:

System of equations is $\begin{array}{l}8n+6m=378\\ 6n+9m=333\end{array}$

(b)

To determine

To write: the augmented matrix.

Answer to Problem 30PPS

Augmented matrix is

  $\left[\begin{array}{l}8\text{ 6 | 378}\\ \text{6 9 | 333}\end{array}\right]$

Explanation of Solution

Calculation:

Place the coefficients of the equations and the constant term into a matrix

  $\begin{array}{l}8n+6m=378\\ 6n+9m=333\end{array}$

  $\left[\begin{array}{l}8\text{ 6 | 378}\\ \text{6 9 | 333}\end{array}\right]$

Therefore, augmented matrix is

  $\left[\begin{array}{l}8\text{ 6 | 378}\\ \text{6 9 | 333}\end{array}\right]$

Conclusion:

Augmented matrix is

  $\left[\begin{array}{l}8\text{ 6 | 378}\\ \text{6 9 | 333}\end{array}\right]$

(c)

To determine

To find: the cost of each uniform and each pom.

Answer to Problem 30PPS

The price of one uniform is $\$39$ and price of one poms is $\$11$ .

Explanation of Solution

Calculation:

To make the first element in the first row a $1$ , multiply first row by $\frac{1}{8}$

  $\left[\begin{array}{l}8\text{ 6 | 378}\\ \text{6 9 | 333}\end{array}\right]$

  $\frac{1}{8}{R}_1$

  $\left[\begin{array}{l}\text{1 }\frac{3}{4}\text{ | }\frac{189}{4}\\ \text{6 9 | 333}\end{array}\right]$

To make the first element in the second row a $0$ , multiply first row by $-6$ and add the result to the second row

  $\left[\begin{array}{l}\text{1 }\frac{3}{4}\text{ | }\frac{189}{4}\\ \text{6 9 | 333}\end{array}\right]$

  $\left(-6\right)R_1+R_2$

  $\left[\begin{array}{l}\text{1 }\frac{3}{4}\text{ | }\frac{189}{4}\\ \text{0 }\frac{9}{2}\text{ | }\frac{99}{2}\end{array}\right]$

To make the second element in the second row a $1$ , multiply second row by $\frac{2}{9}$

  $\left[\begin{array}{l}\text{1 }\frac{3}{4}\text{ | }\frac{189}{4}\\ \text{0 }\frac{9}{2}\text{ | }\frac{99}{2}\end{array}\right]$

  $\frac{2}{9}{R}_2$

  $\left[\begin{array}{l}\text{1 }\frac{3}{4}\text{ | }\frac{189}{4}\\ \text{0 1 | 11}\end{array}\right]$

To make the second element in the first row a $0$ , multiply second row by $\frac{-3}{4}$ and add the result to the first row

  $\left[\begin{array}{l}\text{1 }\frac{3}{4}\text{ | }\frac{189}{4}\\ \text{0 1 | 11}\end{array}\right]$

  $R_1+\left(\frac{-3}{4}\right)R_2$

  $\left[\begin{array}{l}\text{1 0 | 39}\\ \text{0 1 | 11}\end{array}\right]$

Hence the solution to the system of equation is $\left(39,11\right)$ . Hence the price of one uniform is $\$39$ and price of one poms is $\$11$ .

Conclusion:

The price of one uniform is $\$39$ and price of one poms is $\$11$ .