Chapter 5.3, Problem 28E

a)

To determine

To find: The number of the students who take lunches and breakfasts.

Answer to Problem 28E

The number of the students who take the breakfasts and the lunch are $10\text{}$ and $15$ respectively.

Explanation of Solution

Given information: The total number of the students are $50$ .

  

The sum of breakfasts and lunch should be 25 because 25 students have chose dinner.

Hence according to the pictorial representation,

  $\begin{array}{l}Breakfast=10\\ Lunch=15\end{array}$

b)

To determine

To find: The linear equation that shows the numbers of the students who choose lunch and breakfast.

Answer to Problem 28E

The linear equation that shows the numbers of the students who choose lunch and breakfast together is $\{\begin{array}{l}b+l=5\\ -b+l=25\end{array}$ .

Explanation of Solution

Given information: The numbers of students who chose lunch was 5 more than the number who chose breakfast.

  

Let the number of students who choose breakfast and lunch be $b\&l$ respectively.

Hence,

  $b+l=25\mathrm{...}\left(1\right)$

Since the numbers of students who chose lunch was 5 more than the number who chose breakfast. Hence,

  $\begin{array}{l}l=b+5\\ -b+l=5\mathrm{...}\left(2\right)\end{array}$

Hence the system of linear equation:

  $\{\begin{array}{l}b+l=5\\ -b+l=25\end{array}$

c)

To determine

To explain: How can you solve the linear system to check your answer?

Explanation of Solution

Given information: $\{\begin{array}{l}b+l=5\\ -b+l=25\end{array}$

Use elimination method to solve the linear system, add equation (1) and (2) to eliminate b,

  $\{\begin{array}{l}b+l=5\mathrm{...}\left(1\right)\\ -b+l=25\mathrm{...}\left(2\right)\end{array}$

Hence,

  $\begin{array}{l}2l=30\\ l=15\end{array}$

Put $l=15$ in the equation (2),

  $\begin{array}{l}-b+l=25\\ -b=-15+25\\ b=10\end{array}$

Hence,

  $\begin{array}{l}Breakfast=10\\ Lunch=15\end{array}$