Chapter 1, Problem 10PT

a.

To determine

To write:Algebraic expression for total points scored

Answer to Problem 10PT

  $6t+e+3f$

Explanation of Solution

Given information:

Scoring Play	Points
Touchdown	6
Extra point	1
Field goals	3

Number of touchdowns $=t$

Number of extra points $=e$

Number of field goals $=f$

Calculation:

Point for one touchdown $=6$ , one extra point $=1$ , one field goal $=3$

We can write the cost for two adults, four children and one senior in expression as-

  $t(6)+e(1)+f(3)=t(6)+e+f(3)\left[\text{e}\left(\text{1}\right)\text{=e}\text{due to multiplicative identity}\right]$

Hence the expression will be $6t+e+3f$

b.

To determine

Possible combination for total score of $27$

Answer to Problem 10PT

  $t=2,e=3,f=4$

Explanation of Solution

Given information:

Total score $=27$

Calculation:

We can assume any two values and evaluate for third value to obtain a possible combination.

Let total touchdown be $2$ , i.e. $t=2$

And total field goal be $4$ , i.e. $f=4$

We know,

  $\begin{array}{l}6t+e+3f=27\\ \Rightarrow 6·2+e+3·4=27\\ \Rightarrow 12+e+12=27\\ \Rightarrow e+12+12=27\\ \Rightarrow e+24=27\\ \Rightarrow e=27-24\\ \Rightarrow e=3\end{array}$

Therefore a possible combination is two touchdowns, three extra points and four field goals