Answer :
Let's start by defining our variables:
- Let [tex]\( s \)[/tex] represent the number of tokens a game of skee ball requires.
- Let [tex]\( p \)[/tex] represent the number of tokens a game of pinball requires.
From the information given in the problem, we can write two equations based on the total number of tokens used by Tyrone and Erica.
1. Tyrone's token usage: Tyrone played 4 games of skee ball and 6 games of pinball, using a total of 26 tokens. This can be expressed as:
[tex]\[ 4s + 6p = 26 \][/tex]
2. Erica's token usage: Erica played 1 game of skee ball and 9 games of pinball, using a total of 29 tokens. This can be expressed as:
[tex]\[ s + 9p = 29 \][/tex]
We now have a system of linear equations:
[tex]\[ \begin{cases} 4s + 6p = 26 \\ s + 9p = 29 \end{cases} \][/tex]
To solve this system, we can use either substitution or elimination. In this case, let's use the elimination method.
First, let's multiply the second equation by 4 to align the coefficients of [tex]\( s \)[/tex] in both equations:
[tex]\[ 4(s + 9p) = 4 \cdot 29 \][/tex]
[tex]\[ 4s + 36p = 116 \][/tex]
Now we have the modified system of equations:
[tex]\[ \begin{cases} 4s + 6p = 26 \\ 4s + 36p = 116 \end{cases} \][/tex]
Next, we subtract the first equation from the second equation:
[tex]\[ (4s + 36p) - (4s + 6p) = 116 - 26 \][/tex]
[tex]\[ 30p = 90 \][/tex]
Solving for [tex]\( p \)[/tex]:
[tex]\[ p = \frac{90}{30} \][/tex]
[tex]\[ p = 3 \][/tex]
Now, we substitute [tex]\( p = 3 \)[/tex] back into the second original equation to find [tex]\( s \)[/tex]:
[tex]\[ s + 9p = 29 \][/tex]
[tex]\[ s + 9(3) = 29 \][/tex]
[tex]\[ s + 27 = 29 \][/tex]
[tex]\[ s = 29 - 27 \][/tex]
[tex]\[ s = 2 \][/tex]
Thus, we have:
- Every game of skee ball requires [tex]\( 2 \)[/tex] tokens.
- Every game of pinball requires [tex]\( 3 \)[/tex] tokens.
So, filling in the blanks:
Every game of skee ball requires 2 tokens, and every game of pinball requires 3 tokens.
- Let [tex]\( s \)[/tex] represent the number of tokens a game of skee ball requires.
- Let [tex]\( p \)[/tex] represent the number of tokens a game of pinball requires.
From the information given in the problem, we can write two equations based on the total number of tokens used by Tyrone and Erica.
1. Tyrone's token usage: Tyrone played 4 games of skee ball and 6 games of pinball, using a total of 26 tokens. This can be expressed as:
[tex]\[ 4s + 6p = 26 \][/tex]
2. Erica's token usage: Erica played 1 game of skee ball and 9 games of pinball, using a total of 29 tokens. This can be expressed as:
[tex]\[ s + 9p = 29 \][/tex]
We now have a system of linear equations:
[tex]\[ \begin{cases} 4s + 6p = 26 \\ s + 9p = 29 \end{cases} \][/tex]
To solve this system, we can use either substitution or elimination. In this case, let's use the elimination method.
First, let's multiply the second equation by 4 to align the coefficients of [tex]\( s \)[/tex] in both equations:
[tex]\[ 4(s + 9p) = 4 \cdot 29 \][/tex]
[tex]\[ 4s + 36p = 116 \][/tex]
Now we have the modified system of equations:
[tex]\[ \begin{cases} 4s + 6p = 26 \\ 4s + 36p = 116 \end{cases} \][/tex]
Next, we subtract the first equation from the second equation:
[tex]\[ (4s + 36p) - (4s + 6p) = 116 - 26 \][/tex]
[tex]\[ 30p = 90 \][/tex]
Solving for [tex]\( p \)[/tex]:
[tex]\[ p = \frac{90}{30} \][/tex]
[tex]\[ p = 3 \][/tex]
Now, we substitute [tex]\( p = 3 \)[/tex] back into the second original equation to find [tex]\( s \)[/tex]:
[tex]\[ s + 9p = 29 \][/tex]
[tex]\[ s + 9(3) = 29 \][/tex]
[tex]\[ s + 27 = 29 \][/tex]
[tex]\[ s = 29 - 27 \][/tex]
[tex]\[ s = 2 \][/tex]
Thus, we have:
- Every game of skee ball requires [tex]\( 2 \)[/tex] tokens.
- Every game of pinball requires [tex]\( 3 \)[/tex] tokens.
So, filling in the blanks:
Every game of skee ball requires 2 tokens, and every game of pinball requires 3 tokens.
