Question
Members of the pep club were selling raffle tickets for $1.50 and $5. The number of $1.50 tickets sold was two less than four times the number of $5 tickets sold, and the club raised $1,152 from the ticket sales. Let x represent the number of $1.50 tickets sold and let y represent the number of $5 tickets sold. Solve the system of equations to determine how many of each ticket were sold.
1.5x + 5y = 1152
x = 4y – 2
Which one-variable linear equation can be formed using the substitution method?
How many $5 raffle tickets were sold?
Which equation can be used to determine how many $1.50 raffle tickets were sold?
How many $1.50 raffle tickets were sold?
1.5x + 5y = 1152
x = 4y – 2
Which one-variable linear equation can be formed using the substitution method?
How many $5 raffle tickets were sold?
Which equation can be used to determine how many $1.50 raffle tickets were sold?
How many $1.50 raffle tickets were sold?
Answers
GPT-4o mini
To solve the problem, we have the following system of equations:
1. \( 1.5x + 5y = 1152 \)
2. \( x = 4y - 2 \)
### 1. Which one-variable linear equation can be formed using the substitution method?
We can substitute the second equation into the first equation to form a one-variable linear equation. From the second equation, we have:
\[ x = 4y - 2 \]
Substituting this into the first equation gives:
\[ 1.5(4y - 2) + 5y = 1152 \]
### 2. How many $5 raffle tickets were sold?
Now, we can solve this equation for \( y \):
1. Distributing \( 1.5 \):
\[ 6y - 3 + 5y = 1152 \]
2. Combine like terms:
\[ 11y - 3 = 1152 \]
3. Add 3 to both sides:
\[ 11y = 1155 \]
4. Divide by 11:
\[ y = 105 \]
So, **105** $5 raffle tickets were sold.
### 3. Which equation can be used to determine how many $1.50 raffle tickets were sold?
We can use the second equation \( x = 4y - 2 \) to determine how many $1.50 raffle tickets were sold since we have found \( y \).
### 4. How many $1.50 raffle tickets were sold?
Now, substituting \( y = 105 \) into the second equation:
\[ x = 4(105) - 2 \]
Calculating this:
\[ x = 420 - 2 \]
\[ x = 418 \]
So, **418** $1.50 raffle tickets were sold.
### Summary:
- Number of $5 raffle tickets sold: **105**
- Number of $1.50 raffle tickets sold: **418**
1. \( 1.5x + 5y = 1152 \)
2. \( x = 4y - 2 \)
### 1. Which one-variable linear equation can be formed using the substitution method?
We can substitute the second equation into the first equation to form a one-variable linear equation. From the second equation, we have:
\[ x = 4y - 2 \]
Substituting this into the first equation gives:
\[ 1.5(4y - 2) + 5y = 1152 \]
### 2. How many $5 raffle tickets were sold?
Now, we can solve this equation for \( y \):
1. Distributing \( 1.5 \):
\[ 6y - 3 + 5y = 1152 \]
2. Combine like terms:
\[ 11y - 3 = 1152 \]
3. Add 3 to both sides:
\[ 11y = 1155 \]
4. Divide by 11:
\[ y = 105 \]
So, **105** $5 raffle tickets were sold.
### 3. Which equation can be used to determine how many $1.50 raffle tickets were sold?
We can use the second equation \( x = 4y - 2 \) to determine how many $1.50 raffle tickets were sold since we have found \( y \).
### 4. How many $1.50 raffle tickets were sold?
Now, substituting \( y = 105 \) into the second equation:
\[ x = 4(105) - 2 \]
Calculating this:
\[ x = 420 - 2 \]
\[ x = 418 \]
So, **418** $1.50 raffle tickets were sold.
### Summary:
- Number of $5 raffle tickets sold: **105**
- Number of $1.50 raffle tickets sold: **418**