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?

see below

Step-by-step explanation:

1.5x + 5y = 1152    

x = 4y – 2

We can substitute the second equation into the first equation

Which one-variable linear equation can be formed using the substitution method?

1.5(4y-2) +5y = 1152  

Distribute

6y -3 +5y = 1152

Combine like terms

11y-3 = 1152

Add 3 to each side

11y-3+3 = 1152+3

11y = 1155

Divide each side by 11

11y/11 = 1155/11

y = 105

How many $5 raffle tickets were sold?

105 5 dollar tickets were sold

Now we need to find the number of 1.50 tickets

Which equation can be used to determine how many $1.50 raffle tickets were sold?

x = 4y – 2

x = 4(105) -2

   =420-2

   = 418

How many $1.50 raffle tickets were sold?

418    $1.50 tickets were sold

11y = 1155

$5 raffle tickets: 105

x = 4(105) - 2

$1.5 raffle tickets: 418

Step-by-step explanation:

1.5x + 5y = 1152    

x = 4y – 2

1.5(4y - 2) + 5y = 1152

6y - 3 + 5y = 1152

11y = 1155

y = 105

x = 4(105) - 2

x = 418

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.