Question

Hailey sold 20 tickets to the school play for a total of 225$.Early bird tickets were 10$ and regular priced tickets were 15$.Write and sole a system of equations that can be used to find the cost of each type of ticket.

Answer 1

Answer:

The system of equation are $\left \{ x+y=20} \atop {10x+15y=225}} \right.$

The number of early bird tickets are 15 and of regular tickets are 5.

Step-by-step explanation:

Given,

Total number of tickets = 20

Total amount = $225

Solution,

Let the number of early bird tickets be 'x'.

And also let the number of regular tickets be 'y'.

Now total number of tickets is the sum of number of early bird tickets and number of regular tickets.

So framing in equation form, we get;

Total number of tickets = number of early bird tickets + number of regular tickets

$x+y=20\ \ \ \ equation\ 1$

Again, Total amount is the sum of number of early bird tickets multiplied with price of each ticket and number of regular tickets multiplied with price of each ticket.

So framing in equation form, we get;

$10x+15y=225\ \ \ \ \ equation\ 2$

Hence the system of equation are $\left \{ x+y=20} \atop {10x+15y=225}} \right.$

Now we solve the equation by multiplying equation 1 by 10, and get;

$10(x+y)=20\times10\\\\10x+10y=200\ \ \ \ equation\ 3$

Now subtracting equation 3 from equation 2, we get;

$(10x+15y)-(10x+10y)=225-200\\\\10x+15y-10x-10y=25\\\\5y=25\\\\y=\frac{25}{5}=5$

On substituting the value of 'y' in equation 1, we get;

$x+y=20\\\\x+5=20\\\\x=20-5=15$

Hence The number of early bird tickets are 15 and of regular tickets are 5.