Mathematics: Suppose that there are two types of tickets to a show: advance and same-day. advance tickets cost $25 and same-day tickets cost $35. for one performance, there were 45 tickets sold in all, and the total amount paid for them was $1375 . how many tickets of each type were sold?

Suppose that there are two types of tickets to a

trujillo03

10.03.2022

For this case we propose a system of equations:

x: Variable representing the anticipated tickets

y: Variable representing the same day tickets

So:

$x + y = 45\\25x + 35y = 1375$

We clear x from the first equation:

x = 45-y

We substitute in the second equation:

$25 (45-y) + 35y = 1375\\1125-25y + 35y = 1375\\10y = 1375-1125\\10y = 250\\y = 25$

We look for the value of x:

$x = 45-25\\x = 20$

Thus, 20 of anticipated type and 25 of same day type were sold.

20 of anticipated type and 25 of same day type were sold.

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Jstylez1789

10.03.2022

20 advance tickets and 25 same-day tickets.

Step-by-step explanation:

Set up a system of equations.

Let be "a" the number of advance tickets and "s" the number of same-day tickets.

Then:

$\left \{ {{25a+35s=1375} \atop {a+s=45}} \right.$

You can use the Elimination method. Multiply the second equation by -25, then add both equations and solve for "s":

$\left \{ {{25a+35s=1,375} \atop {-25a-25s=-1,125}} \right.\\.............................\\10s=250\\\\s=\frac{250}{10}\\\\s=25$

Substitute s=25 into an original equation and solve for "a":

$a+(25)=45\\\\a=45-25\\\\a=20$

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bearbri4520

31.07.2020

3y (y is the unknown number but it represents one number only... and also 3y represents 3 multiply by y)

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