Question

a basketball team sells tickets that cost $10, $20, or for VIP seats, $30. The team has sold 3241 tickets overall. It has sold 101 more $20 tickets that $10 tickets. The total sales are $59830. How many tickets of each kind have been sold?

Answer 1

x = 1,213

y = 1,314

z = 714

Explanation:

$10 = x

$20 = y

$30 = z

10x + 20y + 30z = 59,830

____________

Solving for x:

Since there 101 more $20 tickets than $10 tickets we are making the equation of:

y = x + 101

First equation (substitute the equation above for y)

x + (x+ 101) + z = 3,241

Subtract 101 from 3,241

2x + z = 3,140

Second equation (substitute y with 'x + 101 ')

10x + 20(x + 101) + 30z = 59,830

10x + 20x + 2,020 + 30z = 59,830

Subtract 2,020 from 59,830

10x + 20x + 30z = 57,810

Add 10x and 20x

30x + 30z = 57,810

Cancel out the 30's and divide 57,810 by 30

x + z = 1,927

Now solve using the first 2 equations

2x - z =3,140 = x + z = 1,927

Subtract z

2x = 3,140 = x + 1,927

Divide x

x = 3,140 = 1,927

Subtract 1,927 from 3,140

x = 1,213

____________________

Solving for y:

y = x + 101

y + 1,213 + 101

y = 1,314

____________________

Solving for z:

1,213 + 1,314 + z = 3,241 (Overall tickets sold)

2,527 + z = 3,241

Subtract 2,527 from 3,241

z = 714

____________________

Check math:

10x + 20y + 30z = 59,830

10(1,213) + 20(1,314) + 30(714) = 59,830