Question

A basketball team sells tickets that cost​ $10, $20,​ or, for VIP​ seats,​ $30. The team has sold 3401 tickets overall. It has sold 176

more​ $20 tickets than​ $10 tickets. The total sales are ​$66,100. How many tickets of each kind have been​ sold?

How many​ $10 tickets were​ sold?----------------

How many​ $20 tickets were​ sold?--------------------

How many​ $30 tickets were​ sold?--------------------

.

Answer 1

Let us consider

X₁: number of tickets sold for $10

X₂: number of tickets sold for $20

X₃: number of tickets sold for $30

The team has sold 3401 tickets overall that is,

X₁ + X₂ +X₃ = 3401

It has sold 176 more $20 tickets than $10 tickets.

That is, X₂ = X₁ + 176

The total sales are $66100

Total sale means number of ticket multiply with price, that is

10X₁ + 20X₂ + 30X₃ = 66100

Plug,  X₂ = X₁ + 176 in both equations that is the first becomes

Equation of total sale becomes,

Divides both side by 10

Now solve both the equations simultaneously that is and  

Multiply first equation by -3 and then add them both

Now add both the equations,

3X₁+3X₃ = 6258

-6X₁ - 3X₃ = -9675

----------------------------

3X₁ - 6X₁ = 6258 - 9675 => -3X₁ = -3417

Divide both sides by -3,

X₁ = 1139

X₂ = X₁ + 176 = 1139 + 176 = 1315

X₁ + X₂ +X₃ = 3401 => 1139 + 1315 + X₃ = 3401 => 2454 + X₃ = 3401

Subtract 2454 from both sides,

X₃ = 947

Therefore, Number of $10 tickets sold = X₁ = 1139

Number of $20 tickets sold = X₂ = 1315

and Number of $30 tickets sold = X₃ = 947