Question

Confused beyond confused

Answer 1

So this does sound quite confusing at first, but let's try to pull the numbers out of this word problem first.

There is a maximum of 1,000 tickets.
The tickets are $8 before and $10 on the day of the show.
They need to sell at least $7,500 worth of tickets.

A) We want to write a system of inequalities, so we translate this into mathy language.

Since the maximum is 1,000, there are no more than 1,000 tickets. Another way of writing this is $T \leq 1000$.

The total price of the tickets must be at least $7,500, (or no less than $7,500). We can write this as $C \geq 7500$.

Now, the key is writing this with the prices of the tickets we are given. Forget about the variables we just named.

Let's say the number of tickets purchased before the show is x, and the number on the day-of is y.

The total number of these tickets can't be more than 1,000, right?

So we can write $x+y \leq 1000$.

If you sell x tickets, since you make $8 for every ticket, you make 8x dollars. If you sell y tickets, since you make $10 for every ticket, you make 10y dollars.

And we need at least $7,500, so we can write $8x+10y \geq 7500$.

These two inequalities model the situation: $x+y \leq 1000$ and $8x+10y \geq 7500$. Don't forget to state that x is the number of tickets sold before the show, and y is the number on the day-of.

B) The club sold 550 tickets before the show, meaning x is 550. Remember, each ticket purchased then costs $8, so thus far, the club made $550⋅8 = $4,400. 

They can sell up to 1,000 – 550 tickets today, so 450 more tickets. Each ticket sold today costs $10, so they can make $450⋅10 = $4,500.

They need to make at least $7,500. They have already made $4,400, and they can make up to $4,500 more, so they can make up to $8,900, which is more than $7,500. So yes, they can meet their goal.