For the opening night at the Opera House, a total of 1000 tickets were sold. Front orchestra seats cost

[tex]200,600\text{ and }200[/tex]

Step-by-step explanation:

GIVEN: For the opening night at the Opera House, a total of [tex]1000[/tex] tickets were sold. Front orchestra seats cost [tex]\$90[/tex] apiece, rear orchestra seats cost  apiece, and front balcony seats cost [tex]\$60[/tex] apiece. The combined number of tickets sold for the front orchestra and rear orchestra exceeded twice the number of front balcony tickets sold by [tex]400[/tex]. The total receipts for the performance were [tex]\$72,000[/tex].

TO FIND: how many tickets of each type were sold.

SOLUTION:

Let the number of Front orchestra seats sold [tex]=\text{x}[/tex]

Let the number of rear orchestra seats sold [tex]=\text{y}[/tex]

Let the number of Front balcony seats sold [tex]=\text{z}[/tex]

total number of tickets sold  [tex]=1000[/tex]

[tex]\text{x}+\text{y}+\text{z}=1000[/tex]

Now,

The combined number of tickets sold for the front orchestra and rear orchestra exceeded twice the number of front balcony tickets sold by [tex]400[/tex].

[tex]\text{x}+\text{y}=2\text{z}+400[/tex]

solving both equations we get,

[tex]\text{z}=200[/tex]

[tex]\text{x}+\text{y}=800[/tex]

total revenue from Front balcony seats [tex]=200\times60[/tex]

                                                                 [tex]=\$12000[/tex]

total revenue from front and rear orchestra [tex]=\$72000-\$12000=$60000[/tex]

total revenue from front and rear orchestra [tex]=90\text{x}+70\text{y}[/tex]

[tex]90\text{x}+70\text{y}=60000[/tex]

[tex]\text{x}+\text{y}=800[/tex]

Solving both equations we get,

[tex]\text{x}=200\text{ y}=600[/tex]

Hence total front orchestra, rear orchestra and front balcony tickets sold are [tex]200,600\text{ and }200[/tex]

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