# Answer to Question #7921 in Algebra for willie turner

Question #7921

a new cruise ship has just launched 3 new ships: the pacafic paradise, the cribbean paradise, and the mediterranean paradise. the cribbean paradise had 33 more deluxe state rooms than the pacific paradise. the mediterranean paradise had 35 fewer deluxe state rooms than three times the number of deluxe state rooms on the pacific paradise. find the number of deluxe state rooms for each ship, if the total number of deluxe state rooms for the three ships is 633

Expert's answer

Let's make denotations:

cribbean paradise has x deluxe rooms

pacific paradise has y deluxe rooms

mediterranean paradise has z deluxe rooms

We know that x+y+z=633

We also know that "cribbean paradise had 33 more deluxe state rooms than the pacific paradise" => x=y+33

"the mediterranean paradise had 35 fewer deluxe state rooms than three times the number of deluxe state rooms on the pacific paradise"=> z+35=3y

Thus, we need to solve the system of equations:

x+y+z=633

x=y+33

z+35=3y => z=3y-35

whence (y+33)+y+(3y-35)=633

5y-2=633

5y=635

y=127

hence x=127+33=160

z=3y-35=381-35=346

cribbean paradise has x deluxe rooms

pacific paradise has y deluxe rooms

mediterranean paradise has z deluxe rooms

We know that x+y+z=633

We also know that "cribbean paradise had 33 more deluxe state rooms than the pacific paradise" => x=y+33

"the mediterranean paradise had 35 fewer deluxe state rooms than three times the number of deluxe state rooms on the pacific paradise"=> z+35=3y

Thus, we need to solve the system of equations:

x+y+z=633

x=y+33

z+35=3y => z=3y-35

whence (y+33)+y+(3y-35)=633

5y-2=633

5y=635

y=127

hence x=127+33=160

z=3y-35=381-35=346

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