x + y + z = 1800 x = 2(y + z) 20x + 35y + 50z = 48,000 where x, y, and z are the numbers of $20 tickets, $35 tickets, and $50 tickets, respectively.
x + y + z = 1800 x = 2(y + z) 20x + 35y + 50z = 48,000 where x, y, and z are the numbers of $20 tickets, $35 tickets, and $50 tickets, respectively.
Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter1: Introduction To Algebra
Section1.5: Translating Sentences Into Equations
Problem 36WE
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Question
Ticket Pricing A theater owner wants to divide an
1800-seat theater into three sections, with tickets costing $20, $35, and $50, depending on the section. He
wants to have twice as many $20 tickets as the sum of
the other kinds of tickets, and he wants to earn $48,000
from a full house. To find how many seats he should
have in each section, solve the system of equations
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