Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Review Exercises - Page 877: 15

Answer

The number of 20th century deaths with war, famine, and tobacco is $124,111,\text{ and }71\text{ million}.$

Work Step by Step

Let us consider that the number of deaths in the 20th century due to war, famine, and tobacco is x, y, and z millions respectively. The number of deaths combing war, famine, and tobacco is 306 million; therefore the equation becomes: $ x+y+z=306$ The difference between the number of deaths between war and famine is 13 million; thus we get: $ x-y=13$ The difference between the number of deaths between war and tobacco is 53 million; therefore, we get: $ x-z=53$ From the equation $ x-y=13$, $ y=x-13$ From the equation $ x-z=53$, $ z=x-53$ Put the value of y and z in equation $ x+y+z=306$ as given below: $\begin{align} & x+\left( x-13 \right)+\left( x-53 \right)=306 \\ & 3x-66=306 \\ & 3x=372 \\ & x=124 \end{align}$ From $ y=x-13$, $\begin{align} & y=124-13 \\ & =111 \end{align}$ From the equation $ z=x-53$ $\begin{align} & z=124-53 \\ & =71 \end{align}$ Hence, the number of deaths in the 20th century due to war, famine, and tobacco is $ x=124,y=111,z=71$ million respectively.
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