Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 1 - Systems of Linear Equations - 1.1 Introduction to Systems of Linear Equations - 1.1 Exercises - Page 11: 66


The airspeed of the first plane is $v_1 = 880$ kilometers per hour. The airspeed of the second plane is $v_2 = 960$ kilometers per hour.

Work Step by Step

We will solve this problem by writing the appropriate system of equations and then find its solution. Let $v_1$ be the speed of the first plane and $v_2$ the speed of the second plane in kilometers per hour. Note that the time of flight of the first plane is $1.5$ hours since it starts flying half an hour latter. Since $v_2$ is greater that $v_1$ by $80$ we will write our first equation: $$v_2-v_1 = 80.$$ Since the planes travel in opposing direction, the distance between them will be the sum of the distance each plane has flown away from the airport. These individual distances are found as multiplying the flight time and velocity so we have for our second equation $$2v_1+1.5v_2=3200$$ Our system reads now \begin{align*} -v_1 + v_2 = &80\\ 2v_1+1.5v_2=&3200 \end{align*} Let us now solve this system. Step 1: Express $v_2$ from the first equation in terms of $v_1$: $$v_2=v_1+80.$$ Step 2: Put this into the second equation and solve for $v_1$: $$2v_1+1.5(v_1+80) = 3200$$ which gives $$2v_1+1.5v_1+120 = 3200$$ and that simplifies to $$3.5v_1=3080.$$ Now we find $$v_1=\frac{3080}{3.5} = 880 \text{ kilometers per hour}$$ and this is the airspeed of the first plane. Step 3: Now find $v_2$ using this calculated value: $$v_2= v_1 + 80 = 960 \text{ kilometers per hour}$$ and this is the airspeed of the second plane.
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