Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 9 - Section 9.1 - Solving Pairs of Linear Equations by Graphing - Exercises - Page 327: 27

Answer

$R_{1}=10$ Ω $R_{2}=4$ Ω

Work Step by Step

To solve this system of equations, we use the graphing method. $R_{1}+R_{2}=14$ $R_{1}-R_{2}=6$ Taking the first equation, we solve for $R_{2}$. $R_{1}+R_{2}=14$ $R_{2}=14-R_{1}$ Find three solutions: For $R_{1}$=2, $R_{2}=14-R_{1}$ $R_{2}=14-2$ $R_{2}=12$ For $R_{1}$=0, $R_{2}=14-R_{1}$ $R_{2}=14-0$ $R_{2}=14$ For $R_{1}$=-2, $R_{2}=14-R_{1}$ $R_{2}=14-(-2)$ $R_{2}=14+2$ $R_{2}=16$ With the three points, $(2,12), (0,14), (-2,16)$, we can graph the straight line that goes through these points. Taking the second equation, we solve for $R_{2}$. $R_{1}-R_{2}=6$ $R_{2}=R_{1}-6$ Find three solutions: For $R_{1}$=2, $R_{2}=R_{1}-6$ $R_{2}=2-6$ $R_{2}=-4$ For $R_{1}$=0, $R_{2}=R_{1}-6$ $R_{2}=0-6$ $R_{2}=-6$ For $R_{1}$=-2, $R_{2}=R_{1}-6$ $R_{2}=-2-6$ $R_{2}=-8$ With the three points, $(2,-4), (0,-6), (-2,-8)$, we can graph the straight line that goes through these points. The intersection point between these two lines is the answer to the system of equations.
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