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: 30

Answer

$i_{1}=1$mA $i_{2}=3$mA

Work Step by Step

To solve this system of equations, we use the graphing method. $3i_{1}+4i_{2}=15$ $5i_{1}-2i_{2}=-1$ Taking the first equation, we solve for $i_{2}$. $3i_{1}+4i_{2}=15$ $4i_{2}=15-3i_{1}$ $i_{2}=\frac{15-3i_{1}}{4}$ Find three solutions: For $i_{1}=2$, $i_{2}=\frac{15-3i_{1}}{4}$ $i_{2}=\frac{15-3(2)}{4}$ $i_{2}=\frac{15-6}{4}$ $i_{2}=\frac{9}{4}=2.25$ For $i_{1}=0$, $i_{2}=\frac{15-3i_{1}}{4}$ $i_{2}=\frac{15-3(0)}{4}$ $i_{2}=\frac{15-0}{4}$ $i_{2}=\frac{15}{4}=3.75$ For $i_{1}=-2$, $i_{2}=\frac{15-3i_{1}}{4}$ $i_{2}=\frac{15-3(-2)}{4}$ $i_{2}=\frac{15+6}{4}$ $i_{2}=\frac{21}{4}=5.25$ With the three points, $(2,2.25), (0,3.75), (-2,5.25)$, we can graph the straight line that goes through these points. Taking the second equation, we solve for $i_{2}$. $5i_{1}-2i_{2}=-1$ $-2i_{2}=-1-5i_{1}$ $i_{2}=\frac{5i_{1}+1}{2}$ Find three solutions: For $i_{1}=2$, $i_{2}=\frac{5i_{1}+1}{2}$ $i_{2}=\frac{5(2)+1}{2}$ $i_{2}=\frac{10+1}{2}$ $i_{2}=\frac{11}{2}=5.5$ For $i_{1}=0$, $i_{2}=\frac{5i_{1}+1}{2}$ $i_{2}=\frac{5(0)+1}{2}$ $i_{2}=\frac{0+1}{2}$ $i_{2}=\frac{1}{2}=0.5$ For $i_{1}=-2$, $i_{2}=\frac{5i_{1}+1}{2}$ $i_{2}=\frac{5(-2)+1}{2}$ $i_{2}=\frac{-10+1}{2}$ $i_{2}=\frac{-9}{2}=-4.5$ With the three points, $(2,5.5), (0,0.5), (-2,-4.5)$, 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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