Question
Verify the third equation in Example 21.5 by substituting the values found for the currents I1I_1 and I3I_3 .
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Final Answer
Substitute for I1I_1 and I3I_3 in 6I1+2I3−45=06I_1 + 2I_3 - 45 = 0 where I1=4.75 AI_1 = 4.75 \textrm{ A} and I3=8.25 AI_3 = 8.25 \textrm{ A}.

Solution video

OpenStax College Physics for AP® Courses, Chapter 21, Problem 34 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. In this problem, we practice checking our answers in solving a circuit. So if we have some answers: here's current I 1 and current I 3 that are given to us in example [21.5], we can substitute those values into each of the equations that we have created to find these answers and the equation should be true with these numbers substituted in. Here's the third equation in the solution which is 6I 1 plus 2I 3 minus 45 equals 0 and if we substitute 4.75 amps in for I 1 and 8.25 amps for I 3 and then do this calculation we should confirm that it does indeed equal 0 and it does!