Explanation Given Information: The given equations are, 3 i 1 + 4 i 2 = 15 , 5 i 1 − 2 i 2 = − 1 Calculation: Consider the provided equations, 3 i 1 + 4 i 2 = 15 ...... (1) And, 5 i 1 − 2 i 2 = − 1 ...... (2) Solve the first equation for i 2 , 3 i 1 + 4 i 2 = 15 4 i 2 = 15 − 3 i 1 Or, i 2 = 15 − 3 i 1 4 ......(3) Find three solutions for this equation, i 1 i 2 = 15 − 3 i 1 4 i 2 1 15 − 3 ( 1 ) 4 = 12 4 3 − 3 15 − 3 ( − 3 ) 4 = 24 4 6 5 15 − 3 ( 5 ) 4 = 0 4 0 The solutions for the first equation in the form of ( i 1 , i 2 ) are ( 1 , 3 ) , ( − 3 , 6 ) , ( 5 , 0 ) . Plot these three points on the graph and connect them with a line as follows, Now, solve the second equation for i 2 , 5 i 1 − 2 i 2 = − 1 − 2 i 2 = − 1 − 5 i 1 2 i 2 = 1 + 5 i 1 Or, i 2 = 1 + 5 i 1 2

Bundle: Elementary Technical Mathematics, Loose-leaf Version, 12th + WebAssign Printed Access Card, Single-Term

12th Edition

ISBN: 9781337890199

Author: Dale Ewen

Publisher: Cengage Learning

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Chapter 9.1, Problem 30E

To determine

To calculate: The value of two currents if the pair of linear equations: 3i1+4i2=15, 5i1−2i2=−1, where i1 and i2 are the two currents in an electric circuit in microamperes (μA).

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Chapter 9.1, Problem 29E

Chapter 9.2, Problem 1E

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Chapter 9 Solutions

Bundle: Elementary Technical Mathematics, Loose-leaf Version, 12th + WebAssign Printed Access Card, Single-Term

Ch. 9.1 - Draw the graphs of each pair of linear equations....Ch. 9.1 - Prob. 2E Ch. 9.1 - Draw the graphs of each pair of linear equations....Ch. 9.1 - Prob. 4E Ch. 9.1 - Prob. 5E Ch. 9.1 - Prob. 6E Ch. 9.1 - Prob. 7E Ch. 9.1 - Prob. 8E Ch. 9.1 - Draw the graphs of each pair of linear equations....Ch. 9.1 - Draw the graphs of each pair of linear equations....

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The value of two currents if the pair of linear equations: 3 i 1 + 4 i 2 = 15 , 5 i 1 − 2 i 2 = − 1 , where i 1 and i 2 are the two currents in an electric circuit in microamperes ( μA ) .

Solution Summary: The author explains how to calculate the value of two currents in an electric circuit in microamperes.

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To calculate: The value of two currents if the pair of linear equations: 3i1+4i2=15, 5i1−2i2=−1, where i1 and i2 are the two currents in an electric circuit in microamperes (μA).

Want to see the full answer?

Chapter 9 Solutions