Simultaneous Equations: Solving [eq2] and Plugging into [eq1]

[camino]

Homework Statement

[eq1]: (Va-Vb)/(2i) = {[Vb-(10<0)]/(4-8i)} + {(Vb)/(6i)}

[eq2]: (Va-Vb)/(2i) = (8<20) - Va

Homework Equations

Note: < is angle

The Attempt at a Solution

Solving [eq2]

(Va-Vb) = 2i[(8<20) - Va]
(Va-Vb) = (2i)(8<20) - (2i)(Va)
-Vb = (2i)(8<20) - (2i)(Va) - Va
Vb = -(2i)(8<20) + (2i)(Va) + Va
Vb = -(16<110) + (2i)(Va) + Va

Then plugging into [eq1]

[Va + (16<110) + (2i)(Va) + Va]/[2i] = {[-(16<110) + (2i)(Va) + Va - (10<0)]/[4-8i]} + {[-(16<110) + (2i)(Va) + Va]/[6i]}

[2Va + (16<110) + (2i)(Va)]/[2i] = {[-(16<110) + (2i)(Va) + Va - (10<0)]/[4-8i]} + {[-(16<110) + (2i)(Va) + Va]/[6i]}

I am stuck at this point. If I could find the LCM of the the 2 fractions on the right and add them together I would be fine, but since there are imaginary numbers I'm not sure if that's even possible. (I tried using the lcm feature on my TI-89, didn't work).

Or if there's a better way to solve this please let me know. Thanks.

 

[berkeman]

Homework Statement

[eq1]: (Va-Vb)/(2i) = {[Vb-(10<0)]/(4-8i)} + {(Vb)/(6i)}

[eq2]: (Va-Vb)/(2i) = (8<20) - Va

Homework Equations

Note: < is angle

The Attempt at a Solution

Solving [eq2]

(Va-Vb) = 2i[(8<20) - Va]
(Va-Vb) = (2i)(8<20) - (2i)(Va)
-Vb = (2i)(8<20) - (2i)(Va) - Va
Vb = -(2i)(8<20) + (2i)(Va) + Va
Vb = -(16<110) + (2i)(Va) + Va

Then plugging into [eq1]

[Va + (16<110) + (2i)(Va) + Va]/[2i] = {[-(16<110) + (2i)(Va) + Va - (10<0)]/[4-8i]} + {[-(16<110) + (2i)(Va) + Va]/[6i]}

[2Va + (16<110) + (2i)(Va)]/[2i] = {[-(16<110) + (2i)(Va) + Va - (10<0)]/[4-8i]} + {[-(16<110) + (2i)(Va) + Va]/[6i]}

I am stuck at this point. If I could find the LCM of the the 2 fractions on the right and add them together I would be fine, but since there are imaginary numbers I'm not sure if that's even possible. (I tried using the lcm feature on my TI-89, didn't work).

Or if there's a better way to solve this please let me know. Thanks.

Perhaps it can be solved in that polar format for the complex numbers, but I would find it easier to solve it in rectangular format. That is, convert each complex number from "magnitude>angle" format into A+jB format. Then to solve, the real parts have to be equal, and the imaginary parts have to be equal. You can convert it back to polar form in the end, if that's the format that the complex number answer is supposed to be in.