solving simultaneous equations

**[nuthing]** · September 2nd 2011, 06:29 PM

Hi, I'm doing electrical systems and I have an issue with solving the simultaneous equations after nodal analysis. can someone please show me the steps, I know it's basic, but I'm pretty desperate.

(-2)((V1-Vth)/80)+((Vth-V1)/20)=0

((V1-24)/20)+((V1-Vth)/20)+((V1-4Vth)/80)=0

any help will be great thanks.

**TKHunny** · September 2nd 2011, 07:01 PM

In this case, I'm tempted to multiply the world by 80.

[-2 * (V1 - Vth)] + [4 * (Vth - V1)] = 0

[4 * (V1 - 24)] + [4 * (V1 - Vth)] + [(V1 - 4Vth)] = 0

SO much nicer. Okay, now what?

**[nuthing]** · September 3rd 2011, 03:29 AM

ok, do I multiply it out or can I take a common factor of 2 out of the first one? (I told you I really needed help :P)

**Siron** · September 3rd 2011, 03:55 AM

You can reform the first equation to:
$-2(v_1-v_{th})+4(v_{th}-v_1)=0 \Leftrightarrow -2(v_1-v_{th})=-4(v_{th}-v_1) \Leftrightarrow v_1-v_{th}=2(v_{th}-v_1)\Leftrightarrow v_1-v_{th}=2v_{th}-2v_1$
$\Leftrightarrow v_1+2v_1=2v_{th}+v_{th} \Leftrightarrow v_1=v_{th}$

So if you use this substitution in the second equation then you have a first degree equation in one variable.

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