How do I solve 2 simultaneous equations using KVL?

[Just_enough]

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So I'm trying to get a better understanding of KVL by looking up practice problem and I'm having some difficulties. I'm attempting to do this problem

my first step is to get an equation for I_a (32v-10i_a+8i_b = 0) an I_b (12i_b+20v-8i_a = 0), adn then solve for either b or a first and plug into the other equation, but this is where the problem occurs. on the site http://fourier.eng.hmc.edu/e84/lectures/ch2/node2.html it got a current for each mesh to be something completely different. I was wondering how they got 4 and 1 for the currents?

 

[NascentOxygen]

Have you checked to see whether their answers satisfy your equations?

 

[Just_enough]

Have you checked to see whether their answers satisfy your equations?

What do you mean? That is a pratice example so I'm assuming their answers and equations are correct and if mine doesn't match up then I'm doing something wrong

 

[NascentOxygen]

Do yours "match up" or not?

 

[Just_enough]

Do yours "match up" or not?

at first, no and the only thing that didnt match up was that i got my 20v to be the opposite sign of what they had

 

[NascentOxygen]

Your equations are identical with theirs. You can multiply everything on both sides by –1 and the equation still holds.

 

[Just_enough]

Your equations are identical with theirs. You can multiply everything on both sides by –1 and the equation still holds.

no not what I mean, on the B side, i have all the other sign matching up except the 20v so multiply by -1 won't make it correct, but that's not my question I just want to know how they got their currents for each of the meshes

 

[NascentOxygen]

If these are your equations:
32v-10i_a+8i_b = 0 and
12i_b+20v-8i_a = 0
then they are indeed identical with those in the worked example. All signs correspond.

You are asking how to solve 2 simultaneous equations?

If we use yours, which I'll write out again neatly:
32 - 10i_a + 8i_b = 0 ... (i)

12i_b + 20 - 8i_a = 0 ... (ii)

Multiply eqn (i) by 4
Multiply eqn (ii) by -5

Then add the corresponding sides of both new equations, and the resultant term for i_a has a coefficient of 0, i.e., i_a disappears from the equation, and with only one unknown in the equation you can solve for it to find the value for i_b.