00:01
So here we have a system of equations.
00:02
Y2 is equal to y1 plus y3.
00:07
5 minus 3y1 minus 5y2 is 0.
00:15
And 10 minus 5i2 minus 7i3 is 0.
00:27
Well, i have three equations, three unknowns.
00:29
And if i substitute, let's see here, let's make everything in terms of 2 and 3.
00:46
So this one tells me that y1 is equal to y2 minus y3.
00:52
So then i can substitute that for this right here.
00:56
I get 5 minus 3 times y2 minus y3 minus 5y2 is equal to 0, and that's equivalent to 5 minus 3y2 minus 5y2, so minus 8y2 minus 3 times minus i3 is plus 3i3 is equal to 0.
01:22
And then i also still have the 10 minus 5i2 minus 7i3 is equal to 0.
01:35
So if i add these two together, or let's multiply the first one times negative 2 and then add it to the other equation, i get negative 10 plus 10, which is 0...