I'll leave the algebra there, transfer it up here, q1* = q2* = a - c over 3b.

先把算式留在那好了,我在这里再抄一遍吧,q1*=q2*=/3b

耶鲁公开课 - 博弈论课程节选

Good, so if I plug q2 = 0 into here, this term disappears and I just get a - c over 2b.

把q2=0代入到算式中,这部分就没有了,只剩/2b

耶鲁公开课 - 博弈论课程节选

What q2 makes this equal to 0 and Katie's answer is solving out the algebra here is that q2 that solves this must be a - c over b.

2为何值时这个算式等于0呢,凯特回答其实就是算出这个的解,即,q2=/b

耶鲁公开课 - 博弈论课程节选

All I'm doing here, nothing particularly exciting, I'm simply plugging this expression in for P and then I multiplied the whole thing out because it was all multiplied by q.

这些计算过程确实很枯燥,我只是把这个算式代入到P中,然后打开整理,因为每一项都乘以了q

耶鲁公开课 - 博弈论课程节选