Let's say you have to go through three or four operations to get a final number, well, do it algebraically.

让我们说,你不得不通过3-4个操作,才能得到最终的数，好吧，用代数方法求解。

麻省理工公开课 - 固态化学导论课程节选

I mean, you may not understand it with algebra before but some of these sorts of pictures have you seen before?

我是说可能代数部分你没看懂,但你们以前见过类似的图像吧

耶鲁公开课 - 博弈论课程节选

I want to show you how easy it is to change the code. But, notice, once I've added this I actually have a problem. This is now an under-constrained problem. I have more unknowns than I have equations.

我加上这个条件后,我马上面临一个问题了,这就是现在是一个非约束性问题了,我的未知数比方程数多,你从代数中可以知道。

麻省理工公开课 - 计算机科学及编程导论课程节选

This is a matter of simple algebra, of taking this and putting it here.

本质上讲,这是简单的代数运算,把这个式子代入上面的式子

耶鲁公开课 - 基础物理课程节选

All right, again this is straightforward kind of algebraic manipulations to get here.

好，必须再提一次这已经是,直截了当的代数运算了。

麻省理工公开课 - 计算机科学及编程导论课程节选

We did something very nerdy, namely, we just played with calculus and algebra.

我们使用了些死板的,微积分和代数

耶鲁公开课 - 博弈论课程节选