We started with solving simple linear equations, then we talked about linear equations with all letters, with some representing constants and one as unknown to be solved.

From there, we talked about transformation of formulae, where from one given formulae, we can rewrite it to make any variable as the subject of the formulae, or, in other words, expressing this variable in terms of the rest.

For example, given 1/u + 1/v = 1/f, we can ask to express v in terms of u and f, or make u the subject, or rewrite it in the form of f = …..

This is is help students get to know better about algebra where we can generalize everything in a linear equation.

We started to talk about quadratic equation where there is one variable whose highest power is 2. For example, x^2 + 3x -5 = 0. Its general form is ax^2 + bx + c = 0, where a, b, c are constants and a <> 0.

For now, we look at a quadratic equation in factored form: (ax+p)(bx+q) = 0. Since we know for P x Q = 0, either P= 0 or Q=0, we do the same: either ax+p = 0 or bx+q = 0. So we have two solutions: x = – p/a or x = – q/b.

Next time, we will talk about quadratic factorization and solving word problem with quadratic equations.

Homework:

Page 17: 3a, 3b, 4a, 4b, 5a, 5c, 5e, 5g

Page 18: 7a, 7b, 10a, 10b, 14a, 14b

Page 19: 18b, 18h, 18i, 20 j i, 20 l i