## Tuesday, July 13, 2010

### Summary of what we have learnt - T3 W03 (12 July - 16 July)

• When we are factorising expressions by grouping we need to take care of the signs ... for instance ... 1 + 2t - k - 2kt = (1+2t) - k(1 + 2t) ... the LEFT hand term (1 + 2t) is the same as 1(1 + 2t) ... hence, when we factorise the expression, there is a COMMON factor of (1+2t) ... leaving us with ... (1+2t)(1 - k) ... some of you have forgotten the (1) in the term (1 - k) !!!
• When we work with algebraic fractions, make sure that you FACTORISE both the NUMERATOR and DENOMINATOR !!!

## Friday, July 9, 2010

### Term 3 Week 2 - 5th to 9th July 2010

Most of the class can understand what was being taught today, but some still have doubts about the equation "undefined" .
QUESTION -
Why 2/0 is undefined when 2÷0=0?

Well, firstly, note that 2/0 is the SAME as 2÷0 & hence, it would still be UNDEFINED!
Anyway, if 2 ÷ 0 is possible, we will be able to see that
letting 2 ÷ 0 = x
0 ( 2 ÷ 0) = 0( x ) .................................................................... convince yourself that 3( x ÷ 3) = x
2 = 0
This is a CONTRADICTION and hence we can see that 2 / 0 is not POSSIBLE ...

Which means that if we have 1÷x, this would be UNDEFINED if x = 0.

QUESTION -
If we have 1÷(x+1), for what value of x would render this undefinable?

For 1÷(x+1), it will be UNDEFINABLE when (x + 1) = 0 ... that means that solving it would make x = -1 ... hence, when x = -1, the function will be undefinable ... Consider the GRAPH of the function y = 1÷(x+1) (using grapher), what do you notice at x = -1?

What about 1÷(2x-3)? Hence, what would be general concept be that would make the expression 1÷(ax+b), undefinable? Does this change if the numerator was something else?

Hence, for functions of the form y = 1÷(ax+b), the DENOMINATOR can never be equal to ZERO ... and this will happen when x = -b ÷ a ... CHECK TO SEE THAT YOU KNOW WHY???

QUESTION -
In order to solve 2x - 5 = 3, we need to find the value of x, that SATISFIES (we discussed this in class today) the equation. Hence, we will have to leave x alone on the left-hand side. That means, we need to remove the constant (-5) and the coefficient (2) ...
How can this be done? I believe that Zhi Qi has a slight issue with this.

Well, when 2x - 5 = 3, we need to REMOVE the 5 on the LEFT ... hence,
2x - 5 + 5 = 3 + 5 ... NOTE that we have ADDED 5 on both sides ... leaving us with 2x = 8 ...
removing the coefficient 2 requires a DIVISION
2x ÷ 2 = 8 ÷ 2 ... leaving us with x = 4 ...

Hence, the EQUATION is SATISFIED when x = 4

CHECKING ... 2 (4 ) - 3 = 5 when is the same as the VALUE as provided in the question!

QUESTION -
When you're trying to SOLVE equations involving algebraic FRACTIONS, what is the general principal involved for questions of this form?

The general principal is to ensure that only 1 fraction is present on both sides of the equal sign!