t^2 + 5t = 0Originally posted by yiha093:2
t =-5t
how solve?
eh, plz explain too![]()
do explain the red partOriginally posted by eagle:t^2 + 5t = 0
t(t+5) = 0
t = 0 or t = -5
Do not cancel the t away or you will lose a solution
Regards,
Eagle
factorise out the t in the equationOriginally posted by yiha093:do explain the red part
how u reach that?![]()
okieOriginally posted by wishboy:factorise out the t in the equation
t² + 5t = 0
t² = t * t
5t = 5 * t
* = multiply
both got t so can "take it out" which means factorise out the t
so...
t² + 5t = 0
t * (t + 5) = 0
therefore,
t = 0 or t + 5 = 0
t = 0 or -5
3x-y=2x+yOriginally posted by yiha093:okie
next question
i have a rectangle
3x-y
3y+4 |----------------- |
| | 2x-3
|------------------|
2x+y
based on wad i c
3x-y=2x+y
2x-3=3y+4
(stuck le)
Find X and Y
?? any1 noes?
3x-y=2x+y ==> x=2y ---(1)Originally posted by yiha093:okie
next question
i have a rectangle
3x-y
|----------------- |
3y+4 | | 2x-3
|------------------|
2x+y
based on wad i c
3x-y=2x+y
2x-3=3y+4
(stuck le)
Find X and Y
?? any1 noes?

thx for ur rrepliesOriginally posted by gLc:im not sure how to do. But here goes.
Since..
3x-y=2x+y
2x-3=3y+4
Thus--> 2x = 3x-2y
-x=-2y
x=2y
2x=3y+4+3
2x=3y+7
2(2y)=3y+7
4y=3y+7
4y-3y=7
So, Y = 7.
3x-(7)=2x+7
3x-2x=7+7
x=14
So Y = 7, X = 14. I think..
txhthxOriginally posted by ^tamago^:
Originally posted by yiha093:thx for ur rreplies
howeever
still dun understand sum parts
how get 2x=3x-2y?
Originally posted by gLc:
im not sure how to do. But here goes.
Since..
3x-y=2x+y
2x-3=3y+4
Thus--> 2x = 3x-2y
-x=-2y
x=2y
....