O-Level Additional Math

[Skyed]

Amath is much easier to get A1 than emath. Not only cos the emath a1 is higher (90+) as compared to Amath, but Amath you keep drilling easy A1. You do enough questions already soon you realize they always test the same thing, and same way to solve.

[Jtutor]

jovially:

Jtutor:

[quote=\"Herbie\"]Hi jtutor, pls check yr pm

Hi Herbie,

As requested, I have just sent the Past Years' O Level Maths Trend Analysis to your email address.

Hope you find them useful.

Cheers!
Jtutor

Hi Jtutor,

May I have a copy too? TIA.

Cheers[/quote]Hi jovially,
I'm currently overseas and will forward you a copy once I'm back. Pls pm me your email address.
Cheers,
Jtutor

[Jtutor]

Hi jovially,


I’m back and have just forwarded you the compilation for your reference.
Hope you will find it useful for your child.

Cheers!
Jtutor

[Herbie]

xy+y+2= 0 and 2x+9y-30. Can advise how ro solve this simult eqn? Tq

[koguma]

Herbie:

xy+y+2= 0 and 2x+9y-30. Can advise how ro solve this simult eqn? Tq

Is there a typo error in the second equation \"2x+9y-30\" ???
Should it be \"2x+9y=30\" ??

If yes, from this equation, makes x the subject,
i.e. x = (30-9y) / 2

Then subsitute x = (30-9y) / 2 into \"xy+y+2= 0\".

[Herbie]

Sorry, part 2 of the qn shld be 2x+9y-3=0

[Herbie]

Sorry, part 2 of the qn shld be 2x+9y-3=0

[Herbie]

xy+y+2= 0 and 2x+9y-30. Can advise how ro solve this simult eqn? Tq

[koguma]

Herbie:

xy+y+2= 0 and 2x+9y-30. Can advise how ro solve this simult eqn? Tq

Think more you type, more you confuse on the part 2
Assume the 2 equations are
xy+y+2= 0 -- (1)
2x+9y-3=0 -- (2)

From (2), make x the subject,
2x = 3-9y
x = (3-9y) / 2 -- (3)

Put (3) into (2),
[(3-9y) / 2][y] + y +2 =0
[(3y-9y^2 ) /2 ]+y +2 = 0
3y - 9y^2 +2y +4 = 0
9y^2 - 5y - 4 =0
(9y+4) (y-1) = 0

either (9y+4) = 0, y =-(4/9)
or (y-1) = 0, y= 1

when y =-(4/9)
x = 3.5

when y =1
x = -3

note: y^2 = square of y

[Herbie]

koguma:

Herbie:

xy+y+2= 0 and 2x+9y-30. Can advise how ro solve this simult eqn? Tq

Think more you type, more you confuse on the part 2
Assume the 2 equations are
xy+y+2= 0 -- (1)
2x+9y-3=0 -- (2)

From (2), make x the subject,
2x = 3-9y
x = (3-9y) / 2 -- (3)

Put (3) into (2),
[(3-9y) / 2][y] + y +2 =0
[(3y-9y^2 ) /2 ]+y +2 = 0
3y - 9y^2 +2y +4 = 0
9y^2 - 5y - 4 =0
(9y+4) (y-1) = 0

either (9y+4) = 0, y =-(4/9)
or (y-1) = 0, y= 1

when y =-(4/9)
x = 3.5

when y =1
x = -3

note: y^2 = square of y

Thanks! Thanks!
Ds tried using y=-xy-2 into eqn 2 and couldn't solve it.