P2 exercise

lgx+lgy=lg3 (1)
lg(3x+y)=1 (2)

From (1): xy=3 ---> y=3/x

lg(3x+y)=1, so 3x+y=10.

3x+3/x=10
3x^2-10x+3=0
(3x-1)(x-3)=0
x=1/3, x=3
when x=1/3, y=9
when x=3, y=1

(and they worked when I checked them).

yes thats what (1) should of been but they made a mistake in the book and made (1)lgx+lgy=1000.
so the eqn would not work if that is the case.

yes thats what (1) should of been but they made a mistake in the book and made (1)lgx+lgy=1000.
so the eqn would not work if that is the case.

well that mistake wasn't made in my book.

from my book:

(1) = lg x + lg y = 1000

2cosxcos50-sinxsin50=sinxcos40+cosxsin40
2cos50-tanxsin50=tanxcos40+sin40
so 2cos50-sin40=tanx(cos40+sin50)
so tan x= (2cos50-sin40)/(cos40+sin50)
so x= tan-1 of all that

this is wrong!

2cos(x+50) = sin (x+40)
2(cosxcos50-sinxsin50) = sinxcos40 + cosxsin40
divide by cosx
2(cos50 - tanxsin50) = tanxcos40 + sin40

because cosx = sin(90 - x) => cos40 = sin50

2cos50 - 2tansxsin50 = tanxsin50 + cos50
2cos50 - cos50 = 3tanxsin50
cos50 = 3tanxsin50
tan50 = 3tanx

tanx = tan50/3
x = 21.665 and 201.665

well, It's different from what the answer say but you must forgive me. It is two in the morning and i've had too many glases of wine.

MB

2cosxcos50-2sinxsin50=sinxcos40+cosxsin40
2cos50-2tanxsin50=tanxcos40+sin40
so 2cos50-sin40=tanx(cos40+2sin50)
so tan x= (2cos50-sin40)/(cos40+2sin50)
so x= tan-1 of all that

this is wrong!

2cos(x+50) = sin (x+40)
2(cosxcos50-sinxsin50) = sinxcos40 + cosxsin40
divide by cosx
2(cos50 - tanxsin50) = tanxcos40 + sin40

because cosx = sin(90 - x) => cos40 = sin50

2cos50 - 2tansxsin50 = tanxsin50 + cos50
2cos50 - cos50 = 3tanxsin50
cos50 = 3tanxsin50
tan50 = 3tanx

tanx = tan50/3
x = 21.665 and 201.665

well, It's different from what the answer say but you must forgive me. It is two in the morning and i've had too many glases of wine.

MB

cos50 = 3tanxsin50
tan50 = 3tanx
MB

You have divided by sin50 to give tan50 - not correct as u know

oops, yes. Damn, I should have spotted that. Admittedly when I posted it I had had 2 bottles of wine in the evening (I know it'ss excessive)


MB

I asked about that question a while ago. There's a printing error in the question, and we were trying to work out what it should have said, but we couldn't. So I'm afraid you'll just have to leave that question. Normally, the methods involve getting rid of the logs and sovling simultaneously like normal.

can anyone do p2 examination style paper pg 189 heinemann book question 1?

i got iT..... PIECE OF CAKE!

lg x + lg y = lg 3

lg ( 3x + y ) = 1


Using laws of logs:

lg x + lg y = lg 3 ----------> lg xy = lg 3

so XY=3 {x= 3/y} and { Y= 3/x}

Substritue in EQ.2:

lg ( 3x+ 3/x) = 1

Since base of logs is 10, then 3x+ 3/x = 10

3x+ 3/x = 10 < Multiply by X?

3x^2 + 3 + 10X=0
Factorising:
(3x-3)(x-1) X=1

XY= 3 so Y=3

I DID IT!!!!
REDO to get the other set!

i got iT..... PIECE OF CAKE!

lg x + lg y = lg 3

1000 = 10^3
1000 does not equal lg3

MB

2cosxcos50-2sinxsin50=sinxcos40+cosxsin40
2cos50-2tanxsin50=tanxcos40+sin40
so 2cos50-sin40=tanx(cos40+2sin50)
so tan x= (2cos50-sin40)/(cos40+2sin50)
so x= tan-1 of all that

hot did u get from tan x= (2cos50-sin40)/(cos40+2sin50)
to x= tan-1 of all that and is it tanx in the end?