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# P2 Logs watch

1. Solve the simultaneous equations:

lgx + lgy = 1000
lg(3x+y) = 1

Ive got the answer of x=1/3 and 3 but the way Ive done is totally wrong, anyone care to explain. I think the book has cocked up on this one somehow..
2. (Original post by imasillynarb)
Solve the simultaneous equations:

lgx + lgy = 1000
lg(3x+y) = 1

Ive got the answer of x=1/3 and 3 but the way Ive done is totally wrong, anyone care to explain. I think the book has cocked up on this one somehow..
i think the actual question should be:

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

therefore using rules of logs you get lgxy = lg3 and thus xy = 3. For the other one you get 10^1 = 3x+y. sub in y=3/x into equation 2 and you get 3x^2-10x+3=0, which factorises to (3x-1)(x-3), i.e. x= 1/3 and 3. Find y using either equation (y= 9 and 1 respectively).
3. (Original post by boygenious)
i think the actual question should be:

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

therefore using rules of logs you get lgxy = lg3 and thus xy = 3. For the other one you get 10^1 = 3x+y. sub in y=3/x into equation 2 and you get 3x^2-10x+3=0, which factorises to (3x-1)(x-3), i.e. x= 1/3 and 3. Find y using either equation (y= 9 and 1 respectively).
Yeh exactly, thats how I got the right answer..
4. This question is flawed. It is from the exam paper at the back of P2 edexcel right? As correctly pointed out you need to edit the question with the 1000 replaced with lg 3. What is the point in setting an impossible question? Bloody edexcel.

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