How would you go about solving:

6 = e^(1+b) + B

2 = e^b +B

where e = euler's number, not an unknown

Is it something to do with subtracting? I just keep getting myself into a dead end.

Thanks

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- June 7th 2006, 10:49 PMfreswoodsimultaneous equations
How would you go about solving:

6 = e^(1+b) + B

2 = e^b +B

where e = euler's number, not an unknown

Is it something to do with subtracting? I just keep getting myself into a dead end.

Thanks - June 7th 2006, 11:45 PMCaptainBlackQuote:

Originally Posted by**freswood**

you rewrite them as:

,

which leaves you with a pair of linear simultaneous equations in and ,

which can be solved in the usual manner.

RonL - June 8th 2006, 12:02 AMfreswood
How would you finish off what you've started?

- June 8th 2006, 12:10 AMCaptainBlackQuote:

Originally Posted by**freswood**

Subtract:

,

so:

,

and substituting back into the second equation:

.

Finally:

RonL - June 8th 2006, 02:00 AMfreswood
Thanks so much! It kinda confused me because I wasn't expecting the answer to be that messy. Whole numbers all the way! :D

- June 8th 2006, 07:39 PMSoroban
Hello, freswood!

Quote:

Solve: .

We have: .

Divide the equations: .

. . Then: .

. . And: .

. . Therefore: .

Substitute into (2): .

. . .

. . Therefore: .