# Two equations to solve simultaneously

## Homework Statement

2 = c(1+sqrt(2)) + d(1-sqrt(2))
5 = c(1+sqrt(2))2 + d(1-sqrt(2))2

## The Attempt at a Solution

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Ive tried finding an expression for c, and replacing c in the other equation, but it gives me the wrong answer... what should i do?

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pasmith
Homework Helper
Show your working. We can't help you if we can't see where you might have gone wrong.

HallsofIvy
Homework Helper
You understand that these are just linear equations, don't you: Ac+ Bd= 2 and A^2c+ B^2d= 5 with $$A= 1+ \sqrt{2}$$ and $$B= 1- \sqrt{2}$$. Yes, you could solve the first equation for c, $$c= (2- Bd)/A$$, and put that into the second equation: A^2[(2- Bd)/A)+ B^2d= A(2- Bd)+ B^2d= 2A- ABd+B^2d= 5 so (B^2- AB)d= 5- 2A and d= (5- 2A)/(B^2- AB)

Personally, I would have multiplied the first equation by A to get A^2c+ ABd= 2A and then subtract that from the second equation: (B^2- AB)d= 5- 2A which immediately gives d= (5- 2A)/(B^2- AB). Similarly, multiply the first equation by B to get ABc+ B^2d= 2B and subtract that from the second equation to get (A^2- AB)c= 5- 2B so that c= (5- 2B)/(A^2- AB).

Now put $$A= 1+\sqrt{2}$$ and $$B= 1- \sqrt{2}$$.

johann1301
Thanks! I got the correct answer using d = (5- 2A)/(B^2- AB). Turns out I've been doing a mistake before i got to this part...

You understand that these are just linear equations, don't you: Ac+ Bd= 2 and A^2c+ B^2d= 5 with $$A= 1+ \sqrt{2}$$ and $$B= 1- \sqrt{2}$$. Yes, you could solve the first equation for c, $$c= (2- Bd)/A$$, and put that into the second equation: A^2[(2- Bd)/A)+ B^2d= A(2- Bd)+ B^2d= 2A- ABd+B^2d= 5 so (B^2- AB)d= 5- 2A and d= (5- 2A)/(B^2- AB)

Personally, I would have multiplied the first equation by A to get A^2c+ ABd= 2A and then subtract that from the second equation: (B^2- AB)d= 5- 2A which immediately gives d= (5- 2A)/(B^2- AB). Similarly, multiply the first equation by B to get ABc+ B^2d= 2B and subtract that from the second equation to get (A^2- AB)c= 5- 2B so that c= (5- 2B)/(A^2- AB).

Now put $$A= 1+\sqrt{2}$$ and $$B= 1- \sqrt{2}$$.
Which, of course, should simplify to...

c = (2 + Sqrt[2])/4

d = (2 - Sqrt[2])/4