# Two equations to solve simultaneously

#### johann1301

1. The problem statement, all variables and given/known data
2 = c(1+sqrt(2)) + d(1-sqrt(2))
5 = c(1+sqrt(2))2 + d(1-sqrt(2))2

3. The attempt at a solution

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

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#### HallsofIvy

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...

#### jimkochanski

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

"Two equations to solve simultaneously"

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