Two equations to solve simultaneously

  • Thread starter johann1301
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In summary, to solve the given linear equations, we can use the values of A and B to find the values of c and d. By substituting c and d into the equations, we get c = (2 + sqrt(2))/4 and d = (2 - sqrt(2))/4.
  • #1
johann1301
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1

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


[/B]
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?

The answer should be c=(sqrt(2)+2)/4
 
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  • #2
Show your working. We can't help you if we can't see where you might have gone wrong.
 
  • #3
You understand that these are just linear equations, don't you: Ac+ Bd= 2 and A^2c+ B^2d= 5 with [tex]A= 1+ \sqrt{2}[/tex] and [tex]B= 1- \sqrt{2}[/tex]. Yes, you could solve the first equation for c, [tex]c= (2- Bd)/A[/tex], 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 [tex]A= 1+\sqrt{2}[/tex] and [tex]B= 1- \sqrt{2}[/tex].
 
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  • #4
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...
 
  • #5
HallsofIvy said:
You understand that these are just linear equations, don't you: Ac+ Bd= 2 and A^2c+ B^2d= 5 with [tex]A= 1+ \sqrt{2}[/tex] and [tex]B= 1- \sqrt{2}[/tex]. Yes, you could solve the first equation for c, [tex]c= (2- Bd)/A[/tex], 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 [tex]A= 1+\sqrt{2}[/tex] and [tex]B= 1- \sqrt{2}[/tex].

Which, of course, should simplify to...

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

d = (2 - Sqrt[2])/4
 

1. What does it mean to solve two equations simultaneously?

Solving two equations simultaneously means finding the values of the variables that satisfy both equations at the same time. This allows us to find the point of intersection between two lines or the solution to a system of equations.

2. How do I solve two equations simultaneously?

To solve two equations simultaneously, you can use either the substitution method or the elimination method. In the substitution method, you solve for one variable in one equation and substitute it into the other equation. In the elimination method, you manipulate the equations to eliminate one variable and then solve for the remaining variable.

3. Can two linear equations have more than one solution?

Yes, two linear equations can have infinitely many solutions. This happens when the two equations represent the same line, and any point on that line is a solution to the system. However, if the two equations represent different lines, they will have one unique solution.

4. What if the two equations have no solution?

If the two equations have no solution, it means that the lines represented by the equations are parallel and will never intersect. This can happen when the slopes of the two lines are equal, but the y-intercepts are different.

5. How do I know if my solution to a system of equations is correct?

To check if your solution is correct, you can substitute the values of the variables into both equations and see if they satisfy both equations. If they do, then your solution is correct. You can also graph the equations and see if the point of intersection matches your solution.

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