SOLUTION: I have to a system of two equations in two unknowns that I have to solve:
1 = C + D
1 = ((1+sqrt(5))/2)*C + ((1-sqrt(5))/2)*D
Hope you can help... :)
Question 892535: I have to a system of two equations in two unknowns that I have to solve:
1 = C + D
1 = ((1+sqrt(5))/2)*C + ((1-sqrt(5))/2)*D
Hope you can help... :) Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! it's a little complicated in the execution, but you follow the same rules for solving systems of equations as you would for a more normal looking problem.
your solutions are:
d = (5-sqrt(5))/10
c = (5+sqrt(5))/10
the execution could get messy so i assigned a = (1+sqrt(5))/2 and i assigned b = (1-sqrt(5))/2
replacing them in the original equation, then the original equation becomes:
1 = c + d
1 = ac + bd
this is a lot easier on the eyes, at least for a while.
using the first equation, i solved for c to get c = 1 - d
i then used 1-d to replace c in the second equation to get:
1 = a(1-d) + bd
simplifying this equation got me:
1 = a - ad + bd
i factored out the d to get:
1 = a - d(a-b)
i subtracted 1 from both sides of this euation and i added d(a-b) to both sides of this equaiton to get:
d(a-b) = a-1
i then divided both sides of this equation by (a-b) to get:
d = (a-1) / (a-b)
this is where i substituted (1+sqrt(5)/2 for a and (1-sqrt(5) for b to get me back to my original expressions.
the equation became.
d = (1+sqrt(5)/2 - 1) / ((1+sqrt(5)/2) + (1-sqrt(5)/2))
the rest is just brute calculations which i will show you in the picture at the bottom.
the end of it all was that i got c = (5+sqrt(5)/10 and i got d = (5-sqrt(5))/10
i verified that these solutions were good by checking the original equation using my calculator and replacing c and d with their values.
all is good as far as i can tell.
once i found d, finding c was easy because c + d = 1 and i could solve for c to get c = 1 - d.
