Question 892535
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.


your solutions are what i showed you up top.



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