CPhill

Compared to Melody and heureka.....I made that WAY too complicated...!!!

But, as Melody has said, "All roads lead to Rome!!!

Very nice, heureka......the additions on the right hand side are numbers in the Fibonacci series..!!!

3-4-5

6-8-10

9-12-15

Notice that the last two are  just multiples of the first....and we could continue this pattern for as long as we wanted to.....

Let's try this again

-2y=6x+12   → y = -3x- 6    (1)

4y-2x= -2   (2)       putting (1) into (2) we have

4(-3x-6)-2x = -2   simplify

-12x -24 - 2x = -2      add 24 to both sides

-14x = 22    divide both sides by -14

x = -22/14 = -11/7

And using (1)  ...  y =  -3(-11/7) - 6 =   33/7 - 6  =  33/7 - 42/7 = -9/7

So our solution is (-11/7, - 9/7)

1/ (2*√2)    multiply top and bottom by √2  ... this gives

√2 / (2*√2*√2) =

√2 / (2 * 2) =

√2 / 4

a + 1/a = 6   multiply through by a

a^2 + 1 = 6a    rearrange

a^2  -6a + 1 = 0    and using the quadratic formula, we find that a = 3 ±√8

And  a^3 + 1/a^3  factors as

(a + 1/a)(a^2 -1 + 1/a^2)      and (a + 1/a) = 6 ...so we have

6(a^2 - 1 + 1/a^2)

If a= 3 + √8 we have

6[ (3 + √8)^2 - 1 + 1 / (3 +√8)^2 ] =

6 [ (9 + 6√8 + 8 - 1 + 1/ (9 + 6√8 + 8 ] =

6 [ 16 + 6√8 + 1/ (17 + 6√8)]    and using the conjugate, (17 - 6√8), we have

6[ 16 + 6√8 + (17  - 6√8) / (289 - 288)]=

6[16 + 6√8 + (17  - 6√8) / 1] =

6[16 + 6√8 + (17  - 6√8)] =

6[ 16 + 17] = 198    

And if a = 3 - √8   we have

6[ (3 - √8)^2 - 1 + 1 / (3 -√8)^2 ] =

6 [ (9 - 6√8 + 8 - 1 + 1/ (9 - 6√8 + 8 ] =

6 [ 16 - 6√8 + 1/ (17 - 6√8)]    and using the conjugate, (17 + 6√8) we have

6[ 16 - 6√8 + (17  + 6√8) / (289 - 288)]=

6[16 - 6√8 + (17  + 6√8) / 1] =

6[16 - 6√8 + (17  + 6√8)] =

6[ 16 + 17] = 198 

Exactly the same result !!!  

And that's the evaluation of  a^3 + 1/a^3......

x^3 = 20x + 7y   (1)

y^3 = 7x + 20y   (2)  subtract (2) from(1)

x^3 - y^3 = 13(x-y)    factor the left side

(x-y)(x^2 + xy + y^2) = 13(x-y)     divide through by (x-y)

x^2 + xy + y^2) = 13   subtract 13 from both sides

x^2 + xy +y^2 - 13 =  0     (3) ....  this is a rotated ellipse....

We need to find the intersection points of (1),(2) and (3)

Here's a graph

ab = 7

a^2b + ab^2 + a + b = 80     factor

a(ab + 1) + b(ab + 1) = 80

(a + b)(ab + 1) = 80

(a+b)(7 + 1) = 80

a + b = 80/8

a + b = 10      square both sides

a^2 + 2ab + b^2 = 100

a^2 +2(7) + b^2 = 100

a^2 + 14 + b2 = 100

a^2 + b^2 = 86

x^2 + 3x + 3/x  + 1/x^2 = 26      multiply through by x^2

x^4 + 3x^3 + 3x + 1 = 26x^2      rearrange

x^4 + 3x^3- 26x^2 + 3x + 1  = 0

This will factor as

(x^2 -4x + 1)(x^2 +7x + 1) = 0

And using the quadratic formula x = (4 ±√12)/2  = 2 ±√3

And  x = (-7±√45)/2  

But 2 +√3   is what we're looking for...so...a = 3, b = 2  and a + b = 5

1/ (√102 + √100) = (√102 - √100)/ 2

And

1/ (√104 + √102) = (√104 - √102)/ 2

So, all fractions would have the same denominators, and all the "intermediate" terms would "cancel" leaving us with

[√10000 - √100]/ 2 =

[100-10] /2 =

90/2 =

45