CPhill

sqrt (3) / 3 = (1/2) s^2 * sqrt (3) / 2       where s is the side of the triangle

1/3 = (1/4)s^2

4/3  = s^2

We have triangle APB

angle PAB = angle PBA = 30

angle APB = 120

Law of Cosines

s^2 = 2x^2 - 2x^2 * cos (120)

s^2 = 2x^2 - 2x^2 *(-1/2)

s^2 = 3x^2

4/3 = 3x^2

4/9 = x^2

2/3 = x

Multiply both sides by 3

3xy + 6x  = x - 2y + z

3xy + 5x = z - 2y

x ( 3y + 5)  = z - 2y

x = [ z -2y ] / [ 3y + 5]

Here's my attempt

The total number of ways of choosing  5 cards from  28 = C(28,5)

Out of  any of the 7 numbers choose any  1   = C(7,1)

We have 4 choices of the same  number, choose any 2 = C(4,2)

Out of the other 6 numbers, choose any 3 = C(6,3)

In  each of these choices, we can  choose any 1 of the 4 numbers = [ C(4,1)]^3 = 4^3

The probability is

[ C (7,1) * C(4,2) * C (6,3) * 4^3 ]  / C ( 28,5)  = 64 /117  ≈  54.7 %  { as Ishiaki found !!! }

40 * 6 * 75 * 12  =

40 * 6 *  [3 *25] * [ 3 *4 ]  =

40 * 6 * [ 3 * 3 ] * [ 25 * 4]  = 

6 * 40 *  9 *  100 =

54 * 4000  = 

54 * 4 * 1000  =

216 * 10^3

3 terminal 0's

Rewrite as rhe second and third equations

xz = x + z →  xz - x = z →  x(z -1) = z  →   x = z / (z -1)

yz = 2 (y+ z) → yz = 2y + 2z  →  yz - 2y = 2z →  y ( z -2)  = 2z →  y = (2z) / (z -2)

Rewrite the first equation. sub for x and y

[ z / ( z -1)]   [(2z) / (z -2)] =   z/ (z -1) + (2z) / ( z -2)

2z^2  =   z (z -2) + 2z (z -1)

2z^2 = 3z^2 - 4z

z^2 - 4z  = 0

z ( z - 4)  = 0

z = 0  (reject because it makes the second equation false )

z = 4

Rearrange as

6x^2 - 21x - 31 = 0

Using the Quad Formula, the largest x value is  [ 21 + sqrt [ 21^2 + 24 *31] ] / 12 ≈ 4.619    { as 3.141 found !!! }

a + 2b = -3  multiply through by -4 ⇒  -4a - 8b = 12   

Rearrange the second equation as  -4a + (2+m) b = k  

This system has infinite solutions when

2 + m = -8

m  = -10

And

k  =12

(x + 1) ( x -1)   =  x^2 + 0x - 1

Let the original amount that Alice had = A   Let the original amount that Bob had = B

1st scenario :  Alice gets n dollars from Bob....her new amt = A + n

Bob loses n dollars.....his new amt times 7 = Alice's new amt

A + n  = 7(B - n)  →    A + n = 7B - 7n  →  A + 8n = 7B    (1)

2nd senario : Bob gets n dollars from Alice....her new amt = A - n

Bob gains n dollars  and alice has 3 times as much as he does

A - n  =  3(B  + n)  → A - n = 3B + 3n  →  A - 4n = 3B  → 2A - 8n = 6B  (2)

Add (1) , (2)

3A = 13B

A / B  =  13  / 3 

Proof :  3A = 13B  →  A = (13/3)B....sub this into  (1)

(13/3)B + 8n  =   7B

8n =(8/3)B    multiply through by 3/8

3n = B

n= B/3

Using (2)

2A - 8(B/3) = 6B

2A = (26/3)B

A = (26/6)B 

A = (13/3)B

A / B  = 13  /  3