CPhill

(2/3)x + (3/5)y =  12

- 3x    + (5/2)y =  6

Multiply the first equation through by   15    and the  second  through by 2

10x + 9y =  180

-6x  + 5y  =  12

Multiply  the first equation  through by  6   and the second through by 10

60x  + 54y  =  1080

-60x + 50y  =    120          add these

104y  =  1200

y  = 1200  /  104  =  150 / 13

And

-6x + 5(150/13)  = 12

-6x =  12 - 750/13

-6x = ( 156 - 750) / 13

x = (156 - 750) / (-6 *13)   = 99 / 13

Area  of large rectangle =  4 * 5  = 20

Area of larger right triangle at right =  (1/2) 4 * 5 =  10

Area of smaller right triangle at left  = (1/2) 3 * 4 =  6

Area of polygon   =

20  - 10  - 6   =     4

Let the point  we are looking for  be  ( x  ,  (1/sqrt (2)) (x^2 -18) )

Using the square of the distance formula

D^2  = x^2  + [ (1/sqrt (2))(x^2 -18)]^2   

D^2 =  x^2  + (1/2) ( x^2 - 18)^2        take the derivative of  this and  set to 0

D^2'  =   2x  + 2x ( x^2 - 18)  = 0

2x ( x^2 - 18 + 1)  =  0

2x ( x^2 - 17)  = 0

The second factor set = 0  is what we want

x^2  - 17   =  0

x^2  = 17

x = sqrt (17)

y = -1/sqrt (2)

The shortest distance is

sqrt [  (sqrt (17))^2  + ( -1/sqrt (2))^2 ] =  sqrt [ 17 + 1/2 ]  =  sqrt (17.5) ≈   4.183 units

sin 45 - cos 30  = 

[sqrt 2]  / 2   -  [ sqrt 3 ]  / 2  =

[ sqrt (2)  - sqrt (3) ]  /  2

a0  =  4

a3 =   6

So

a2  + 2a1  =  6     (1)

a1  + 2a0 =  a2    →  a1  + 2*4  = a2   →   a1 - a2 =  -8    (2)

Add (1)  and (2)

3a1   =  -2

a1  =  -2/3

a2  =  6 - 2a1 =  6 - (2)(-2/3)  =  22/3

a4  =  a3 + 2a2

a4  =  6 + 2(22/3)  =   62/3

a5 =  62/3 + 2(6)  =    98/3

0       1        2            3        4           5  

4     -2/3     22/3        6      62/3       98/3

Let us suppose that N tickets were sold

Then

[ .75N*20  +  .25N* 15 ]  / N  =

N [ 15  + 3.75 ] / N   =

15 + 3.75  =

$18.75  =  average  selling price

Let's suppose that such integers exist.....we can write

5 ( 5a + 3b)  =  1

(5a + 3b)  =   1  / 5

But  (5a + 3b)   must be an integer, not a fraction

So......no such integers exist

See my answer under your original post

Let the cost of the bill  =  C

Will paid (1/3)C

Mike and Sue together paid  (2/3) C

Mike paid   (2/7) (2/3)C

Sue  paid (5/7)(2/3)C 

And

What Mike paid + $6  = What Sue paid   .....so....

(2/7) (2/3)C + 6 =  (5/7)(2/3)C

(4/21)C  + 6 =  (10/21)C

(10/21)C - (4/21)C  = 6

(6/21)C  = 6

C =  6(21/6)  = $ 21

Proof

Will paid (1/3)(21)  =$7

Mike paid (2/7) (2/3) (21)  = $4

Sue paid $10

See my answer under your original post