AM Circles

Mean, in this case, is the sum of the circles by the number of circles. Therefore $x-y = \frac{(5+49+16+y)-(3+24+23+x)}{4}$ $4(x-y) = (y-x) + 20$ $4(x-y) = -(x-y) + 20$ $5(x-y) = 20$ $(x-y) = 4$

x = (5 + 49 + y + 16)/4

y = (3 + x + 24 + 23)/4

4x = 5 + 49 + y + 16

4y = 3 + x + 24 + 23

4x - 4y = 5 + 49 + y + 16 - 3 - x - 24 - 23

5x - 5y = 20

x - y = 4

$x = \frac{49+5+16+y}{4} = \frac{70+y}{4}$ , $y= \frac{3+24+23+x}{4} = \frac{50+x}{4}$

$x= \frac{70+\frac{50+x}{4}}{4}$

$4x = 70 + \frac{50+x}{4}$

$16x = 280 + 50+x$

$15x = 330$

$x= \frac{330}{15} = 22$

$y = \frac{50+22}{4} = \frac{72}{4} =18$

$x-y = 22-18 = \boxed{4}$

4x = 70 + y

4y = 50 + x

4x - y = 70 (I) -x + 4y= 50 (II)

4×(I) + (II) = 15x =330

x = 22

4×(22) - y = 70; (I)

y = 18

x - y = 4

Begin by setting up a system of equations
$x=\frac{49+16+5+y}{4}$
$x=\frac{70+y}{4}$
$4x-y=70$
$y=\frac{24+23+3+x}{4}$
$y=\frac{50+x}{4}$
$x-4y=50$
Setup system of linear equations
$\begin{vmatrix} 4 & -1 \\ -1 & 4 \end{vmatrix} \begin{vmatrix} x \\ y \end{vmatrix} = \begin{vmatrix} 70 \\ 50 \end{vmatrix}$
Induce Cramer's rule:
$x = \frac{\begin{vmatrix} 70 & -1 \\ 50 & 4 \end{vmatrix}}{\begin{vmatrix} 4 & -1 \\ -1 & 4 \end{vmatrix}} = \frac{330}{15}=22$
$y = \frac{\begin{vmatrix} 4 & 70 \\ -1 & 50 \end{vmatrix}}{\begin{vmatrix} 4 & -1 \\ -1 & 4 \end{vmatrix}} = \frac{270}{15}=18$
$x-y =22-18 =4$

We are given $4x = y + 5 + 49 + 16$ $4y = x + 3 + 24 + 23$ Subtracting these two equations: $4(x-y) = (y-x) + (5-3) + (49-24) + (16-23)$ This results in $5(x-y) = (5-3) + (49-24) + (16-23)$ $x-y = \frac{(5-3) + (49-24) + (16-23)}{5} = 4$ The surprise in this problem is that $x-y$ is not equal to the mean of the differences.

by taking mean of the four white circles and four grey circles,compute the value of y or x and compare both the equations you will get one value by putting that value back in the first equation you will get the other value then you will find x-y.