Answer:
[tex]y-x = 16[/tex]
Step-by-step explanation:
Given
Set 1: (9,5,y,2,x)
Set 2: (8,x,4,1,3)
Required
(y - x)
First the mean values of set 1 and set 2 has to be calculated
For set 1
[tex]Mean _1 = \frac{(9+5+y+2+x)}{5}[/tex]
Collect like terms
[tex]Mean _1 = \frac{9+5+2+y+x}{5}[/tex]
[tex]Mean _1 = \frac{16+ y+x}{5}[/tex]
For set 2
[tex]Mean _2 = \frac{(8+x+4+1+3)}{5}[/tex]
Collect like terms
[tex]Mean _2 = \frac{8+4+1+3+x}{5}[/tex]
[tex]Mean _2= \frac{16+ x}{5}[/tex]
Given that the mean of set 1 is twice the mean of set 2;
[tex]Mean_1 = 2Mean_2[/tex]
[tex]\frac{16+ y+x}{5} =2 * \frac{16+x}{5}[/tex]
Multiply both sided by 5
[tex]5 * \frac{16+ y+x}{5} = 5 * 2 * \frac{16+x}{5}[/tex]
[tex]16+ y+x = 2 * (16+x)[/tex]
Open bracket
[tex]16+ y+x = 32 + 2x[/tex]
Subtract 16 from both sides
[tex]16+ y+x- 16 = 32 + 2x - 16[/tex]
[tex]16 - 16 + y+x = 32 - 16 + 2x[/tex]
[tex]y+x = 16 + 2x[/tex]
Subtract 2x from both sides
[tex]y+x-2x = 16 + 2x-2x[/tex]
[tex]y-x = 16[/tex]
