Question 1203745
A man has $250,000 invested in three properties. 

One earns 12%,  Let investment in this one be $ x

second at  10% Let investment in this one be $y

and the amount invested at 8% is twice that invested at 12%.

Third one 8%.  So amount invested in this one is $ 2x
 
 $250,000 invested in three properties. 
x+y+2x =250000
3x+y = 250 000--------------------------------(1)

His annual income from the properties is $23,900
Interest equation

12%x +10%y + 8%(2x)= 23 900
0.12x+0.10y +0.16x =23900
0.28x + 0.10y - 23900----------------------------(2)

Divide (1) by 10
0.30x+0.1y =25000
0.28x + 0.10y - 23900    (from (2)  Subtract

0.02x =1100
x = 1100/0.02
x =55 000

2x = 110 000

3x+y =250 000
3*55000+y = 250 000

165 000 +y =250 000
y =85000

12%property $ 55 000
10%property $     85 000
8%property $ 110 000

Check  

12%*55000+10%*85000+8%*110000 =23,900

you can now calculate  the annual income from each property?
12%property $ 
10%property $
8%property $