Answer:
Step-by-step explanation:
let x be part at 8%, y be part at 10% and z be part at 15%
(1) x+y+z=6300
the amount of money invested at 15% was $100 more than the amounts invested at 8% and 10% combined. so
(2) z=x+y+100
The total annual income from these investments was $766. so
0.08x+0.1y+0.15z=766 or
(3) 8x+10y+15z=76600
substitute (2) into (1)
x+y+x+y+100=6300
(4) 2x+2y=6200
substitute (2) inot (3)
8x+10y+15(x+y+100)=76600
(5) 23x+25y=75100
solving (4) n (5)
x=1200
y=1900
z=1200+1900+100=3200
Answer:
Step-by-step explanation:
Let the part at 8% be x; at 10% be y; at 12% be z.
Total investment is $6300:
x + y + z = 6300
Total annual income is $766:
0.08x + 0.10y + 0.12z = 766
The amount of money invested at 15% was $100 more than the amounts invested at 8% and 10% combined:
z - x - y = 100
Solving the three equations with x, y and z:
x= $1200
y= $1900
z= $3200
