Question 144949
let x="amount @ 3%", y="amount @ 4%", and z="amount @ 6%"


"A man divides $10000 among 3 investments" __ x+y+z=10000


"His annual income from the first two investments is $80 less than his income from the third investment"
__ (3%)x+(4%)y=(6%)z-80


"his total income is $460 per year" __ (3%)x+(4%)y+(6%)z=460


substituting __ [(6%)z-80]+(6%)z=460 __ adding 80 __ (12%)z=540 __ dividing by 12% __ z=4500


substituting __ x+y+(4500)=10000 __ subtracting 4500 __ x+y=5500 __ subtracting x __ y=5500-x


substituting __ (3%)x+(4%)y=(6%)(4500)-80 __ (3%)x+(4%)y=190 __ dividing by 4% __ .75x+y=4750


substituting __ .75x+(5500-x)=4750 __ subrtacting 5500 __ -.25x=-750 __ dividing by -.25 __ x=3000


substituting __ (3000)+y=5500 __ subtracting 3000 __ y=2500