A man invested one-third of his money at 5%, half of his money at 3%, and the rest at 4.5%. If his total annual investment income was $470, how much had he invested?

Answer:[tex]\$12,000[/tex]Step-by-step explanation:Letx ----> the total amount of money to be investedx/3 ----> amount of money invested at 5%x/2 ----> amount of money invested at 3%The rest of the money is x-(x/2 + x/3)=x/6x/6 ----> amount of money invested at 4.5%we know thatThe interest earned by the amount of money at 5% plus the interest earned by the amount of money at 3% plus the interest earned by the amount of money at 4.5% must be equal to $470 so[tex](0.05)\frac{x}{3} +(0.03)\frac{x}{2}+(0.045)\frac{x}{6}=470[/tex]Solve for xMultiply by 6 both sides to remove the fraction[tex]0.10x +0.09x+0.045x=470[/tex][tex]0.10x +0.09x+0.045x=2,820[/tex][tex]0.235x=2,820[/tex][tex]x=\$12,000[/tex]