Respuesta :
Answer:
11,500 at 2% and $6,000 at 12%.
Step-by-step explanation:
Let x represent amount invested at 2% and y represent amount invested at 12%.
We have been given that Theodore invests a total of $17,500 in two accounts. We can represent this information in an equation as:
[tex]x+y=17500...(1)[/tex]
[tex]x=17500-y...(1)[/tex]
We are also told that he invested $17,500 in two accounts paying 2% and 12% annual interest, respectively. After one year, the total interest was $950.00.
Interest earned at 2% in one year would be [tex]0.02x[/tex].
Interest earned at 12% in one year would be [tex]0.12y[/tex]
[tex]0.02x+0.12y=950...(1)[/tex]
Upon substituting equation (1) in equation (2), we will get:
[tex]0.02(17500-y)+0.12y=950[/tex]
[tex]350-0.02y+0.12y=950[/tex]
[tex]350+0.10y=950[/tex]
[tex]350-350+0.10y=950-350[/tex]
[tex]0.10y=600[/tex]
[tex]\frac{0.10y}{0.10}=\frac{600}{0.10}[/tex]
[tex]y=6000[/tex]
Therefore, Theodore invested $6,000 at 12%.
Upon substituting [tex]y=6000[/tex] in equation (1), we will get:
[tex]x=17500-6000[/tex]
[tex]x=11500[/tex]
Therefore, Theodore invested $11,500 at 2%.
