Let x represent amount invested at 12% and y represent amount invested at 6%.
We have been given that $5400 is invested, part of it at 12% and part of it at 6%.
We can represent this information in an equation as:
[tex]x+y=5400...(1)[/tex]
[tex]y=5400-x...(1)[/tex]
We are also told that for a certain year, the total yield is $504.00. This means that amount of interest for one year was $504.00. We can represent this information in an equation as:
[tex]0.12x+0.06y=504...(2)[/tex]
Upon substituting equation (1) in equation (2), we will get:
[tex]0.12x+0.06(5400-x)=504[/tex]
[tex]0.12x+324-0.06x=504[/tex]
[tex]0.06x+324=504[/tex]
[tex]0.06x+324-324=504-324[/tex]
[tex]0.06x=180[/tex]
[tex]\frac{0.06x}{0.06}=\frac{180}{0.06}[/tex]
[tex]x=3000[/tex]
Therefore, $3000 was invested at 12%.
Upon substituting [tex]x=3000[/tex] in equation (1), we will get:
[tex]y=5400-3000=2400[/tex]
Therefore, $2400 was invested at 6%.
