Question

Sam invested $1800, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $24 more than twice the income from the 6% investment. How much did he invest at each rate?

Answer 1

Sam invested $1200 at 8% and $600 at 6%

Explanation:

A linear system with two equations can be modeled from the information given. Let 'x' be the amount invested at 8% and 'y' the amount invested at 6%

The sum of both amounts invested at an 8% rate and 6% rate equals 1800

`x+y = 1800`

The yearly income on the 8% investment was $24 more than twice the income from the 6%

`0.08x = 2*(0.06)y + 24`

Solving the linear system:

`0.08x = 2*(0.06)y + 24\\x = 1.5y +300\\\\(1.5y +300)+y = 1800\\2.5y=600\\y=550.75\\x=1.5*(600) +300\\x=1200`

Sam invested $1200 at 8% and $600 at 6%