Respuesta :
Given:
Total investment is, T (P) = $21,000.
First type of interest is, r(1) = 4%=0.04.
Second type of interest is, r(2) = 8.5%=0.085.
Annual interest amount is, T (I) = $1,515.00.
The objective is to find the amount invested in each type of interest.
Consider the amount invested in first type as x and the amount invested in second type as y.
Then the equation of total investment can be written as,
[tex]x+y=21000\ldots\ldots..(1)[/tex]Now, the equation of total interest amount can be written as,
[tex]\begin{gathered} r(1)x+r(2)y=T(I) \\ 0.04x+0.085y=1515\ldots\ldots\ldots\ldots.(2) \end{gathered}[/tex]Multiply equation (1) by 0.04.
[tex]\begin{gathered} 0.04x+0.04y=0.04(21000) \\ 0.04x+0.04y=840\ldots\ldots\ldots\ldots(3) \end{gathered}[/tex]On solving equation (2) and (3),
[tex]\begin{gathered} 0.04x+0.085y=1515 \\ \frac{0.04x+0.04y=840}{0.045y=675} \\ y=\frac{675}{0.045} \\ y=15000 \end{gathered}[/tex]Substitute the value of y in equation (1) to find x.
[tex]\begin{gathered} x+y=21000 \\ x+15000=21000 \\ x=21000-15000 \\ x=6000 \end{gathered}[/tex]Hence, the amount invested at 4% is $6000 and the amount invested at 8.5% is $15000.
