Question

Joy invests a total of $32,000 in two accounts paying 2% and 14% annual interest, respectively. How much was invested in each account if, after one year, the total interest was $2,980.00. $ was invested at 2% and $ was invested at 14%.

Answer 1

We have the following system of equations, given the information on the problem:

`\begin{gathered} x+y=32000 \\ 0.02x+0.14y=2980 \end{gathered}`

where 'x' represents the money in the account with 2% annual interest and 'y' represents the money in the account with 14%.

We can solve for x the first equation to get the following:

`x=32000-y`

doing the susbtitution on the second equation using this expression, we get:

`\begin{gathered} 0.02(32000-y)+0.14y=2980 \\ \Rightarrow640-0.02y+0.14y=2980 \\ \Rightarrow0.12y=2980-640=2340 \\ \Rightarrow y=(2340)/(0.12)=19500 \\ y=19500 \end{gathered}`

now that we know the value of y, we can use it to find the value of x:

`\begin{gathered} x=32000-19500=12500 \\ \Rightarrow x=12500 \end{gathered}`

therefore $12500 was invested at 2% and $19500 was invested at 14%