Question

Larry Mitchell invested $16,000 advance at 7% annual simple interest and the rest at 2% annual simple interest if his total yearly interest from both accounts was $920 find the amount invested at each rate

Answer 1

To solve this question we have to set and solve a system of equations.

Let S be the amount (in dollars) that Larry Mitchell invested in the 7% account, and T be the amount (in dollars) that he invested in the 2% account, then we can set the following system of equations:

`\begin{gathered} S+T=16000, \\ 0.07S+0.02T=920. \end{gathered}`

Solving the first equation for T we get:

`T=16000-S\text{.}`

Substituting T=16000-S in the second equation we get:

`0.07S+0.02(16000-S)=920.`

Applying the distributive property we get:

`0.07S+320-0.02S=920.`

Adding like terms we get:

`0.05S+320=920.`

Subtracting 320 from the above equation we get:

`\begin{gathered} 0.05S+320-320=920-320, \\ 0.05S=600. \end{gathered}`

Now, dividing by 0.05 we get:

`\begin{gathered} (0.05S)/(0.05)=(600)/(0.05), \\ S=12000. \end{gathered}`

Finally, substituting S=12000 in T=16000-S we get:

`\begin{gathered} T=16000-12000, \\ T=4000. \end{gathered}`

Therefore, Larry invested $12,000 in the 7% account and $4,000 in the 2% account.

Answer: Larry invested $12,000 in the 7% account and $4,000 in the 2% account.