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1 year ago

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A total of $7000 is invested: part at 8% and the remainder at 12%. How much is invested at each rate if the annual interest is $

580?

1 answer:

Ivahew [28]1 year ago

4 0

We invested $7000 invested in two accounts, we don't know how much exactly is in each account, therefore we will assign variables to these amounts. The amount invested in the "8%" account is "x", while the other one is "y". The sum of these two numbers must be equal to $7000, therefore:

$x+y=7000$

The sum of interest of each account should be equal to 580, therefore:

$0.08\cdot x+0.12\cdot y=580$

We need to solve the system in order to determine the values.

$\begin{gathered} \begin{cases}x+y=7000 \\ 0.08x+0.12y=580\end{cases} \\ \begin{cases}-0.12x-0.12y=-840 \\ 0.08x+0.12y=580\end{cases} \\ -0.04x=-260 \\ x=\frac{-260}{-0.04}=6500 \\ \end{gathered}$

Now we can replace the value of "x" in the first equation to determine the value of "y".

$\begin{gathered} 6500+y=7000 \\ y=7000-6500 \\ y=500 \end{gathered}$

The values are 6500 in the 8% account and 500 in the 12% account.

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Then integrate term-by-term to get

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and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

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