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Savatey [412]
11 months ago
13

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?
Mathematics
1 answer:
Ivahew [28]11 months 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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Diano4ka-milaya [45]

Answer:

B

Step-by-step explanation:

Starting with 7,000, after 0 years there will be no increase so you still have 7,000.

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After 1 year: 7350

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Add that to previous population.

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At this point so far the yearly populations have been (7000, 7350, 7718)

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Step-by-step explanation:

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