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Alborosie
3 years ago
15

In one year, Michael earned $6300 as a work study in college. He invested part of the money at 9% and the rest at 7%. If he rece

ived a total of $493 in interest at the end of the year, how much was invested at 7%? How much was invested 9%?
Mathematics
1 answer:
Nitella [24]3 years ago
3 0
Let x represent invested money at 9% and y at 7%
Now we write system:
x+y = 6300
1.09*x + 1.07* y = 6300 + 493 = 6793

1.09x + 1.09y = 6867
Here we multiplied first equation with 1.09 and now we will subtract second from first:

0.02y = 74
y = 3700
x = 6300-3700 = 2600

Michael invested 2600 at 9% and 3700 at 7%
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Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
lyudmila [28]

Answer:

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Step-by-step explanation:

We can solve this system just by summing each side of the equation:

So, both left sides will be sum, and both right sides too.

The resulting expression will be:

(3-y)+(3-y) = 6+21

Ordering and solving both sides:

6-2y=27\\-2y=27-6\\y=\frac{21}{-2}

Hence, the value to the system is -\frac{21}{2}

It's important to combine both equation, because the exercise is asking for the solution of the system.

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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