Conner invests $8600 in two different accounts. the first account paid 7 %, the second account paid 12 % in interest. at the end of the first year he had earned $812 in interest. how much was in each account?
$ at 7 %
$ 12 %

Part 1) The amount invested in the first account at 7% was $4,400

Part 2) The amount invested in the second account at 12% was $4,200

Step-by-step explanation:

we know that

The simple interest formula is equal to

where

I is the Final Interest Value

P is the Principal amount of money to be invested

r is the rate of interest  

t is Number of Time Periods

Let

x------> the amount invested in the first account at 7%

(8,600-x) -----> the amount invested in the second account at 12%

in this problem we have

substitute in the formula above

so

therefore

The amount invested in the first account at 7% was $4,400

The amount invested in the second account at 12% was $4,200

$4,400 at 7%

$4,200 at 12%

Step-by-step explanation:

Let the amount invested in 7% be x, so,

amount invested in 12% would be "8600 - x"

We can now write an equation and solve for x:

Thus, the amount invested in 12% is  8600 - 4400 = 4200

So,

$4,400 at 7%

$4,200 at 12%

correct answer:

a + bi

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