Question

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 %

$ ___at 12 %

Answer 1

Answer:

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

$I=P(rt)$

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

$t=1\ year\\ P1=\ x\\ P2=\ 8,600-x\\ I=\ 812\\r1=0.07\\r2=0.12$

substitute in the formula above

$812=x(0.07*1)+(8,600-x)(0.12*1)$

$812=0.07x+1,032-0.12x$

$0.12x-0.07x=1,032-812$

$0.05x=220$

$x=\ 4,400$

so

$8,600-x=\ 8,600-\ 4,400=\ 4,200$

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

Answer 2

Answer:

$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:

$0.07(x)+0.12(8600-x)=812\\0.07x+1032-0.12x=812\\-0.05x=-220\\x=\frac{-220}{-0.05}=4400$

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

So,

$4,400 at 7%

$4,200 at 12%