Answer:
The amount invested at 4% is $2,200 and the amount invested at 6% is $8,400
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 at 4%
(3x+1,800) -----> the amount invested at 6%
we have that
The interest earned by the account at 4% plus the interest earned by the amount at 6% must be equal to $592
so
The linear equation that represent this situation is
![0.04x+0.06(3x+1,800)=592](https://tex.z-dn.net/?f=0.04x%2B0.06%283x%2B1%2C800%29%3D592)
solve for x
![0.04x+0.18x+108=592](https://tex.z-dn.net/?f=0.04x%2B0.18x%2B108%3D592)
![0.22x=592-108\\0.22x=484\\x=\$2,200](https://tex.z-dn.net/?f=0.22x%3D592-108%5C%5C0.22x%3D484%5C%5Cx%3D%5C%242%2C200)
![3x+1,800=3(2,200)+1,800=\$8,400](https://tex.z-dn.net/?f=3x%2B1%2C800%3D3%282%2C200%29%2B1%2C800%3D%5C%248%2C400)
therefore
The amount invested at 4% is $2,200 and the amount invested at 6% is $8,400