George and a total of $592 in simplest interest from two separate accounts. In an account earning 6% interest, George invested $1800 more than three times the amount he invested in an account earning 4%. How much did he invests in each account

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calculista calculista · Answer 1 · 2020-03-30T04:43:09+02:00

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

[tex]I=P(rt)[/tex]

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

[tex]0.04x+0.06(3x+1,800)=592[/tex]

solve for x

[tex]0.04x+0.18x+108=592[/tex]

[tex]0.22x=592-108\\0.22x=484\\x=\$2,200[/tex]

[tex]3x+1,800=3(2,200)+1,800=\$8,400[/tex]

therefore

The amount invested at 4% is $2,200 and the amount invested at 6% is $8,400