$6,000 dollars is invested in two different accounts earning 3% and 5% interest. At the end of one year, the two accounts earned $220 in interest. How much money was invested at 3%? $2,000 $3,000 $4,000

money invested at 3% is $4000

Step-by-step explanation:

let the money invested in 3% account be x

then money invested in 5% account = 6000 - x

we know interest = p*r*t/100

p is the principle amount invested

here t = 1 year

interest on 3% money

interest= x*3*1/100 = 3x/100

interest on 5% money

interest= (6000-x)*5*1/100

Total interest earned on both the account

3x/100 + 5(6000-x)/100 = (3x + 30,000 - 5x)/100 =( 30000-2x)/100

given that total interest earned = $220

thus

( 30000-2x)/100 = 220

=> 30000-2x = 220*100= 22,000

=> -2x = 22,000- 30,000

=> x = -8000/-2 = 4000

Thus, money invested at 3% = x = $4000

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step-by-step explanation:

exa:

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3+11

14

step-by-step explanation:

from this example, we can see that subtracting with integers is the same as adding the opposite. we can solve using a number line or by using the saying "keep, change, change" to us solve. it is important to remember that we can subtract a negative and end up with a larger number than we started with.