you invested $22,000 in two accounts paying 4% and 9% annual interest. if the total interest earned for the year was $1180, how much was invested at each rate

Amount invested at 4% is $16,000

Amount invested at 9% is  $6,000

Step-by-step explanation:

Let one vestment be x

if total investment is $22,000 then

other investment will  $22,000 - x

simple interest earned in any year is given by

SI = p*r*t/100

where SI is the interest earned

t is the time period of investment

r is the rate of annual interest

interest on one account 4%

p = x

t = 1 year

SI = x*4*1/100 = 4x/100

interest on one account 9%

p = 22,000 - x

t = 1 year

SI = (22,000 - x)*9*1/100 = (198000 - 9x)/100

it is given that total interest earned was $1180

thus sum of SI calculated for the 9% and 4% investment will be equal to 1180

4x/100 +  (198000 - 9x)/100 = 1180

=> (4x+198000 - 9x)/100 = 1180

=> 198000 - 5x = 1180*100

=> -5x = 118,000 - 198000

=> -5x = -80,000

=> x = -80,000/-5 = 16,000

Thus,

Amount invested at 4% is $16,000

Amount invested at 9% is $(22,000 - 16,000) = $6,000

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hard

step-by-step explanation: