Investment/intrest

Question

Larry Mitchell invested part of his $11,000 advance at 3% annual simple interest and the rest at 9% annual simple interest. If his total yearly interest from both accounts was $750​,

find the amount invested at each rate

The amount invested at 3% is $

​

The amount invested at 9% is $

Answer 1

Let x = $ invested at 3%

y = $ invested at 9%.

x + y = 11000 (total of $)

.03x = .09y = 750 (total of interest, I = prt)

Multiply 2nd equation by 100 (to clear the decimals--my preference)

3x+9y=75000

Divide each term of that equation by 3 (to make the numbers smaller--my preference)

x+3y = 25000

x+y = 11000

Subtract x+y = 11000 from the x + 3y = 25000:

2y = 14000

y = 7000 (divide both sides by 2)

x = 4000 (because x + y = 11000)

Check:

4000 + 7000 = $11000

.03(4000) + .09(7000) = 120 + 630 = $750