Question

Larry Mitchell invested part of his $23,000 advance at 3% annual simple interest and the rest at 7% annual simple interest.His total yearly interest from both accounts was $810 find the amount invested at each rate

Answer 1

Answer:

$20,000 into the account earning 3% interest

$3000 into the account earning 7% interest

Step-by-step explanation:

let "a" be the account that earns 3% annual simple interest

let "b" be the account that earns 7% annual simple interest

Create two equations to represent the total investment and interest:

a + b = 23,000

0.03a + 0.07b = 810

When two or more equations have the same variables, it is called a "system" and you can solve for the variables.

Solve using the method substitution.

Rearrange one of the equation so that one variable is isolated

a + b = 23,000

a = 23,000 - b

Substitute "a" for 23,000 - b in the other equation

0.03a + 0.07b = 810

0.03(23,000 - b) + 0.07b = 810    Distribute over the bracket

690 - 0.03b + 0.07b = 810      Combine like terms (with the variable "b")

690 + 0.04b = 810       Start isolating "b" by subtracting 690 on both sides

0.04b = 120        Divide both sides by 0.04

b = 3000         Investment in 7% interest account

Substitute "b" for 3000 in the simpler equation to find "a"

a + b = 23,000

a + 3000 = 23,000     Subtract 3000 from both sides to isolate "a"

a = 20,000        Investment in 3% interest account

Therefore Larry Mitchell invested $20,000 into the account earning 3% annual simple interest and $3000 into the account earning 7% annual simple interest.