Theodore invests a total of 23,500 in two accounts

Question

Theodore invests a total of 23,500 in two accounts paying 5% and 9% annual interest, respectively. How much was invested in eqxh account if, after 1 year, the total interest was 1,955

Answer 1

$4000 was invested at 5%

$19500 was invested at 9%

A system of equations in two unknowns, x and y will work here

x = amount invested at 5%

y = amount invested at 9%

Equation 1 for the total amount invested

x + y = 23500

Equation 2 for the total interest earned in a year according to the rates given

.05x + .09y = 1955

I will use the Substitution Method with Equation 1 below to solve the problem

x + y = 23500

y = 23500 - x

Substitute this quantity for y in Equation 2

0.05x + 0.09y = 1955

0.05x + 0.09(23500 - x) = 1955

0.05x + 2115 - 0.09x = 1955

Combine like term

-0.04x + 2115 = 1955

Subtract 2115 from both sides of the equation

-0.04x = -160

Divide both sides by -0.04x

x = 4000

y = 23500 - x

y = 23500 - 4000 = 19500

Checking with Equation 2

0.05x + 0.09y = 1955

0.05(4000) + 0.09(19500) =1955

200 + 1755 = 1955

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