contestada

A person invested ​$7700 for 1​ year, part at ​8%, part at ​9%, and the remainder at ​12%. The total annual income from these investments was ​$811. The amount of money invested at 12​% was ​$1100 more than the amounts invested at 8​% and ​9% combined. Find the amount invested at each rate.

Respuesta :

raphealnwobi

The amount invested at 8% was $1400, $1900 was invested at 9% and $4400 was invested at 12% making a total annual income of $811.

What is an equation?

An equation is an expression that shows the relationship between two or more numbers and variables.

Let x represent the amount of money invested at 8%, y represent the amount of money invested at 9% and z represent the amount of money invested at 12%

x + y + z = 7700   (1)

Total interest is:

(x * 1 * 0.08) + (y * 1 * 0.09) + (z * 1 * 0.12) = total interest

0.08x + 0.09y + 0.12z = 811    (

Also:

z = 1100 + x + y    (3)

From equation 1, 2 and 3:

x = 1400, y = 1900 and z = 4400

T

The amount invested at 8% was $1400, $1900 was invested at 9% and $4400 was invested at 12% making a total annual income of $811.

Find out more on equation at: https://brainly.com/question/2972832

#SPJ1

Otras preguntas