System of Equations Word Problem

Question

A sum of $10,000 is invested in 3 accounts that pay 3%,4%, and 5% interest.Twice as much money is invested in the account paying 5%, as in the account paying 3% and the total annual interest from all three accounts is $421.

Set up a system of equations and solve.

Answer 1

Assume: amounts are invested for 1 year at simple interest

Let x = amount invested at 3%

y = amount invested at 4%

z = amount invested at 5%

x + y + z = 10000

z = 2x

0.03x + 0.04y + 0.05z = 421

Replace z by 2x to get:

3x + y = 10000

0.13x + 0.04y = 421

Multiply the first equation by -0.04:

-0.12x - 0.04y = -400

0.13x + 0.04y = 421

Add the equations to obtain: 0.01x = 21

x = $2100

z = 2x = $4200

y = 10000 - x - y = $3700

Answer 2

10000=p(.03)+

2p(.05)+

p(.04)=int=421

a)p=2477

b)2p=4954

c)p=2477

=9908

off $92

Would of done better with imfo about c).