Question

Ann invested $61,000 into two accounts. The first account pays an annual interest rate of 5% and the second account pays an annual interest rate of 7%. After one year, Ann earned $3430 in interest. How much was invested in each account? Round your answer to the nearest cent, if necessary. [For full credit, set a system of linear equations and solve the system]

Answer 1

Answer:

2nd Account has $19,000

1st Account has $42,000

Step-by-step explanation:

First let's split the money invested.

x is money in account 1

y is money in account 2

1st equation

$61,000=x+y

5%=0.05

7%=0.07

2nd equation

0.05*x+0.07*y=$3,430

x=61,000-y

substitute into 2nd equation

0.05(61,000-y)+0.07y=3430

Solve.

3050-0.05y+0.07y=3430

0.02y=3430-3050

0.02y=380

y=380/0.02

y=$19,000

There are 2(two) ways to solve for the 1st account, but I will do it the long way.

y=61,000-x

substitute into 2nd equation

0.07(61,000-x)+0.05x=3430

Solve.

4270-0.07x+0.05x=3430

-0.02x=3430-4270

-0.02x=-840

x=-840/-0.02

x=$42,000

The short way...

x=61,000-y

x=61,000-19,000

x=$42,000