117k views
3 votes
Samantha invests a total of $25,000 in two accounts paying 4% and 11% annual interest, respectively. How much was invested in each account if, after one year, the total interest was $2,120.00.

User Jun Hsieh
by
5.5k points

1 Answer

4 votes

Answer:

$9000 was invested into the first account for 4% interest.

$16,000 was invested into the second account for 11% interest.

Step-by-step explanation:

Let "a" be the money invested in the first account , 4%

let "b" be the money invested in the second account , 11%

The equation for the total interest is:

0.04a + 0.11b = 2120

I converted 4% and 11% into decimal form by dividing by 100.

The equation for total investment is:

a + b = 25,000

Solve the system between these two equations:

Rearrange a + b = 25000 to isolate one of the variables. I will isolate for "b".

a + b = 25000

b = 25000 - a

Substitute b = 25000 - a into the other equation

0.04a + 0.11b = 2120

0.04a + 2750 - 0.11a = 2120 Use distributive property over brackets

2750 - 0.07a = 2120 Combine the like terms with variable "a"

-0.07a = -630 Subtract 2750 from both sides

a = -630/-0.07 Divide both sides by -0.07

a = 9000

Substitute a = 9000 into an equation to find "b" . Choose a simpler equation

b = 25000 - a

b = 25000 - 9000

b = 16000

Therefore $9000 was invested into the account paying 4% interest. $16,000 was invested into the account paying 11% interest.

User Bgraham
by
5.6k points