college algebra help

Question

Shaun invests $5800 in two different accounts. The first account paid 7 %, the second account paid 9 % in interest. At the end of the first year he had earned $490 in interest. How much was in each account?

$. ______ at 7%

$ ______ at 9%.

Answer 1

7x = (58-x)9 = .490

-2x = 522-490

2x =32

x= 16, $1600 invested at 7%

58-16 = 42, $42000 invested at 9%

7(16) + 9(42) = 112+378 =490

490/5800 = 8.448% average return on both

9-8.448 = 0.552

8.448-7 = 1.448

1.448 is 2.623

the 9% amount invested is about 2.623 times the 7% amount invested

5800-x = 2.623x

3.623x = 5800

x= 5800/3.623 = about 1600

5800-1600 = 4200

9(42)+7(16) = 378 + 112 = 490= total return

Answer 2

Hello, thank you for taking the time to post your question!

The key on this type of question is finding a way to relate the variables to one another algebraically! So on this one we have

x + y = 5800    because there is $5800 total invested between the two accounts

0.07x + 0.09y = 490  since between the 7% rate and 9% rate he earns a total of $490 in interest

I would plug y = (5800 – x) into the second equation and then solve for the value of x! when I do that I get x = 1600, meaning that there is $1600 invested in the 7% first account

Which means that y = 5800 – 1600 = 4200, meaning that there is $4200 invested in the 9% second account

I hope that helps you get moving in a better direction on this type of question! Feel free to reach out if you have any additional questions beyond that :)