A woman invests a total of $20,000 in two accounts, one paying 3.5% and the other paying 5.5% simple interest per year. Her annual interest is $820. How much did she invest at each rate?

Answer 1

1. Total investment is x + y
2. interest in first account is x times 3.5% = 0.035x
3. interest in second account is y times 5.5% = 0.055y
4. the total interest is given as $820
5. We write the system as
6. x + y = 20,000
7. 0.035x + 0.055y = 820
8. Eliminate one of the terms. Let's do x:
9. 35* (x + y = 20,000)
10. 1,000* (0.035x + 0.055y = 820)
11. The new system becomes:
12. 35x + 35y = 700,000
13. 35x + 55y = 820,000
14. Subtract and solve for y:
15. 35x + 35y = 700,000
16. - 35x - 55y = - 820,000
17. - 20y = - 80,000
18. - 20y / - 20 = - 80,000 / - 20
19. y = 4,000
20. Plug in value for y to solve for x:
21. x + (4,000) = 20,000
22. Isolate for y:
23. x + (4,000) ( - 4,000) = 20,000 ( - 4,000)
24. x = 16,000
25. Explain the result:
26. Woman invests $16,000 at the rate of 3.5% and $4,000 at the rate of 5.5%.

*Why am I multiplying the first line by 35 and the second by 1,000? Because to eliminate the term for x, we need to make the terms for x in both lines equal to each other. I used 35 as the common term. 35 times x = 35x, and 1,000 times 0.035x = 35x.

Answer 2

Here's the substitution method:

x + y = 20000

then y = 20000 - x

.035x + .055(20000 - x) = 820 etc. Not so bad.

Answer 3

x + y = 20000

.035x + .055y = 820

It is a toss up whether to use substitution or elimination; I would choose elimination

Multiply the first equation by 7 and the second equation by 200.

Then subtract to eliminate the x.