How do you solve these investment problems?

Question

Olivia invested $7800 in two accounts, one earning 4% simple interest and the other earning 5% simple interest. If the total interest earned in one year is $350, how much was invested in each account?

Tonya invested $3500 more in account A than account B. Account A earned 3% simple interest and account B earned 2% simple interest. If the total interest in one year is $255, how much was invested in each account?

Answer 1

The working equation for these is

INTEREST EARNED = SUM (amount in acct) x (interest %)

With Olivia, If X is invested at 4 %, the balance (7800-X) is invested at 5%.   So your equation becomes

$ 350 = 0.04 X + 0.05 (7800-X)

Tonya's problem isn't much harder.  If B is the amount in acct B, the amount in account A is B+3500.  You should be able to set it up and solve.  

Answer 2

Olivia invested $7800 in two accounts, one earning 4% simple interest and the other earning 5% simple interest. If the total interest earned in one year is $350, how much was invested in each account?

let x be the 4% account, then the 5% account is 7800 - x, that's how you get this to one unknown variable.

so 0.04x + 0.05 (7800-x) = 350

0.04x + 390 - 0.05x = 350

390 = 350 + 0.01x, 0.01x = 40, x = 4000, so the 4% account had 4000 dollars, the 5% account had 3800 dollars

--------------------------------------...

Tonya invested $3500 more in account A than account B. Account A earned 3% simple interest and account B earned 2% simple interest. If the total interest in one year is $255, how much was invested in each account?

so again, let's get it to one unknown: A = 3500 + B, so

0.03(3500 + B) + 0.02B = 255

so 105 + 0.03B +0.02B =255

0.05B = 150

B = 3000

so A = 6500