$6,000 dollars is invested in two different accounts earning 3% and 5% interest. At the end of one year, the two accounts earned $220 in interest.

How much money was invested at 3%?

$2,000
$3,000
$4,000

Question

$6,000 dollars is invested in two different accounts earning 3% and 5% interest. At the end of one year, the two accounts earned $220 in interest.

How much money was invested at 3%?

$2,000
$3,000
$4,000

hero · Answer

Hint:

x + y = 6000
.03x + .05y = 220

anonymous · Answer

thanks... your my hero

anonymous · Answer

lol xD

hero · Answer

If you can let me know what the answer is, I'll give you a medal.

anonymous · Answer

okay(: , i'm trying to set it up... 
.03x+.05y=220
then i moved y to the other side like .03x=220-.05y
then i got x=7333.3333-1.666y

anonymous · Answer

but i have a feeling i did something wrong

anonymous · Answer

am i doing it right?

hero · Answer

Not exactly

anonymous · Answer

what am i doing wrong?

hero · Answer

Multiply the second equation by 100

hero · Answer

Let me know what you get afterwards

anonymous · Answer

.03x=220-.05y  is this what you mjean by second equation?

hero · Answer

1st Equation
x + y = 6000

2nd Equation
.03x + .05y = 220

I don't understand what you're confused about here.

hero · Answer

Multiply the second equation by 100 and let me know what you get afterwards

anonymous · Answer

3x-5y=22000

anonymous · Answer

3x-y=4400

anonymous · Answer

good so far?