SOLUTION: martha has $6000 to invest. she puts x dollars of this money into a savings account that earns 3% per year, and with the rest, she buys a certificate of deposit that earns 6% per y

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: martha has $6000 to invest. she puts x dollars of this money into a savings account that earns 3% per year, and with the rest, she buys a certificate of deposit that earns 6% per y      Log On


   

Question 181917This question is from textbook algebra 1
: martha has $6000 to invest. she puts x dollars of this money into a savings account that earns 3% per year, and with the rest, she buys a certificate of deposit that earns 6% per year. suppose at the end of one year, martha has a total of $6315. how much money did martha invest in each account? This question is from textbook algebra 1

Answer by mangopeeler07(462) About Me  (Show Source):
You can put this solution on YOUR website!
x---money invested at 3%
y---money invested at 6%

1x+1y=6000----------b/c she has 6000 to invest
1.03x+1.06y=6315-------------b/c of the percent of investment and the final amount of money

solve eq. 1 for x
x=6000-y

plug that in to the second eq.
1.03(6000-y)+1.06y=6315

Distribute
6180-1.03y+1.06y=6315

Combine like terms
.03y=135

Solve for y
y=4500

x=6000-y
x=6000-4500
x=1500

Check answers:
x+y=6000
1500+4500=6000-------true
1.03(1500)+1.06(4500)=6315-----true
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Answers:
money invested at 3%----$1500
money invested at 6%----$4500