SOLUTION: Susan invests 2 times as much money at 8% as she does at 6%. If her total imterest after 1 year is $1,100. How much does she have invested at each rate? I made an equation 2x+

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Question 1053825: Susan invests 2 times as much money at 8% as she does at 6%. If her total imterest after 1 year is $1,100. How much does she have invested at each rate?
I made an equation 2x+ y =1100 and 0.08x + 0.06y= $1100 but when i solve i get a negative number so i mnow my formulas are nit correct
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39616)   (Show Source): You can put this solution on YOUR website!
Your first equation is wrong. Think why!

How much was the total investment?
How much was the interest for the one year?
These are different numbers and different quantities.

Susan invests 2 times as much money at 8% as she does at 6%.
How should this be symbolized? THIS IS A RATIO.
Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!

Susan invests 2 times as much money at 8% as she does at 6%. If her total imterest after 1 year is $1,100. How much does she have invested at each rate?
I made an equation 2x+ y =1100 and 0.08x + 0.06y= $1100 but when i solve i get a negative number so i mnow my formulas are nit correct
The correct approach is:

Let the amount invested at 6% be S

Then amount invested at 8% = 2S

Do you understand this?

Interest earned from 6% investment: .06S

Interest earned from 8% investment: .08(2S), or .16S

We then get the following TOTAL INTEREST AMOUNT equation: .06S + .16S = 1,100  

Solve this for S, the amount invested at 6%

Double or multiply value of S by 2 to get amount invested at 8%.

It's that easy....nothing complex! 

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