SOLUTION: A woman invests $8000 -Part of it at 11% and the remainder at 12% -Her yearly total interest for the two investments is $930 How much did she invest at each rate?

Question 833974: A woman invests $8000
-Part of it at 11% and the remainder at 12%
-Her yearly total interest for the two investments is $930
How much did she invest at each rate?
Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
Use the formula for simple interest
(1) I = prt
where I is the interest (only) earned on an investment (or bank deposit) at the annual rate of r (decimal, not a %) for a period of t years.
For this problem we have two investment that we will call x and y. The total interest earned on the total for one year is the sum of each or
(2) x*0.11*1 + y*0.12*1 = 930
We are also given that the total investment (x+y) is $8000 whiuch gives us a second equation
(3) x + y = 8000
We need to solve (2) and (3) as simultaneous equations. You can use substitution or elimination. Let's do both.
To use substitution, solve (3) for x and put it into (2). This gives us
(4) x = 8000 - y and (2) becomes
(5) 0.11*(8000 - y) + 0.12*y = 930 or
(6) 880 - 0.11*y + 0.12*y = 930 or
(7) 0.01*y = 930 - 880 or
(8) y = 100*50 or
(9) y = 5000
Then from (3) we get
(10) x = 3000
Using elimination (one technique used by computers), we "fix" the coefficient of one of the variables to be the same in both of the given equations. Let's work with x. My first step is to multiply (2) by 100 to get
(11) 11*x + 12*y = 93000 (Look mom, no decimals)
and multiply (3) by 11 to get
(12) 11*x + 11*y = 88000
Now subtract (12) from (11) to get
(13) 0 + y = 5000 or
(14) y = 5000 and from (3) we get
(15) x = 3000
Let's use (2) to check our answers.
Is (3000*0.11*1 + 5000*0.12*1 = 930)?
Is (330 + 600 = 930)?
Is (930 = 930)? Yes
Answer: $3000 was invested at 11% and $5000 was invested at 12%.

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