SOLUTION: Solve this application problem using a system of equations: 6000 is invested, part of it at 20% and part of it at 4%. For a certain year, the total yield is $800.00. How much was i

Question 1152879: Solve this application problem using a system of equations: 6000 is invested, part of it at 20% and part of it at 4%. For a certain year, the total yield is $800.00. How much was invested at 20%? How much was invested at 4%?
Found 2 solutions by ankor@dixie-net.com, MathTherapy:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
6000 is invested, part of it at 20% and part of it at 4%.
For a certain year, the total yield is $800.00.
How much was invested at 20%? How much was invested at 4%?
:
Let a = amt invested at 20%
the total amt invested is $6000, therefore
(6000-a) = amt invested at 4%
:
.20a + .04(6000-a) = 800
.20a + 240 - .04a = 800
.20a - .04a = 800 - 240
.16a = 560
a = 560/.16
a = $3500 invested at 20%
and
6000 - 3500 = $2500 invested at 4%
:
:
Check this by finding the actual amt earned from each investment
.20(3500) = 700
.04(2500) = 100
----------------
earned total: 800
Answer by MathTherapy(10557) (Show Source): You can put this solution on YOUR website!
Solve this application problem using a system of equations: 6000 is invested, part of it at 20% and part of it at 4%. For a certain year, the total yield is $800.00. How much was invested at 20%? How much was invested at 4%?

Let amount invested at 4% and 20% be F and T, respectively
Then we get: F + T = 6,000 ---- eq (i)
Also, .04F + .2T = 800 -------- eq (ii)
.2F + .2T = 1,200 ------ Multiplying eq (i) by .2 ----- eq (iii)
.16F = 400 ----- Subtracting eq (ii) from eq (iii)
F, or amount invested at 4% =
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