Aperson invested $6700 for one year, part at 8%, part at 10%, and the remainder at 12%. the total annual income from these investments was $716. the amount of money invest at 12% was $300 more than the amount at 8% and 10% combined. find the amount invested at each rate.

The amount invested at 8% rate is $1,200

The amount invested at 10% rate is $2,000

The amount invested at 12% rate is $3,500

Step-by-step explanation:

step 1

Let

x-----> the amount invested at 8% rate

y-----> the amount invested at 10% rate

z-----> the amount invested at 12% rate

----> equation A

----> equation B

substitute equation A in equation B

-----> equation C

we know that

The simple interest formula is equal to

where

I is the Final Interest Value

P is the Principal amount of money to be invested

r is the rate of interest  

t is Number of Time Periods

in this problem we have

substitute in the formula above

substitute equation A

 -----> equation D

step 2

Solve the system of equations

x+y=3,200 -----> equation C

 -----> equation D

Solve the system by graphing

The solution is the point (1,200,2,000)

see the attached figure

Find the value of z

therefore

The amount invested at 8% rate is $1,200

The amount invested at 10% rate is $2,000

The amount invested at 12% rate is $3,500

2 and 2 ÷3

Step-by-step explanation:

The computation is shown below;

=

Now if we divide 3 in the 9 so it would be provide 2 times having 2 places remaining

So,

Or 2 and 2 ÷3