Mr. may invested $21,000, part at 8% and the rest at 6%. if the annual income for from both investments were equal find the amount invested at each rate

the amount of money invested at the rate of 8% is $9000

the amount of money invested at the rate of 6%. Is $12000

Step-by-step explanation:

Let x represent the amount of money invested at the rate of 8%.

Let y represent the amount of money invested at the rate of 6%.

Mr. May invested $21,000, part at 8% and the rest at 6%. This means that

x + y = 21000

The formula for simple interest is expressed as

I = PRT/100

Where

P represents the principal

R represents interest rate

T represents time

Considering the investment at the rate of 8%,

P = x

R = 8

T = 1

I = (x × 8 × 1)/100 = 0.08x

Considering the investment at the rate of 6%,

P = y

R = 8

T = 1

I = (y × 8 × 1)/100 = 0.06y

If the annual income from both investments were equal, it means that

0.08x =0.06y - - - - - -1

Substituting x = 21000 - y into equation 1, it becomes

0.08(21000 - y)=0.06y

1680 - 0.08y = 0.06y

0.06y + 0.08y = 1680

0.14y = 1680

y = 1680/0.14 = 12000

Substituting y = 12000 into

x = 21000 - y

x = 21000 - 12000

x = 9000

the answer is d. (1.2, -4.7) on edgn

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