# I need an equation to solve the following problem.Al invested a part of $800 at 9% per annum and the rest at 12% per annum. After one year the total interest earned was $79.50. How much did Al...

I need an equation to solve the following problem.

Al invested a part of $800 at 9% per annum and the rest at 12% per annum. After one year the total interest earned was $79.50. How much did Al invest at each rate?

*print*Print*list*Cite

I am going to give you the equation and solve for the values invested at each interest rate.

Let the amount invested by Al at the interest rate of 9% per annum . Therefore the amount invested at the interest rate of 12% per annum is 800-x.

So the interest earned is x*.09 + (800 - x)* .12. This is given as $79.5

So the equation we get is x*.09 + (800 - x)* .12 = $79.5

=> .09x + 800*.12 - .12x = 79.5

=> (.12 - .09x) = .12*800 - 79.5

=> .03x = 16.5

=> x = 16.5 / .03 = 550

**Therefore the amount invested at the interest rate of 9% is $550 and $250 is invested at the interest rate of 12%**

The total investment = %800.

Let the two Parts of $800 be P and (800-P) dollars invested in 9% and 12% respectively.

Then the the interest at the end of the year for investment P is PNR/100 = P*1*9/100 = 9P/100.

Similarly the interest earned on the investment (800-P) dollars = (800-P)NR/100 = (800-P)*1*12/100 = 12(800-P)/100.

Therefore the sum of the interests of two investments = 9P/100+12(800-P)/100 = (9P+9600-12P)/100 = (9600-3P)/100

But the above interest is given to be $79.5.

Therefore we have the equation:

(9600 - 3P)/100 = $79.5.....(1).

We can now solve for P.

Multiply both sides of (1) by 100:

9600 - 3P = 79.5(100) = 7950.

9600 -7950 = 3P

1650 = 3P.

1650/3 = 3P/3.

550 = P.

Therefore investments are : $550 in 9% interest and $(800-550) = $250 in 12%.

Tally: 550*9/100 + 250*12/100 = 49.5 + 30 = 79.5.

The intare