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...

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
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