# A total of \$12,000 was invested in two stocks paying 6% and 8% annual interest, respectively. At the end of the year, the total interest earned from the investments was \$890. How much was invested at each rate? \$3,500 was invested at 6% annual interest, while \$8,500 was invested at 8% annual interest.

Let a be the amount invested at 6% and b be the amount invested at 8%. The total amount invested is \$12,000, and the total interest earned is \$890.

.06a+.08b=890. This is the total interest earned in 1 year. 6% of a (.06a) and 8% of b (.08b) where the total interest is \$890. Multiplying both sides of this equation by 50 yields 3a+4b=44,500.

a+b=12,000 This is the total amount invested.

We now have two linear equations with two unknowns. There are a number of ways to solve this (graphing, substitution, guess and check), but we can use linear combinations (also known as the multiplication and addition method or the elimination method).

3a+4b=44,500
a+b=12,000. Multiply both sides by 3.

3a+4b=44,500
3a+3b=36,000. Subtract.

b=8,500

Then a=12,000-8,500=3,500

Check: a+b=3,500+8,500=12,000

.06(3500)+.08(8500)=890