Systems of equations word problem

Question

Steve had invested $12,000, part at 6% annual interest and the rest at 12% annual interest. If his total income from both investments is 1,140 per year, how much is invested at 6%?

Answer 1

Let a = the amount invested at 6%.

Let b = the amount invested at 12%.

We know that a + b = 12000 (eq1)

The interest income for a is I=prt or 0.06a.

The interest income for b is I=prt or 0.12b.

So 0.06a + 0.12b = 1140 (eq2)

 

Let's solve eq1 for a so we can use that in eq2.

a + b = 120000

a = 12000 - b

 

Substituting into eq2:

0.06(12000 - b) + 0.12b = 1140

720 - 0.06b + 0.12b = 1140

720 + 0.06b = 1140

0.06b = 1140 - 720

0.06b = 420

b = 7000

 

Substituting back into eq1:

a + 7000 = 12000

a = 12000 - 7000 = 5000

 

There is an easier way ... instead of solving eq1 for a, we can solve eq1 for b.