Question

Sean invested $3,000 into two accounts. One account paid 4% interest and the other paid 8% interest. He earned 5% interest on the total investment. How much money did he put in each account?

Answer 1

1st investment with 8% is $750

2nd investment with 4% is 2250

Explanation:

Here we have two equations with two unknowns. We know that amount 1, x, plus amount 2, y, is 3000. i.e.,

x + y = 3000

We also know that, when you add the interest earned on amount 1 to the interest earned on amount 2, the total interest is 5% of 3000, or 150.

i.e.

.08x + .04y = 150.

Now we use substitution so that each equation has only one variable. We learned in the first equation that x + y = 3000. When we subtract x from both sides to solve for y, we can see that y = 3000 - x. We then replace the y in the second equation with the right hand side of the one we just solved for y:

.08x + .04(3000 - x) = 150. Or:

.08x + 120 - .04x = 150

.04x = 30

Divide both sides by .04 to solve for x, and now we have

x = 750.

Now we simply plug this into our first equation and solve for y:

x + y = 3000

750 + y = 3000

y = 2250

So, $750 was placed in the account earning 8%, and $2,250 was placed in the account earning 4%.