Investment in two accounts

Question

Wyatt invests $9700 in two different accounts. The first account paid 12 %, the second account paid 2 % in interest. At the end of the first year, he had earned $784 in interest. How much was in each account?

$_____ at 12% interest

$_____ at 2% interest

Answer 1

Investment in two accounts: Wyatt invests $9700 in two different accounts. The first account paid 12%, the second account paid 2% in interest. At the end of the first year, he had earned $784 in interest. How much was in each account? $_____ at 12% interest and $_____ at 2% interest

Detailed Solution:

First Account - Amount: $x & Interest: $0.12x Second Account - Amount: $y & Interest: $0.02y

Amount: x + y = 9700 ==> Equation 1

Interest: 0.12x + 0.02y = 784 ==> Equation 2

In TI-84 Plus, create 2 x 3 Matrix A of the above Equations, do the math: Matrix ==> Math ==> rref([A])

rref([A]) gives you the solution: x = 5900 and y = 3800

First Account - Amount: x = $5900 & Interest: $0.12(5900) = $708

Second Account - Amount: y = $3800 & Interest: $0.02(3800) = $76

Check:

Amount: x + y = 9700 ==> 5900 + 3800 = 9700 ==> 9700 = 9700

Interest: 0.12x + 0.02y = 784 ==> 0.12(5900) + 0.02(3800) = 784 ==> 784 = 784

Answer 2

Let a be amount at 12% and b be amount at 2 %

You know

a + b = 9700 (money invested)

.12a + .02b = 784 (money earned)

You can use substitution or elimination to solve the simultaneous linear equations.