Question

$11,335

is invested, part at 9%
and the rest at 6%
. If the interest earned from the amount invested at 9%
exceeds the interest earned from the amount invested at 6%
by $865.35
, how much is invested at each rate? (Round to two decimal places if necessary.)

Answer 1

Amount invested at 9% = $6832.15; Amount invested at 6% = $4502.85

Explanation:

We'll need a system of equations to find the amounts invested at both 9% and 6% and we'll need to convert the percentages to decimals (i.e., 0.09 and 0.06):

• Let x represent interest earned from the 9% investment
• Let y represent interest earned from the 6% investment
• Let 0.09x represent the amount invested at 9%
• Let 0.06y represent the amount invested at 6%

First equation in system:

We know that the interest earned from the 9% investment is $865.35 greater than the interest earned from the 6% investment.

Thus, one equation we can use is x = y + 865.35

Second equation in system:

We also know that the amount invested at 9% plus the amount invested at 6% equals $11335.00

Thus, our second equation is 0.09x + 0.06y = 11335

Step 1: We can use substitution to solve. We must plug in x from the first equation for x in the second equation to solve for y:

0.09(y + 865.35) + 0.06y = 11335

0.09y + 77.8815 + 0.06y = 11335

0.15y + 77.88 = 11335

0.15y = 11257.12

y = 75047.46667

y = 75047.47

Step 2: Now we can plug in 75047.47 for y into any of the two equations in our system to solve for y. Because the first equation is simpler, let's use this one:

x = 75047.47 + 865.35

x = 75912.82

Step 3: We can now find the amount invested by multiplying x by 0.09 and y by 0.06:

0.09x = 0.09(75912.82) = 6832.1538 = $6832.15

0.06y = 0.06(75047.47) = 4502.8482 = $4502.85

Optional Step 4: We can check that our numbers are correct by substituting 75912.82 for x and 75047.47 for y in the two equations in our system:

Checking solutions for first equation:

75912.82 = 75047.47 + 865.35

75912.82 = 75912.82

Checking solutions for second equation:

0.09(75912.82) + 0.06(75047.47) = 11335

6832.1538 + 4502.8482 = 11335

6832.15 + 4502.85 = 11335

11335 = 11335

Answer 2

• 9% : $10,303
• 6% : $1032

Explanation:

You want the amounts invested at 9% and 6% if they total $11,335 and the interest from the 9% investment exceeds the interest from the 6% investment by $865.35.

Setup

If we let x represent the amount invested at 9%, the amount invested at 6% is (11335-x). The relation between the values of interest earned is ...

(x)·9% -(11335-x)·6% = 865.35

Solution

0.15x -680.10 = 865.35 . . . . . simplify

0.15x = 1545.45 . . . . . . . . add 680.10

x = 10303 . . . . . . . . . . . divide by 0.15

11335 -x = 1032

$10,303 was invested at 9%; $1,032 at 6%.

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Additional comment

It is convenient to use a single variable, and to let it represent the amount invested at the higher rate. If you use a system of 2 equations, you can effectively do this by substituting for the variable representing the amount invested at the lower rate. Using this approach keeps all of the numbers positive, tending to reduce errors.