Jennifer deposits a total of $10,000 across three accounts - a checking account, a savings account, and a CD account. The annual interest rates for the accounts are 2%, 4%, and 10%, respectively. After 1 year, the total interest earned across all the accounts is $560. Also, the total amount initially invested across both the checking and savings account is $4000 greater than the amount initially invested into the CD account. How much money was invested into each account initially?

Let the initial amounts invested in the checking, savings, and CD accounts be denoted by:

- $ x $ for the checking account,
- $ y $ for the savings account,
- $ z $ for the CD account.

We are given the following information:

1. The total deposit is $10,000:
   \[
   x + y + z = 10,000 \tag{1}
   \]
2. The interest rates are 2%, 4%, and 10% for the checking, savings, and CD accounts, respectively. The total interest earned after 1 year is $560:
   \[
   0.02x + 0.04y + 0.10z = 560 \tag{2}
   \]
3. The total amount invested in the checking and savings accounts combined is $4,000 more than the amount invested in the CD account:
   \[
   x + y = z + 4,000 \tag{3}
   \]

Now, let's solve the system of equations.

### Step 1: Substitute equation (3) into equation (1)
From equation (3), we know that $ x + y = z + 4,000 $. Substituting this into equation (1):
\[
z + 4,000 + z = 10,000
\]
Simplifying:
\[
2z + 4,000 = 10,000
\]
\[
2z = 6,000
\]
\[
z = 3,000
\]

### Step 2: Substitute $ z = 3,000 $ into equation (3)
Substitute $ z = 3,000 $ into equation (3):
\[
x + y = 3,000 + 4,000 = 7,000 \tag{4}
\]

### Step 3: Substitute $ z = 3,000 $ into equation (2)
Substitute $ z = 3,000 $ into equation (2):
\[
0.02x + 0.04y + 0.10(3,000) = 560
\]
Simplifying:
\[
0.02x + 0.04y + 300 = 560
\]
\[
0.02x + 0.04y = 260 \tag{5}
\]

### Step 4: Solve the system of equations (4) and (5)
From equation (4), we know $ x + y = 7,000 $. We can express $ x $ as:
\[
x = 7,000 - y \tag{6}
\]
Substitute equation (6) into equation (5):
\[
0.02(7,000 - y) + 0.04y = 260
\]
Simplifying:
\[
0.02(7,000) - 0.02y + 0.04y = 260
\]
\[
140 - 0.02y + 0.04y = 260
\]
\[
140 + 0.02y = 260
\]
\[
0.02y = 120
\]
\[
y = \frac{120}{0.02} = 6,000
\]

### Step 5: Solve for $ x $
Now that we know $ y = 6,000 $, substitute this into equation (6):
\[
x = 7,000 - 6,000 = 1,000
\]

### Final Answer:
- $ x = 1,000 $ (initial investment in the checking account),
- $ y = 6,000 $ (initial investment in the savings account),
- $ z = 3,000 $ (initial investment in the CD account).

Would you like more details or have any questions?

### 5 Related Questions:
1. How would the solution change if the interest rates for the accounts were different?
2. What if the total interest earned was different, say $600? How would that affect the investments?
3. How can you express the total interest as a function of the initial investments and interest rates?
4. If you knew the final amounts after 1 year, how could you backtrack to the initial investments?
5. What is the significance of higher interest rates for long-term investment planning?

### Tip:
Always try to express variables in terms of each other to reduce the number of unknowns when solving systems of equations.

Find out how to solve a problem involving the distribution of $10,000 across three accounts with varying interest rates. Learn to use algebra and linear equations to determine the initial investments, given the interest rates and total interest earned after 1 year. Ideal for high school students in Grades 9-12.

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