Math Problem Statement

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?

Solution

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

  • xx for the checking account,
  • yy for the savings account,
  • zz 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+4,000x + y = z + 4,000. Substituting this into equation (1): z+4,000+z=10,000z + 4,000 + z = 10,000 Simplifying: 2z+4,000=10,0002z + 4,000 = 10,000 2z=6,0002z = 6,000 z=3,000z = 3,000

Step 2: Substitute z=3,000z = 3,000 into equation (3)

Substitute z=3,000z = 3,000 into equation (3): x + y = 3,000 + 4,000 = 7,000 \tag{4}

Step 3: Substitute z=3,000z = 3,000 into equation (2)

Substitute z=3,000z = 3,000 into equation (2): 0.02x+0.04y+0.10(3,000)=5600.02x + 0.04y + 0.10(3,000) = 560 Simplifying: 0.02x+0.04y+300=5600.02x + 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=7,000x + y = 7,000. We can express xx as: x = 7,000 - y \tag{6} Substitute equation (6) into equation (5): 0.02(7,000y)+0.04y=2600.02(7,000 - y) + 0.04y = 260 Simplifying: 0.02(7,000)0.02y+0.04y=2600.02(7,000) - 0.02y + 0.04y = 260 1400.02y+0.04y=260140 - 0.02y + 0.04y = 260 140+0.02y=260140 + 0.02y = 260 0.02y=1200.02y = 120 y=1200.02=6,000y = \frac{120}{0.02} = 6,000

Step 5: Solve for xx

Now that we know y=6,000y = 6,000, substitute this into equation (6): x=7,0006,000=1,000x = 7,000 - 6,000 = 1,000

Final Answer:

  • x=1,000x = 1,000 (initial investment in the checking account),
  • y=6,000y = 6,000 (initial investment in the savings account),
  • z=3,000z = 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.

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Math Problem Analysis

Mathematical Concepts

Algebra
Systems of Linear Equations

Formulas

Simple Interest Formula: Interest = Principal × Rate × Time
Substitution Method for Solving Linear Equations

Theorems

Basic Properties of Linear Equations

Suitable Grade Level

Grades 9-12