Problem 193 - Elementary Algebra

Question

In the following exercises, translate to a system of equations and solve the system. Daniela invested a total of \(\$ 50,000,\) some in a certificate of deposit (CD) and the remainder in bonds. The amount invested in bonds was \(\$ 5000\) more than twice the amount she put into the CD. How much did she invest in each account?

Step-by-Step Solution

Answer

Daniela invested \(15000\) in the CD and \(35000\) in bonds.

Step 1: Define Variables

Let \(x\) represent the amount invested in the CD. Let \(y\) represent the amount invested in bonds.

Step 2: Write the Equations

Based on the problem, two relationships can be written as equations:1. The total amount invested: \(x + y = 50000\)2. The amount invested in bonds: \(y = 2x + 5000\)

Step 3: Substitute Equation 2 into Equation 1

Substitute \(y = 2x + 5000\) into \(x + y = 50000\): \(x + (2x + 5000) = 50000\)

Step 4: Solve for \(x\)

Combine like terms and solve for \(x\): \(x + 2x + 5000 = 50000\)\(3x + 5000 = 50000\)\(3x = 45000\)\(x = 15000\)

Step 5: Solve for \(y\)

Substitute \(x = 15000\) back into the equation \(y = 2x + 5000\): \(y = 2(15000) + 5000\)\(y = 30000 + 5000\)\(y = 35000\)

Step 6: State the Solution

Daniela invested \(15000\) in the CD and \(35000\) in bonds.

Key Concepts

linear equations

Linear equations are equations between two variables that result in a straight line when plotted on a graph. They follow the form: \(ax + b = c\). In this exercise, we have two amounts invested: one in a certificate of deposit (CD) and the other in bonds. These investments are represented by our variables, \(x\) and \(y\). The relationship between the total amount invested and each individual investment is expressed using two linear equations.
Example:
- Total investment equation: \(x + y = 50000\)
- Bonds investment equation: \(y = 2x + 5000\)
This set of linear equations helps us determine how much Daniela invested in each account.

substitution method

The substitution method is used to solve a system of linear equations by substituting one equation into another. Start by solving one of the equations for one variable, then substitute that expression into the remaining equation. This reduces the number of variables, making it easier to solve.
Steps:

• Identify the given equations: \(x + y = 50000\) and \(y = 2x + 5000\).
• Solve one equation for one variable.
• Substitute this expression into the second equation. This creates a single equation with one unknown.
• Solve for the unknown variable.
• Substitute back to find the other variable.

Example in our exercise:
- Solve equation \(y = 2x + 5000\) for \(y\).
- Substitute this into the first equation: \(x + (2x + 5000) = 50000\).

investment problem

An investment problem often requires figuring out how different amounts of money are distributed in various financial accounts. It's common to set up a system of equations to represent each part of the total investment. In this scenario, Daniela's total investment is split between a CD and bonds with a specific relationship between them.
Key considerations:

• Total Amount Invested: The sum of all investments equals a given total, here, \(\text{\textdollar}50,000\).
• Relationship Between Investments: The problem states that the bonds amount is \(\text{\textdollar}5000\) more than twice the CD amount.
• Formulate Equations: This relationship is translated into linear equations.
  Having these equations allows us to use algebraic methods to find the investment amounts. For instance, in our exercise:
  - \(x + y = 50000\)
  - \(y = 2x + 5000\).

algebraic solution

An algebraic solution involves manipulating equations to solve for unknown variables. Using algebraic techniques allows us to systematically find precise values.
Steps in our exercise:

• From Step 3: Substituting \(y = 2x + 5000\) into \(x + y = 50000\):
  \(x + (2x + 5000) = 50000\).
• Combine like terms: \(3x + 5000 = 50000\).
• Solve for \(x\):
  Subtract 5000 from both sides: \(3x = 45000\).
  Divide by 3: \(x = 15000\).
• Substitute \(x\) back into the second equation:
  \(y = 2(15000) + 5000\).
  Simplify: \(y = 35000\).

Our final solution states that Daniela invested \(\text{\textdollar}15,000\) in the CD and \(\text{\textdollar}35,000\) in bonds.
Algebraic solutions are powerful in making sure the calculations are correct and verifiable.