Problem 196 - Elementary Algebra

Question

In the following exercises, translate to a system of equations and solve the system. Jen and David owe \(\$ 22,000\) in loans for their two cars. The amount of the loan for Jen's car is \(\$ 2000\) less than twice the amount of the loan for David's car. How much is each car loan?

Step-by-Step Solution

Answer

The loan for David's car is \$8000\, and the loan for Jen's car is \$14000\.

Step 1: Define the Variables

Let x be the amount of the loan for David's car and y be the amount of the loan for Jen's car.

Step 2: Set Up the Equations

Use the information from the problem to set up the equations. The total loan amount equation is:en: x + y = 22000The second equation, based on the relationship between the loans, is:en: y = 2x - 2000

Step 3: Substitute y to Solve for x

Substitute the value of y from the second equation into the first equation:en: x + (2x - 2000) = 22000Combine like terms:en: 3x - 2000 = 22000Add 2000 to both sides:en: 3x = 24000

Step 4: Solve for x

Divide both sides by 3:en: x = 8000

Step 5: Solve for y

Substitute the value of x back into one of the original equations to solve for y. Using the equation y = 2x - 2000:en: y = 2(8000) - 2000y = 16000 - 2000y = 14000

Step 6: Verify the Solution

Check the values of x and y in the first equation:x + y = 220008000 + 14000 = 22000Both conditions of the problem are satisfied.

Key Concepts

linear equations

Linear equations are mathematical expressions that represent straight lines. Each equation is formed by equating a linear polynomial to a constant. In our example, we have two linear equations:

• Equation 1: \( x + y = 22000 \)
• Equation 2: \( y = 2x - 2000 \)

These equations help us investigate the relationships between different variables. Since the highest power of the variables is 1, they are called linear equations.

substitution method

The substitution method is a technique used to solve a system of equations. Here's how it works:

• Solve one equation for one variable
• Substitute this variable into the other equation
• Solve for the second variable

In our example, we solve the second equation \( y = 2x - 2000 \) for \( y \) and substitute it into the first equation \( x + y = 22000 \). This reduces the system to a single equation with one variable, which we can solve easily.

solving equations

Solving equations involves finding the values of variables that make the equation true. Let's look at our example:

• Combine like terms: \( x + 2x - 2000 = 22000 \)
• Simplify to get \( 3x - 2000 = 22000 \)
• Add 2000 to both sides to isolate the term \( 3x \): \( 3x = 24000 \)
• Divide both sides by 3 to find \( x \): \( x = 8000 \)

After finding \( x \), we substitute it back into the second equation to find \( y \): \( y = 2(8000) - 2000 \), which simplifies to \( y = 14000 \). Then, we verify our solution by checking that \( x + y = 22000 \).

algebra problems

Algebra problems involve manipulating mathematical symbols and expressions to find unknown values. To solve these problems more easily:

• Identify what you're solving for and define the variables
• Set up equations based on the problem's information
• Use suitable methods to solve the equations (like substitution or elimination)
• Always recheck your answers

In our case, defining \( x \) and \( y \), setting up the equations \( x + y = 22000 \) and \( y = 2x - 2000 \), using the substitution method to solve for \( x \) and \( y \), and finally verifying our solution, allow us to confidently conclude that the car loans for Jen and David are \( \(14000 \) and \)8000 respectively.