Problem 54 - College Algebra with Modeling and Visualization

Question

If possible, solve the system of linear equations and check your answer. $$ \begin{aligned} &100 x+200 y=300\\\ &200 x+100 y=0 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \( x = -1 \), \( y = 2 \).

Step 1: Write the Equations

The system of equations given is:1. \( 100x + 200y = 300 \)2. \( 200x + 100y = 0 \)

Step 2: Simplify the Equations

Divide the first equation by 100:\( x + 2y = 3 \)Divide the second equation by 100:\( 2x + y = 0 \)

Step 3: Solve for y in Terms of x from the Second Equation

From the equation \( 2x + y = 0 \), solve for \( y \):\( y = -2x \)

Step 4: Substitute y in Terms of x into the First Equation

Substitute \( y = -2x \) into the first simplified equation \( x + 2y = 3 \):\( x + 2(-2x) = 3 \)This simplifies to:\( x - 4x = 3 \)

Step 5: Solve for x

Combine like terms:\( -3x = 3 \)Divide both sides by \(-3\):\( x = -1 \)

Step 6: Find y Using the Value of x

Substitute \( x = -1 \) into \( y = -2x \):\( y = -2(-1) = 2 \)

Step 7: Verify the Solution

Substitute \( x = -1 \) and \( y = 2 \) back into the original equations:1. \( 100(-1) + 200(2) = 100 \) correct as \( 300 \)2. \( 200(-1) + 100(2) = 0 \) correct as \( 0 \)Both equations hold true, confirming the solution is correct.

Key Concepts

Solving Systems of Equations

Solving systems of equations involves finding values for variables that satisfy all equations in the system simultaneously. This means that the solution must make each equation true when the variables are substituted back into them. In our exercise, we started with two linear equations:

• \( 100x + 200y = 300 \)
• \( 200x + 100y = 0 \)

Each equation represents a line in a coordinate plane, and solving the system means finding the point where these lines intersect.Typically, systems of linear equations can be solved using various methods such as graphing, substitution, or elimination. The choice of method may depend on the specific system you are dealing with. In our case, we used the substitution method. This can be particularly useful when one of the equations can be easily solved for one variable.

Substitution Method

The substitution method is a straightforward approach to solve systems of equations. It involves solving one equation for one variable, then substituting that expression into another equation to find the other variable. Here’s a breakdown of how this method was applied:1. **Solve One Equation for One Variable**: In the system, the second equation \( 2x + y = 0 \) was easily solved for \( y \), yielding \( y = -2x \). This expression defines \( y \) in terms of \( x \).
2. **Substitute the Expression in the Other Equation**: We substituted \( y = -2x \) into the first equation. The first equation became \( x + 2(-2x) = 3 \), simplifying to \( x - 4x = 3 \).This method transforms the system into one equation with just one variable, making it simpler to solve. It allowed us to find \( x = -1 \), which can then be used to determine the corresponding value of \( y \).

Simplification of Equations

Simplification is an important step in the process as it makes equations easier to manipulate and solve. By dividing each equation in our system by 100, we streamlined the expressions without changing the solutions. The original equations:

• \( 100x + 200y = 300 \) simplified to \( x + 2y = 3 \)
• \( 200x + 100y = 0 \) simplified to \( 2x + y = 0 \)

This didn’t alter the solutions but made the arithmetic considerably simpler.Here’s why this matters:- **Reduces Complexity**: Fewer numbers and simpler coefficients make the equation easier to solve manually.- **Leads to Correct Solutions**: Keeping track of calculations is easier, reducing the risk of errors.By simplifying, we are able to more smoothly apply the chosen substitution method and quickly solve for the variables \( x \) and \( y \). Ensuring the system remains balanced and accurate during simplification is key to finding the right solution.