Problem 21 - Intermediate Algebra

Question

For Problems 19-48, solve each system by using either the substitution or the elimination-by-addition method, whichever seems more appropriate. (Objective 2) $$ \left(\begin{array}{l} x=5 y+7 \\ 4 x+9 y=28 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \( x = 7 \) and \( y = 0 \).

Step 1: Identify the appropriate method

The system of equations given is \( x = 5y + 7 \) and \( 4x + 9y = 28 \). Since the first equation is already solved for \( x \), substitution is the more appropriate method for solving this system.

Step 2: Substitute the expression

Substitute \( x = 5y + 7 \) from the first equation into the second equation. This gives us: \( 4(5y + 7) + 9y = 28 \).

Step 3: Simplify and solve for \( y \)

Expand the expression: \( 20y + 28 + 9y = 28 \). Combine like terms: \( 29y + 28 = 28 \). Subtract 28 from both sides: \( 29y = 0 \). Divide by 29: \( y = 0 \).

Step 4: Substitute back to find \( x \)

Use \( y = 0 \) in the equation \( x = 5y + 7 \). Substitute 0 for \( y \): \( x = 5(0) + 7 \), which simplifies to \( x = 7 \).

Step 5: Verify the solution

Substitute \( x = 7 \) and \( y = 0 \) back into the original second equation to verify the solution: \( 4(7) + 9(0) = 28 \). Simplifying gives \( 28 = 28 \), which confirms the solution is correct.

Key Concepts

Substitution Method

The substitution method is a popular way to solve a system of equations, especially when one equation is already solved for one variable. In our exercise, the equation \( x = 5y + 7 \) already isolates \( x \), making substitution an ideal approach.
Here's how it works:

• First, solve one of the equations for one variable if it isn't already.
• Second, substitute this expression into the other equation. This removes one variable and allows you to solve for the other.
• Once the value is found, substitute it back into the first equation to find the value of the initial variable.

For example, substituting \( x = 5y + 7 \) into \( 4x + 9y = 28 \) gives us a single equation with just \( y \). Solving this simplifies the process and helps us find that \( y = 0 \). The method is straightforward and helps to break down the problem into simpler, manageable pieces.

Elimination Method

The elimination method, also known as the addition method, involves eliminating one of the variables by adding or subtracting equations. This approach is particularly useful when both equations are in standard form.
How the elimination method works:

• Align two equations one above the other, making sure like terms are in the same column.
• Manipulate one or both equations so that adding or subtracting them removes one variable.
• Once a variable is eliminated, solve the resulting equation for the remaining variable.
• Substitute the discovered value back into one of the original equations to find the other variable.

This method is powerful when equations are in a format that allows quick elimination through addition or subtraction. Though we didn't use it in the original problem due to substitution being more direct, understanding this method expands your problem-solving toolkit.

Algebraic Solutions

Algebraic solutions involve using algebraic operations to find the values of variables that satisfy all equations in a system. These solutions rely on a clear understanding of algebraic operations, such as combining like terms and using properties of equality.
To achieve an algebraic solution:

• Identify which method (substitution or elimination) simplifies the calculation.
• Use algebra steps systematically to isolate and solve for each variable.
• Verify solutions by substituting them back into the original equations to ensure consistency.

Algebraic solutions ensure accuracy and clarity in finding answers as they provide a structured format for solving any system of linear equations. Students not only find the solutions but also verify them, ensuring a comprehensive understanding of how variables interact within the system.