Problem 35 - 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) $$ \left(\begin{array}{l} y=\frac{2}{3} x-4 \\ 5 x-3 y=9 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \( x = -1 \) and \( y = -\frac{14}{3} \).

Step 1: Express y in terms of x (Substitution method)

The first equation is already solved for y: \( y = \frac{2}{3}x - 4 \). We will use this expression for y in the second equation.

Step 2: Substitute y in the second equation

Take the expression for \( y \) from Step 1 and substitute it into the second equation:\[ 5x - 3 \left( \frac{2}{3}x - 4 \right) = 9 \]

Step 3: Simplify the equation

Distribute \(-3\) across the terms inside the parentheses:\[ 5x - 2x + 12 = 9 \]Combine like terms:\[ 3x + 12 = 9 \]

Step 4: Solve for x

Subtract 12 from both sides to isolate the term with x:\[ 3x = -3 \]Then divide by 3 to find \( x \):\[ x = -1 \]

Step 5: Substitute x back to find y

With \( x = -1 \), substitute back into the expression for y:\[ y = \frac{2}{3}(-1) - 4 \]\[ y = -\frac{2}{3} - 4 \]\[ y = -\frac{2}{3} - \frac{12}{3} \]\[ y = -\frac{14}{3} \]

Step 6: Verify the solution

Substitute \( x = -1 \) and \( y = -\frac{14}{3} \) into the second original equation. Check:\[ 5(-1) - 3\left(-\frac{14}{3}\right) = 9 \]\[ -5 + 14 = 9 \] which is true, confirming our solution is correct.

Key Concepts

Substitution Method

The substitution method is one of the primary strategies used to solve systems of linear equations. This method involves expressing one variable in terms of another and then substituting this expression into the other equation. It systematically reduces a system of equations to a single variable equation that can be easily solved.

In our original system, we have:

• The first equation already expressed in terms of \( y \), which is \( y = \frac{2}{3}x - 4 \).
• By replacing \( y \) with \( \frac{2}{3}x - 4 \) in the second equation, we transform the system into a single equation with one variable, which is \( x \).

This approach simplifies the problem dramatically, making it manageable to solve for the unknown variable. After finding one variable, you can easily substitute back to find the other.

Elimination Method

The elimination method, also known as the addition method, is an alternative approach for solving systems of equations. This method focuses on eliminating one of the variables by adding or subtracting equations from each other.

Here's how it typically works:

• Multiply or divide one or both equations to align coefficients if needed.
• Add or subtract the equations to cancel out one of the variables.
• Solve the resulting equation for the remaining variable.

Once a variable is eliminated, the system simplifies to a single equation. You solve for the remaining variable and then back-substitute to find the other variable. Although this method wasn't used in the original example, it is important to understand both methods as it allows for flexibility depending on the given system.

Algebraic Solutions

Algebraic solutions are a systematic way of solving equations using algebraic manipulations. These solutions are essential because they provide a clear, logical process for finding the values of unknown variables in a system of equations.

In the context of the substitution method used in the original example:

• We expressed one variable in terms of another, simplifying the two-variable system into a single equation.
• This process involved substitution, algebraic simplification, and solving the linear equation to find \( x = -1 \).
• Subsequently, we substituted back to find \( y = -\frac{14}{3} \).
• Verification steps ensured accuracy by checking solutions against the original equations.

Algebraic solutions not only guide us to find the correct answers but also help us verify them, ensuring they satisfy all original conditions provided in the system.