Problem 23 - Intermediate Algebra

Question

Solve each system by elimination. See Examples 3 and 4 $$ \left\\{\begin{array}{l} 2 s+3 t=-8 \\ 2 s-3 t=-8 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \(s = -4\) and \(t = 0\).

Step 1: Understand the System

We have a system of equations:\[\begin{align*}2s + 3t &= -8 \2s - 3t &= -8\end{align*}\]We are going to solve this system using the elimination method.

Step 2: Align Equations

Observe that both equations are already aligned with terms in columns (i.e., terms of \(s\) and \(t\)) matching vertically. Both equations have the coefficient of \(s\) as 2.

Step 3: Eliminate a Variable

Since the coefficients in front of \(s\) are the same, we can eliminate \(s\) by subtracting the second equation from the first:\[\begin{align*}(2s + 3t) - (2s - 3t) &= -8 - (-8)\end{align*}\]This simplifies to:\[\begin{align*}6t &= 0\end{align*}\]Divide both sides by 6 to solve for \(t\):\[t = 0\]

Step 4: Solve for Remaining Variable

Substitute \(t = 0\) back into one of the original equations. Let's choose the first equation:\[2s + 3(0) = -8\]This simplifies to:\[2s = -8\]Divide both sides by 2 to solve for \(s\):\[s = -4\]

Step 5: Verify Solution

Substitute \(s = -4\) and \(t = 0\) into the second equation to verify:\[2(-4) - 3(0) = -8\]This also simplifies to:\[-8 = -8\]Both equations are satisfied, verifying our solution is correct.

Key Concepts

Elimination Method

The elimination method is a powerful technique used to solve systems of linear equations by removing one variable. In general, the objective is to either add or subtract the equations in the system to cancel out one of the variables. This simplifies the problem and allows the solution to focus on finding values for the remaining variable.

Here's how the process works:

• Align the Equations: Ensure that corresponding terms are aligned in a column. This alignment helps in easy addition or subtraction.
• Eliminate a Variable: Use addition or subtraction to eliminate one variable. The key is to have the coefficients of the variable you want to eliminate match in absolute value. In some cases, you might need to multiply one or both equations to achieve this.
• Simplify and Solve: Once one variable is eliminated, you are left with a single equation with one variable, which is simpler to solve.

In this specific exercise, we immediately observed that the coefficients of the \(s\) terms were equal, allowing us to easily subtract one equation from the other to eliminate \(s\) and thus simplify our work.

Solving Linear Systems

Solving linear systems involves finding the values of the variables that satisfy all equations at the same time. A system of equations is a set of two or more linear equations with the same variables. There are various methods to solve these systems, such as substitution, elimination, or using matrices.

The strategy is to reduce the system to simpler forms until you isolate each variable. In most algebra problems, solutions are found where graphs of lines intersect, representing the set of solutions of the equations:

• The Substitution Method: Solve for one variable in terms of the other and substitute back into the other equation.
• The Elimination Method: Add or subtract equations to eliminate one variable, as demonstrated in this exercise.
• Graphical Method: Plot both equations on a graph and identify the intersection point. While not used here, it's a great visual tool.

In our scenario, applying the elimination method reduced our system from two unknowns to just one, enabling quicker and cleaner calculations.

Algebraic Solutions

Once the system has been simplified using a method like elimination, the next step is to find the algebraic solutions, which are the specific values of the variables that satisfy all the equations. Let's see how this is done:

• Solve the Simplified Equation: Once one variable is eliminated, solve the remaining equation. In our exercise, solving the equation \(6t = 0\) led to \(t = 0\).
• Substitute to Find Other Variables: Once one variable is known, substitute it back into one of the original equations to find the other variable. We substituted \(t = 0\) into the original equation to get \(s = -4\).
• Verification: Always substitute all variables back into the original equations to ensure the solution satisfies all the conditions. Doing this confirms the correctness of your solution.

This exercise used algebraic solutions to guarantee that the linear system was solved correctly, showing that learning to navigate these kinds of problems involves methodically applying algebraic skills.