Problem 56 - Algebra 2

Question

For each system, choose the method of solving that seems easier to use. Explain why you made each choice. \(\left\\{\begin{aligned} 2 m+3 n &=12 \\\\-5 m+n &=-13 \end{aligned}\right.\)

Step-by-Step Solution

Answer

The substitution method seems easier for this system as the coefficient of 'n' in the second equation is 1, making it easy to isolate and substitute.

Step 1: Assess the given system of equations

Start by looking at the coefficients of the variables in each equation. Notice that the coefficient of the variable 'n' in the second equation is 1, which makes it a good candidate for the substitution method, as we can easily isolate 'n'.

Step 2: Choose a suitable method for solving the system

In this case, the substitution method seems easier because if 'n' is isolated in the second equation, it can then be substituted into the first equation, allowing us to solve for 'm' then 'n' in turn.

Key Concepts

System of Linear Equations

Understanding a system of linear equations is the starting block for solving them effectively. Essentially, a system of equations is a set of two or more equations with the same variables. The solution to the system is the set of values that satisfies all equations simultaneously.

For example, the system given in the exercise \(2m + 3n = 12\) and \( -5m + n = -13\) consists of two linear equations with two variables, \(m\) and \(n\). There are several methods to solve such systems, including graphing, substitution, elimination, and matrix methods. The choosing of a method often depends on the structure of the equations and the coefficients of the variables. It's crucial to pick the most efficient strategy to simplify the solving process.

In the provided system, one can notice that \(n\) has a coefficient of \(1\) in the second equation, hinting at a potential starting point for a particular solving approach—namely, the substitution method.

Substitution Method

The substitution method is a powerful algebraic tool used to solve systems of equations. This method involves solving one of the equations for one variable in terms of the other and then substituting this expression into the other equation. The beauty of this approach is that it reduces the system to a single variable equation, which is typically easier to solve.

In the exercise, step 1 involves assessing the system and identifying that the second equation has the variable \(n\) with a coefficient of \(1\), which makes isolation straightforward. By isolating \(n\), you can express \(n\) as \(n = -13 + 5m\). Step 2 then involves substituting this expression into the first equation, replacing \(n\) with \( -13 + 5m\), which simplifies to a one-variable equation in terms of \(m\). Once \(m\) is found, substitute its value back into the expression for \(n\) to find the solution to the system.

Algebraic Methods

Algebraic methods encompass a variety of techniques used to solve systems of equations, including the substitution method discussed above. Other methods include the elimination method, where you eliminate one variable by adding or subtracting equations, and the matrix method, involving linear algebra and the use of row operations to solve systems.

Each method has its own set of rules and suitability depending on the situation. For instance, elimination is often useful when the coefficients of one variable are the same or opposites. Matrix methods are more efficient for larger systems. Understanding these methods requires a grasp of algebraic principles and the ability to manipulate equations effectively.

When deciding which algebraic method to use, consider the complexity of the system, the coefficients of the variables, and your own familiarity with the techniques. Often, the goal is to simplify the system to reduce the potential for error and to make the calculation more manageable.