Problem 37 - Algebra 1

Question

Use linear combinations to solve the system of linear equations. $$\begin{aligned}&2 m-4=4 n\\\&m-2=n\end{aligned}$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(m = 4\) and \(n = 2\)

Step 1: Isolate a variable in the second equation

We can isolate n in the second equation \(m-2=n\) by rewriting it as \(n = m - 2\)

Step 2: Substitute \(n = m - 2\) in the first equation

Substitute \(n = m - 2\) in the first equation \(2m - 4 = 4n\) to get a linear equation with only m: It now becomes \(2m - 4 = 4(m - 2)\)

Step 3: Solve for m in the linear equation

Solving the equation \(2m - 4 = 4(m - 2)\) simplifies to \(2m = 4m - 8\), which simplifies further to \(2m = 8\). Finally, we see that \(m = 4\)

Step 4: Substitute m into the original equation to find n

Now that we have found the value of m, we substitute it into the equation \(n = m - 2\) which we rearranged at first. This yields \(n = 4 - 2 = 2\)

Key Concepts

Understanding Linear Combinations

Linear combinations are an essential concept in solving systems of linear equations. A linear combination involves adding or subtracting equations after multiplying them by constants. This technique helps eliminate one variable, making it simpler to solve for the remaining variable.

For example, if you have two equations:

• 2x + 3y = 6
• 4x - y = 5

You can multiply the second equation by 3 to match the coefficients of y in both equations:

• 12x - 3y = 15

Adding these equations eliminates y:

• 14x = 21

Now, you can easily solve for x. This process makes finding the solution to systems of equations more straightforward. Linear combinations are powerful for simplifying complex algebraic problems into manageable steps.

Solving Systems of Equations

A system of equations consists of two or more equations with the same set of variables. The goal is to find the set of variable values that satisfy all equations in the system simultaneously.

In this exercise, the system is:

• 2m - 4 = 4n
• m - 2 = n

Different methods exist for solving these systems, such as graphing, substitution, and elimination.

Graphing involves drawing each equation on a graph to find the intersection point, representing the solution. Elimination, like linear combinations, removes one variable to solve for the other.

Substitution, on the other hand, makes use of rearranging one equation to express a variable in terms of another and then substituting it into another equation. Each method provides a structured way to deduce the solution effectively.

The Substitution Method in Action

The substitution method is particularly helpful when one of the equations is easily rearranged. Here's how it works using our system:

• Start with the second equation, \(m - 2 = n\).
• Rearrange to isolate n, giving \(n = m - 2\).
• Substitute this expression for n into the first equation: \(2m - 4 = 4(m - 2)\).
• Now, solve for m:
  • Expand to get \(2m - 4 = 4m - 8\).
  • Rearrange, leading to \(2m = 8\).
  • Solve, and you find \(m = 4\).
• Finally, substitute m back into the isolated equation for n:
  • \(n = 4 - 2\)
  • Giving \(n = 2\).

The substitution method simplifies complex systems by breaking them down into smaller, easily manageable steps, ensuring accurate solutions.