Problem 7 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the system by the method of elimination. Label each line with its equation. $$\left\\{\begin{array}{c} 2 x+y=5 \\ x-y=1 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 2\) and \(y = 1\).

Step 1: Set Up the Equations

The system of equations to solve is as follows: \(2x + y = 5\) and \(x - y = 1\). Label the first equation as Eq1 and the second one as Eq2.

Step 2: Elimination Process

The aim is to eliminate one of the variables to obtain a simpler equation. In this case, add Eq1 and Eq2 to eliminate variable \(y\). Doing so results in \( 2x + y + x - y = 5 + 1 \) which simplifies to \( 3x = 6 \). Label this as Eq3.

Step 3: Solving for x

Now, solve Eq3 for the variable \(x\) by dividing both sides by 3. This yields \( x = 6/3 = 2\).

Step 4: Solve for y

Substitute \(x = 2\) into Eq1 to get \( 2(2) + y = 5 \) which simplifies to \( 4 + y = 5\). Solving for \(y\) by subtracting 4 from both sides results in \( y = 1\).

Key Concepts

Elimination Method

The elimination method is a popular technique for solving systems of linear equations. It involves strategically eliminating one variable so that you can solve for the other variable directly. In our exercise, this method is used effectively to simplify the equations.
Here is how it works:

• Align the equations one beneath the other, such as:
  1. Eq1: \(2x + y = 5\)
  2. Eq2: \(x - y = 1\)
• Choose which variable to eliminate. In this case, we aim to eliminate \(y\) by adding Eq1 and Eq2.
• Perform the addition: \( (2x + y) + (x - y) = 5 + 1 \). Notice that \(+y\) and \(-y\) cancel each other out, leaving us with \(3x = 6\).

The power of the elimination method lies in its ability to simplify the problem, turning it into a more straightforward equation which can quickly yield the value of one of the variables. Once one variable is determined, substitute it back to find the other. The elimination method often requires rearranging or even multiplying equations to align variables for easy elimination.

Linear Equations

Linear equations form the foundation of algebra and are essential in mathematics. They appear as straight lines when graphed, which is why they're called 'linear'. A system of linear equations consists of two or more linear equations with the same variables.
In the given exercise, the equations are:

• \(2x + y = 5\)
• \(x - y = 1\)

Linear equations have a few key features:

• The highest power of the variable is always 1, ensuring a constant rate of change or slope in the graph.
• When solved, they reveal the point of intersection between lines; this point represents the solution to the system.
• Using methods like substitution or elimination, you can efficiently determine the variable values satisfying all equations in the system.

Understanding linear equations allows students to solve problems ranging from simple financial calculations to complex physics equations, making them a vital mathematical skill.

Substitution Method

Although not the primary method used in this exercise, the substitution method is another fundamental technique for solving systems of equations that can be highly useful. This approach involves substituting one equation into another after isolating a variable.
Here's how it works:

• Select one of the equations and solve for one variable. For instance, solve Eq2 \((x - y = 1)\) for \(x\), leading to \(x = y + 1\).
• Substitute this expression for \(x\) into the other equation (Eq1), resulting in \(2(y + 1) + y = 5\).
• Simplify and solve for \(y\), then use this value to find \(x\) using the expression derived earlier.

The substitution method is particularly useful when one of the coefficients is either 1 or -1 since it makes isolating and solving for a variable straightforward. It can also be advantageous in systems involving non-linear equations or when equations are not easily manipulable for elimination.