Problem 2 - Understanding Elementary Algebra with Geometry

Question

$$\text { In Exercises } 1-14, \text { solve the system of equations using the elimination method.}$$ $$\left\\{\begin{array}{l} x-y=4 \\ x+y=6 \end{array}\right.$$

Step-by-Step Solution

Answer

x = 5, y = 1

Step 1 - Write Down the System of Equations

Given the system of equations: \(\begin{cases} x - y = 4 \ x + y = 6 \end{cases}\)

Step 2 - Add the Equations

Add the two equations together to eliminate the variable \(y\): \( (x - y) + (x + y) = 4 + 6 \) This simplifies to: \( 2x = 10 \)

Step 3 - Solve for x

Divide both sides of the equation by 2 to solve for \(x\): \( x = \frac{10}{2} \) \( x = 5 \)

Step 4 - Substitute x Value into One of the Original Equations

Substitute \(x = 5\) into the first equation \(x - y = 4\): \( 5 - y = 4 \) Solve for \(y\): \( y = 5 - 4 \) \( y = 1 \)

Step 5 - Write Down the Solution

The solution to the system of equations is: \(x = 5\) and \(y = 1\)

Key Concepts

solving systems of equations

When solving systems of equations, we aim to find the values of the variables that satisfy all the equations in the system. These equations can represent various real-world scenarios where multiple conditions must be met simultaneously. For instance, in the given exercise, we have two linear equations with variables x and y:

• \( x - y = 4 \)
• \( x + y = 6 \)

We are tasked with finding values for x and y that make both equations true. There are different methods to solve these systems, such as the elimination method and the substitution method. Both techniques offer a systematic approach to arrive at the correct solution.

elimination method

The elimination method involves adding or subtracting the equations to cancel out one of the variables. This makes it easier to solve for the remaining variable. Let's see how this works step by step:

• First, write down the system of equations: \( \begin{cases} \ x - y = 4 \ x + y = 6 \ \)
• Next, add the two equations together: \( (x - y) + (x + y) = 4 + 6 \) . The y terms cancel out, simplifying to: \( 2x = 10 \)
• Finally, solve for x by dividing both sides by 2: \( x = 5 \)

This elimination process makes the solution more manageable by reducing the system to a simpler equation with one variable.

substitution method

The substitution method involves solving one of the equations for one variable and substituting that expression into the other equation. This method can be particularly useful when one of the equations is simple enough to solve for one variable directly. In the provided exercise, after determining \( x = 5 \) using the elimination method, we can substitute this value back into one of the original equations to find y:

• Substitute \( x = 5 \) into the first equation: \( 5 - y = 4 \)
• Solve for y: \( y = 1 \)

By using substitution, we confirm the values of both variables, ensuring they satisfy the original system of equations.

linear equations

Linear equations are equations of the first degree, meaning they involve variables raised only to the power of one. The general form of a linear equation in two variables is: \( ax + by = c \), where a, b, and c are constants. In the provided exercise, both equations are linear:

• \( x - y = 4 \)
• \( x + y = 6 \)

These equations graph as straight lines. The solution to the system is the point where these two lines intersect. In our example, the lines intersect at the point (5, 1), which means the values \( x = 5 \) and \( y = 1 \) solve both equations.