Problem 24 - Prealgebra

Question

For each of the following equations, complete the given table. $$2 x-y=6$$ $$\begin{array}{l|l} \hline x & y \\ \hline 1 & \\ \hline & 6 \\ \hline-6 & \\ \hline & -6 \end{array}$$

Step-by-Step Solution

Answer

Fill the table with (1, -4), (6, 6), (-6, -18), and (0, -6).

Step 1: Solve for y using the first equation

Given the equation is \(2x - y = 6\). We need to solve for \(y\) in terms of \(x\).Rearranging the equation gives:\[ y = 2x - 6 \]This formula can be used to find the missing \(y\) values in the table when \(x\) is given.

Step 2: Calculate y for x = 1

Substitute \(x = 1\) into the equation \(y = 2x - 6\):\[ y = 2(1) - 6 = 2 - 6 = -4 \]So, \(y = -4\) when \(x = 1\). The first row in the table is (1, -4).

Step 3: Solve for x when y = 6

Given \(y = 6\), use the equation \(y = 2x - 6\):\[ 6 = 2x - 6 \]Add 6 to both sides:\[ 12 = 2x \]Divide both sides by 2:\[ x = 6 \]Thus, when \(y = 6\), \(x = 6\). The second row in the table is (6, 6).

Step 4: Calculate y for x = -6

Substitute \(x = -6\) into the equation \(y = 2x - 6\):\[ y = 2(-6) - 6 = -12 - 6 = -18 \]So, \(y = -18\) when \(x = -6\). The third row in the table is (-6, -18).

Step 5: Solve for x when y = -6

Given \(y = -6\), use the equation \(y = 2x - 6\):\[ -6 = 2x - 6 \]Add 6 to both sides:\[ 0 = 2x \]Divide both sides by 2:\[ x = 0 \]Thus, when \(y = -6\), \(x = 0\). The fourth row in the table is (0, -6).

Key Concepts

Table Completion

Completing a table for a given linear equation is a straightforward task once you understand the relationship between the variables. Imagine the table as a way to organize potential values for your variables. Here, our task is to find the missing values in rows using the equation \(2x - y = 6\). This equation expresses a linear relationship between variables \(x\) and \(y\). Let's break down how to complete each row:

• For every known value of \(x\) or \(y\), substitute it into the equation.
• Use algebraic manipulation to solve for the unknown variable.
• Write the found value in your table under the correct column.

This process turns an incomplete table into a complete one, showing the relationship between your inputs (\(x\)) and outputs (\(y\)). Remember, each row corresponds to a specific point defined by the equation, revealing different aspects of the line represented by the equation.

Variable Substitution

Variable substitution is a key method to solving for unknowns in equations. In our case, we often start with a known variable value, say \(x = 1\). We then substitute this value into the rearranged equation \(y = 2x - 6\).

This process involves:

• Inserting the known value into the formula: \(y = 2(1) - 6\).
• Carrying out the arithmetic operations: \(2 - 6 = -4\).

Substituting given values simplifies the problem into basic arithmetic, making it easy to calculate the unknowns. This technique is very handy as it can be applied to any point on the line, provided the equation is correctly rearranged earlier. By mastering variable substitution, you become adept at transforming complex algebra into solvable puzzles.

Equation Solving

Solving equations like \(2x - y = 6\) involves isolating the variable to find its value. Let's consider how to handle such equations to determine unknowns effectively:

- **Rearrangement**: This is your first task. For instance, if you need to solve for \(y\) in terms of \(x\), rearrange the equation. From \(2x - y = 6\), you add \(y\) to both sides and subtract 6: \(y = 2x - 6\).
- **Solving for a Specific Variable**: For situations where you know the value of one variable, substitute it and solve for the other. For example, if \(y = 6\), substitute into the rearranged equation: \(6 = 2x - 6\). Then process:

• Add 6 to each side: \(12 = 2x\).
• Divide by 2 to isolate \(x\): \(x = 6\).

Equation solving is about breaking down and isolating variables to find their values. With practice, this technique becomes speedy and intuitive, equipping you to tackle an array of mathematical challenges effortlessly.