Problem 25 - Prealgebra

Question

For each of the following equations, complete the given table. $$y=6 x-1$$ $$\begin{array}{c|c} x & y \\ \hline-1 & \\ \hline & 5 \\ \hline & -13 \\ \hline 0 & \end{array}$$

Step-by-Step Solution

Answer

Complete table: (x, y) pairs are (-1, -7), (1, 5), (-2, -13), (0, -1).

Step 1: Calculate y when x = -1

Given the equation \( y = 6x - 1 \), substitute \( x = -1 \) into the equation. \[ y = 6(-1) - 1 \]Calculate:\[ y = -6 - 1 = -7 \]Thus, when \( x = -1 \), \( y = -7 \).

Step 2: Calculate x when y = 5

Rearrange the given equation to solve for \( x \) when \( y = 5 \).Set \( y = 5 \):\[ 5 = 6x - 1 \]Add 1 to both sides:\[ 6 = 6x \]Divide by 6:\[ x = 1 \]Thus, when \( y = 5 \), \( x = 1 \).

Step 3: Calculate x when y = -13

Use the same steps to solve when \( y = -13 \).Set \( y = -13 \):\[ -13 = 6x - 1 \]Add 1 to both sides:\[ -12 = 6x \]Divide by 6:\[ x = -2 \]So, when \( y = -13 \), \( x = -2 \).

Step 4: Calculate y when x = 0

Substitute \( x = 0 \) into the equation:\[ y = 6(0) - 1 \]Calculate:\[ y = 0 - 1 = -1 \]Therefore, when \( x = 0 \), \( y = -1 \).

Key Concepts

Table of Values

A table of values is a simple yet powerful tool used to understand the relationship between variables in an equation, such as in the linear equation \( y = 6x - 1 \). It helps us visualize how changing one variable affects the other. In this equation, we have two variables, \( x \) and \( y \). The table includes given values for \( x \) or \( y \) and requires us to calculate and fill in the missing corresponding values.To use a table of values:

• List different values for one variable, like \( x \).
• Use the equation to determine the corresponding \( y \) values for those \( x \) values.
• If the \( y \) value is known, work backwards to find the \( x \) value.

The goal is to complete the table by solving the equation for each set of values, allowing you to see whether a pattern or relationship exists.

Solving for x and y

When working with linear equations, solving for \( x \) and \( y \) is a fundamental skill. The idea is to find the specific value of one variable that satisfies the equation given the value of the other variable.For example:

• To find \( y \) when \( x = -1 \), substitute \( x = -1 \) into \( y = 6x - 1 \), giving you \( y = 6(-1) - 1 = -7 \).
• To find \( x \) when \( y = 5 \), rearrange the equation to \( x = \frac{y+1}{6} \), and replace \( y \) with 5: \( x = \frac{5+1}{6} = 1 \).

This process is critical as it often involves basic algebra, like rearranging the equation and performing arithmetic operations. Ultimately, solving for \( x \) and \( y \) in a linear equation helps us understand how they depend on each other.

Substitution Method

The substitution method is a straightforward technique used to solve equations, especially useful when dealing with a system of equations, but also handy in completing tables of values.Here is how it works:

• Identify the equation, such as \( y = 6x - 1 \).
• If you have a known value for one of the variables (say \( x \)), substitute that value into the equation to find the other variable (\( y \)).
• When you have a known value for \( y \) and need \( x \), rearrange the equation first, then substitute the known \( y \) value.

For example, to find \( y \) when \( x = 0 \), substitute 0 for \( x \) in the equation: \( y = 6 \times 0 - 1 = -1 \). Using substitution allows you to systematically find solutions and fill out tables of values efficiently. It's all about replacing one variable with a known value to determine the unknown in an equation.