Problem 53 - Introductory Algebra for College Students

Question

find five solutions of each equation. Select integers for \(x,\) starting with \(-2\) and ending with \(2 .\) Organize your work in a table of values. $$y=8 x-5$$

Step-by-Step Solution

Answer

The table of values for the equation \(y=8x-5\) from \(x=-2\) to \(x=2\) is: \[\begin{tabular}{|c|c|} \hline x & y \\ \hline -2 & -21 \\ -1 & -13 \\ 0 & -5 \\ 1 & 3 \\ 2 & 11 \\ \hline \end{tabular}\]

Step 1: Substituting \(x = -2\)

Replace \(x\) with -2 in the equation \(y = 8x - 5\) to find the corresponding \(y\) value: \(y = 8(-2) - 5 = -16 - 5 = -21\).

Step 2: Substituting \(x = -1\)

Replace \(x\) with -1 in the equation \(y = 8x - 5\) to find the corresponding \(y\) value: \(y = 8(-1) - 5 = -8 - 5 = -13\).

Step 3: Substituting \(x = 0\)

Replace \(x\) with 0 in the equation \(y = 8x - 5\) to find the corresponding \(y\) value: \(y = 8(0) - 5 = 0 - 5 = -5\).

Step 4: Substituting \(x = 1\)

Replace \(x\) with 1 in the equation \(y = 8x - 5\) to find the corresponding \(y\) value: \(y = 8(1) - 5 = 8 - 5 = 3\).

Step 5: Substituting \(x = 2\)

Replace \(x\) with 2 in the equation \(y = 8x - 5\) to find the corresponding \(y\) value: \(y = 8(2) - 5 = 16 - 5 = 11\).

Key Concepts

Substitution Method

The substitution method is a powerful technique used to solve linear equations. It involves replacing variables with specific values to find the solutions to an equation. In our given exercise, we start by selecting values for the variable \(x\). By substituting these values one by one into the given equation \(y = 8x - 5\), we can easily compute the corresponding \(y\) values.

Here's how it works step-by-step:

• First, choose a value for the unknown variable \(x\).
• Next, replace the variable \(x\) in the equation with the chosen value.
• Then, perform the arithmetic operations according to the equation.
• Finally, this calculation provides you with the corresponding \(y\) value, which is a solution to the equation.

By following these steps for different selected values of \(x\), you can create a set of (x, y) pairs that satisfy the equation. It's a systematic way to explore and visualize solutions.

Integer Solutions

Integer solutions refer to solutions of an equation where the values of the variables are integers. In the context of linear equations, like in our exercise, this means that both \(x\) and \(y\) must be integer values. Working with integers is often easier because they are whole numbers, and it avoids complications involving decimals or fractions.

When you substitute integer values for \(x\) into a linear equation, you can quickly determine the corresponding integer \(y\) values if the equation permits it. In our task, beginning from \(-2\) up to \(2\), each substitution results in a different integer value for \(y\), providing us a neat set of integer solutions like \((-2, -21), (-1, -13), (0, -5), (1, 3), (2, 11)\). This makes it particularly useful for students to practice basic arithmetic operations and develop a deeper understanding of solving equations.

Table of Values

A table of values is a useful tool to organize and present the results obtained from an equation clearly and systematically. It helps in visualizing the relationship between variables. In our exercise, we use a table to list integer values of \(x\) ranging from \(-2\) to \(2\), along with their respective \(y\) values calculated from the equation \(y = 8x - 5\).

Here’s how to construct a table of values:

• Start by creating two columns, one for \(x\) and one for \(y\).
• List the chosen integer values for \(x\) down the first column.
• Then write the corresponding \(y\) values next to each \(x\) value based on the substitutions.

Such a table provides an organized way to display solutions, making it easier to see patterns or verify correctness. It also serves as a reference for checking work and a visual aid that complements graphing the equation on the coordinate plane.