Problem 26 - Prealgebra

Question

For each of the following equations, complete the given table. $$y=3 x+1$$ $$\begin{array}{c|c} x & y \\ \hline-2 & \\ \hline & -2 \\ \hline & 4 \\ \hline 0 & \end{array}$$

Step-by-Step Solution

Answer

For \( x = -2, y = -5 \); \( y = -2, x = -1 \); \( y = 4, x = 1 \); \( x = 0, y = 1 \).

Step 1: Plug in x = -2

To find the value of \( y \) when \( x = -2 \), substitute \( x = -2 \) into the equation \( y = 3x + 1 \). Calculate: \( y = 3(-2) + 1 = -6 + 1 = -5 \). Therefore, when \( x = -2 \), \( y = -5 \).

Step 2: Solve for x when y = -2

Substitute \( y = -2 \) into the equation \( y = 3x + 1 \) to find \( x \): \( -2 = 3x + 1 \). Subtract 1 from both sides to get \( -3 = 3x \). Divide both sides by 3 to solve for \( x \): \( x = -1 \). Thus, when \( y = -2 \), \( x = -1 \).

Step 3: Solve for x when y = 4

Substitute \( y = 4 \) into the equation \( y = 3x + 1 \) to find \( x \): \( 4 = 3x + 1 \). Subtract 1 from both sides to get \( 3 = 3x \). Divide both sides by 3 to solve for \( x \): \( x = 1 \). Thus, when \( y = 4 \), \( x = 1 \).

Step 4: Plug in x = 0

To find the value of \( y \) when \( x = 0 \), substitute \( x = 0 \) into the equation \( y = 3x + 1 \). Calculate: \( y = 3(0) + 1 = 0 + 1 = 1 \). Therefore, when \( x = 0 \), \( y = 1 \).

Key Concepts

Solving Linear Equations

A linear equation is a fundamental mathematical expression which represents a line on a graph. When solving linear equations, the goal is to find the value of the unknown variable. In our exercise, we are dealing with the linear equation \( y = 3x + 1 \). Here, \( x \) is the independent variable, which means you can choose different values for it, and \( y \) is the dependent variable, which depends on the chosen \( x \) value. The equation essentially says that for any value of \( x \), multiply it by 3, then add 1 to find the corresponding \( y \). Solving equations like this often involves substituting the given value into the equation in order to find the unknown variable.

Using the Substitution Method

The substitution method is a powerful tool to find specific variable values when given another variable. It involves replacing the unknown variable in an equation with its value to solve for another unknown.
In our exercise, we have four situations:

• When \( x = -2 \), plug it into the equation to find \( y \): \( y = 3(-2) + 1 = -5 \).
• When \( y = -2 \), substitute into the equation to find \( x \): solve \( -2 = 3x + 1 \) resulting in \( x = -1 \).
• When \( y = 4 \), use the equation to find \( x \): solve \( 4 = 3x + 1 \) giving \( x = 1 \).
• When \( x = 0 \), find \( y \): \( y = 3(0) + 1 = 1 \).

Substitution is helpful because it directly replaces the variable with the number provided, simplifying the equation.

Understanding Equation Tables

Equation tables are an excellent method for visualizing how changing one variable affects another in a linear equation.
In these tables, a set of coordinate pairs are listed, showing the \( x \) value and corresponding \( y \) value derived from substituting into the equation \( y = 3x + 1 \).
For example, when the exercise provides a table and asks for missing values, you apply substitution:

• Fill in the missing \( y \) for a given \( x \), like \( x = -2 \).
• Determining the missing \( x \) when a \( y \) is given, for instance \( y = -2 \).

This process helps to see how inputs are related to outputs, forming a straightforward pattern which is crucial for solving linear equations.

Exploring Coordinate Pairs

Coordinate pairs, often expressed as \((x, y)\), represent points on a Cartesian plane. Each pair comes from substituting values into the equation and then solving.
Using our example equation \( y = 3x + 1 \):

• If \( x = -2 \), then the coordinate pair is \((-2, -5)\).
• If \( y = -2 \), solving gives \( x = -1 \), so the pair is \((-1, -2)\).
• For \( y = 4 \), solving leads to \( x = 1 \), resulting in the pair \((1, 4)\).
• When \( x = 0 \), the value of \( y \) is 1, giving the pair \((0, 1)\).

Each of these coordinate pairs corresponds to a point where the equation intersects with the grid. They are fundamental in plotting the behavior and position of the linear equation in a graph.