Problem 39 - Introductory and Intermediate Algebra for College Students 4th

Question

Determine whether each ordered pair is a solution of the given equation. $$y=-4 x \quad(-5,-20),(0,0),(9,-36)$$

Step-by-Step Solution

Answer

The pair (-5,-20) is not a solution to the equation. The pairs (0,0) and (9,-36) are solutions to the equation.

Step 1: Substitute the first pair into the equation

Replace \(x\) with -5 and \(y\) with -20 in the equation \(y = -4x\). Hence we get -20 = -4(-5). Solve the right side, we get -20 = 20. Since both sides are not equal, (-5,-20) is not a solution to the equation.

Step 2: Substitute the second pair into the equation

Repeat the process with the second pair (0,0). Replace \(x\) with 0 and \(y\) with 0 in the equation \(y = -4x\), hence we get 0 = -4*0. After simplifying it results in 0 = 0. Since both sides are equal, (0,0) is indeed a solution to the equation.

Step 3: Substitute the third pair into the equation

Finally, with the third pair (9,-36), substitute \(x\) = 9 and \(y\) = -36 into the equation \(y = -4x\). Hence we get -36 = -4*9. After simplifying we get -36 = -36. Since both sides are equal, (9,-36) is indeed a solution to the equation.

Key Concepts

Understanding Ordered Pairs in Equations

An ordered pair, often written as \((x, y)\), is a way to represent two related numbers or variables. In the context of linear equations, these pairs represent points on a coordinate plane where the first number is the \(x\)-coordinate and the second is the \(y\)-coordinate. Let's consider this in relation to the equation in our exercise: \(y = -4x\).

To determine if an ordered pair is a solution to this equation means checking if substituting \(x\) and \(y\) from the pair into the equation results in a true statement. Thus, given a pair like \((-5, -20)\), plug \(-5\) for \(x\) and \(-20\) for \(y\). If both sides of the equation are equal after substitution, then the pair is a solution.

This approach is crucial in algebra to understand whether a certain point, represented by the ordered pair, falls on the line described by the equation. Each ordered pair offers a potential solution, and verifying it helps confirm or deny its validity.

Performing Solution Checking

Solution checking is the process of verifying if a given ordered pair satisfies an equation. This process involves substituting the values from the ordered pair into the equation and ensuring that both sides remain equal after simplification.

For example, consider the equation \(y = -4x\). To check if the ordered pair \((0, 0)\) is a solution, replace \(x\) with \(0\) and \(y\) with \(0\). Substituting these values, the equation becomes:
- Left side: \(y = 0\)
- Right side: \(-4 \times 0 = 0\)

Since both sides equal \(0\), \((0, 0)\) is indeed a solution. This means that the point \((0, 0)\) lies on the line described by the equation.

Solution checking is a key skill in algebra, as it allows you to verify potential solutions to equations and better understand the relationships between variables. It's a systematic and reliable way to ascertain if a given point is part of the line or curve described by the equation.

Explaining the Substitution Method

The substitution method is a reliable technique for solving linear equations that involves replacing variables with specific values. Here, we use the values from given ordered pairs to test against the equation.

In the equation \(y = -4x\), suppose we need to check the ordered pair \((9, -36)\). Substitution guides us to:

• Replace \(x\) with \(9\)
• Replace \(y\) with \(-36\)

Performing these substitutions transforms the equation into \(-36 = -4 \times 9\). Simplifying the right side gives \(-36 = -36\), confirming that the pair \((9, -36)\) is a valid solution.

This method not only helps in verifying whether the pair is a solution but also reinforces how equations work, showing how changes in the input \(x\) result in changes in the output \(y\). Understanding the substitution method is fundamental for solving equations, as it lays the groundwork for dealing with more complex algebraic problems.