Problem 46 - Introductory and Intermediate Algebra for College Students 4th

Question

Determine whether each ordered pair is a solution of the given equation. $$x+5 y=0 \quad(0,0),\left(1, \frac{1}{5}\right),\left(2,-\frac{2}{5}\right)$$

Step-by-Step Solution

Answer

The ordered pairs (0,0) and \(2, -\frac{2}{5}\) are solutions to the given equation.

Step 1: Understand the Equation

First, it is needed to understand the given equation, which is \(x + 5y = 0\). An ordered pair will be solution of this equation if, substituted into the equation the result gives a true assertion.

Step 2: Checking the first ordered pair

The first ordered pair is (0,0). When these values are applied to the equation, \(x + 5y = 0 + 5*0 = 0\), which exactly matches the given equation, so (0,0) is a solution.

Step 3: Checking the second ordered pair

The second pair is \(1, \frac{1}{5}\). Plug these values into the equation, \(x + 5y = 1 + 5*\frac{1}{5} = 1 + 1 = 2\), which does not hold true to the equation \(x + 5y = 0\). Hence \((1, \frac{1}{5})\) is not a solution.

Step 4: Checking the third ordered pair

Now apply the third ordered pair, \(2, -\frac{2}{5}\). If we substitute these into the equation, \(x + 5y = 2 + 5*(-\frac{2}{5}) = 2 - 2 = 0\), it turns out to be 0 and matches to the given equation, therefore (2, -2/5) is also a solution.

Key Concepts

Understanding Ordered Pairs

In algebra, an ordered pair is a fundamental concept. It consists of two elements arranged in a particular order, usually written in the form

• An ordered pair is commonly denoted as \((x, y)\), where \(x\) is the first component and \(y\) is the second.
• These elements can represent a point on a coordinate plane.In the context of linear equations, ordered pairs are potential solutions to the equation.This means that when you substitute the values of \((x, y)\) into the equation, they should satisfy the equation for the pair to be considered a solution.

An ordered pair will belong to the solution set of the equation if substituting both values results in a true equality.

Exploring Linear Equations

Linear equations are those which graph as straight lines on a coordinate plane. They are usually in the form \(ax + by = c\).

• In this equation, \(a\), \(b\), and \(c\) represent constants while \(x\) and \(y\) are variables.
• In the given exercise, the equation is \(x + 5y = 0\), which means that for every change in \(x\), there is a fixed corresponding change in \(y\) that keeps the equation balanced.

Understanding the structure of linear equations is crucial for determining whether a given ordered pair is a solution. This specific type of equation illustrates a direct relationship between \(x\) and \(y\). It implies that each x-coordinate has a unique y-coordinate that makes the equation true.

Solution Verification Process

To verify if an ordered pair is a solution of a linear equation, substitute the values of \((x, y)\) into the equation.

• If the equation holds true (both sides of the equation are equal) after the substitution, then the ordered pair is indeed a solution.
• In the given exercise, you substitute each ordered pair into \(x + 5y = 0\).Each pair requires its own verification process.

For example:- With the ordered pair \((0, 0)\), substituting into the equation gives you \(0 + 5(0) = 0\), which is true.- Checking \((1, \frac{1}{5})\), the equation becomes \(1 + 5(\frac{1}{5}) = 2\), and that does not equal zero, hence it is not a solution.- Lastly, \((2, -\frac{2}{5})\) checks out because \(2 + 5(-\frac{2}{5}) = 0\), confirming it is a solution.Always make sure to follow this verification step to confirm solutions in algebra problems.