Problem 33 - Elementary Algebra

Question

Is the given value a solution to the linear equation? $$ -x+3=-18 ; x=-15 $$

Step-by-Step Solution

Answer

No, \(x=-15\) is not a solution to the equation.

Step 1: Understand the Problem

We need to determine if the suggested value, \(x = -15\), satisfies the given linear equation \(-x + 3 = -18\). This means that substituting \(-15\) for \(x\) in the equation should hold true for the equation to be a solution.

Step 2: Substitute the Value

Substitute \(x = -15\) into the equation. This results in the expression \(-(-15) + 3\).

Step 3: Simplify the Expression

Simplify the expression \(-(-15) + 3\) to detect any arithmetic errors. We get \(15 + 3\).

Step 4: Calculate the Result

Calculate \(15 + 3\) to find the resulting value. \(15 + 3 = 18\).

Step 5: Compare with the Right Side of the Equation

Compare the result, which is \(18\), with the right side of the equation \(-18\). Since \(18 eq -18\), the given value \(x = -15\) does not satisfy the equation.

Key Concepts

Linear Equation

A linear equation is an algebraic expression composed of variables and constants raised to the first power. It can be expressed in the form \(ax + b = c\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is a variable. Linear equations have straightforward solutions and are graphically represented as straight lines.
Linear equations have only one solution for each given expression, which is why it's crucial to understand if a particular value of \(x\) satisfies the equation. These equations are fundamental in mathematics because they model various real-life situations, such as calculating distances, speeds, or financial profits. The simplicity of linear equations makes them a great starting point for learning algebra.

Evaluation of Equations

The evaluation of equations involves checking if the chosen values for variables satisfy the equation. This is done by substituting the values into the equation and simplifying the expression. In the case of the linear equation \[-x + 3 = -18\], we evaluate if \(x = -15\) is a solution.
The process involves substituting \(-15\) into the equation for \(x\), leading to \(-(-15) + 3 = 15 + 3 = 18\).
This result should equal the right side of the equation, \(-18\), for \(x = -15\) to be the correct solution. If the left and right sides do not equal, the value is not a solution. Evaluating equations is necessary to confirm or reject potential solutions.

Algebraic Substitution

Algebraic substitution is a method used to replace variables in an equation with their numerical values. This process is essential for evaluating if a given value satisfies an equation.
In the provided exercise, after understanding that we need to test \(x = -15\), substitution involves replacing \(x\) in the equation \(-x + 3 = -18\) with \(-15\). This produces the expression \(-(-15) + 3\).
Substituting effectively changes an equation from one with variables to a simple numeric statement that can be solved easily.

• Think of substitution as swapping the placeholder (variable) with the actual number being tested.
• Ensure the replacement is accurate to prevent mistakes in following evaluations.

Mastering substitution solidifies understanding of how equations work and aids in verifying potential solutions.

Problem-Solving Steps

Problem-solving in algebra typically follows a structured set of steps to find the solution.
The process starts by understanding the problem, which is determining whether a specific value satisfies an equation.
Steps involved include:

• Comprehending the problem: Grasp what the equation represents and what value needs testing.
• Substitution: Insert the given value into the equation.
• Simplification: Resolve the equation by simplifying the expression after substitution.
• Comparison: Compare the result of the simplification with the right side of the equation to verify the solution.

Each step is critical and builds on the former to ensure no detail is overlooked. Systematic problem-solving ensures clarity and accuracy, leading to the correct conclusion.