Problem 4 - Intermediate Algebra

Question

Check to determine whether 2 is a solution of the following equations. a. \(\frac{x}{2}+\frac{4}{x+2}=x\) b. \(\frac{x+2}{x-2}+\frac{1}{x^{2}-4}=1\)

Step-by-Step Solution

Answer

2 is a solution for equation (a) but not for (b), as it's outside the domain.

Step 1: Substitute x = 2 in Equation (a)

First, we substitute \( x = 2 \) into the left-hand side of the equation: \( \frac{x}{2} + \frac{4}{x+2} \).\Substituting gives us: \( \frac{2}{2} + \frac{4}{2+2} = 1 + \frac{4}{4} = 1 + 1 = 2 \).\The right-hand side of the equation is also \( x \), which becomes \( 2 \).\Since both sides equal 2, \( x = 2 \) is a solution for equation (a).

Step 2: Identify the Domain of Equation (b)

Equation (b) is \( \frac{x+2}{x-2} + \frac{1}{x^2-4} = 1 \). We first check if \( x = 2 \) is in the domain.\The expressions \( x-2 \) and \( x^2-4 \) should not be zero.\Plug in \( x = 2 \); both \( x-2 \) and \( x^2-4 \) become zero.\Therefore, \( x = 2 \) is not in the domain of equation (b).\Thus, \( x = 2 \) cannot be a solution for equation (b).

Key Concepts

Solution of Equations

Every equation consists of two parts separated by an equal sign: the left side and the right side. To find the solution, we need these two sides to have equal values when a particular number is substituted for the variable. There are different strategies to ascertain if a number is indeed a solution to an equation. One of these is the substitution method. After finding a potential solution, substitute it into the equation. If both sides of the equation calculate to the same number, you have a valid solution. If they do not match, the number is not a solution. In the provided exercise, substituting 2 into equation (a) yields equal values on both sides, confirming 2 as a solution. For equation (b), the attempted substitution results in terms that are undefined, indicating that 2 is not a valid solution for this equation.

Substitution Method

The substitution method is a straightforward strategy to determine if a number can be a solution to an equation. This involves different steps:

• Take the candidate number and substitute it into the equation wherever the variable appears.
• Simplify the expressions on both sides.
• Compare the results from both sides. If they are equal, the number is a solution; if not, it isn't.

Let's consider equation (a) from the exercise. When we put 2 in place of x, we simplify both sides. If both result in the same number, like they did here with both sides equalling 2, the substitution confirms it's a solution. However, caution is needed, especially when variables appear in denominators. This can lead to terms that are undefined if the substitution results in dividing by zero, as was evident in equation (b).

Domain of a Function

The domain of a function refers to all the possible input values (usually x) that allow the function to generate valid outputs. Mathematically, the domain constraints are often applied where functions become undefined, such as division by zero. In equation (b) of the exercise, when substituting 2 for x, the term \(x-2\) went to zero, and \(x^2-4\) also equaled zero, making parts of the equation undefined. Thus, 2 is not in the domain of the function described by equation (b). To determine the domain practically:

• Look for fractions, ensuring denominators do not equal zero.
• For even roots, resolve to ensure the radicand is non-negative.
• Consider real-life constraints; sometimes negative values may not be practical, like in length measurements.

Understanding the domain helps clarify not just where an equation holds true but also where potential solutions cannot exist.