Problem 31 - College Algebra

Question

Exercises \(31-50\) contain equations with variables in denominators. For each equation, a. Write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation. $$ \frac{4}{x}=\frac{5}{2 x}+3 $$

Step-by-Step Solution

Answer

The solution to the equation is \(x = -\frac{1}{3}\). Remember, \(x\) cannot equal 0, which doesn't conflict with our solution.

Step 1: Identify the Restrictions

The denominators of the equation are \(x\) and \(2x\). The restrictions occur where this denominator is equal to 0. For \(x\), this is where \(x=0\), and for \(2x\) this is where \(x=0\). Hence, the only restriction is \(x \neq 0\)

Step 2: Equate Denominators

Before solving the equation, one should ensure the same denominator for each term. So we multiply the second fraction by 2, resulting in \(\frac{4}{x}=\frac{10}{2x}+3\) which simplifies to \(\frac{4}{x}=\frac{5}{x}+3\).

Step 3: Solve the Equation

Clear the fractions by multiplying through by \(x\), leading to \(4 = 5 + 3x\).

Step 4: Solve for x

Rearrange the equation to get the variable alone by subtracting 5 from both sides getting \(-1 = 3x\). Divide each side of the equation by 3, so \(x = -\frac{1}{3}\).

Step 5: Check for Restrictions

The solution \(x = -\frac{1}{3}\) should be checked against the restrictions from Step 1. Since our solution is \(x = -\frac{1}{3}\) and the restriction is \(x \neq 0\), our solution is valid.

Key Concepts

Equations with Variables in Denominators

Equations where the variable is present in the denominator can present unique challenges. This means that solving them requires attention to certain details that might not be necessary in other types of equations. When you encounter such equations, start by identifying all the denominators involved.
For example, an equation like \( \frac{4}{x} = \frac{5}{2x} + 3 \) has two denominators: \( x \) and \( 2x \). An important initial step in managing these equations is understanding the nature of the variables and how they affect the equation when placed in the denominator. Pay close attention to these spots, as they define crucial elements of your solution process.
Do note that a variable in the denominator means that the equation is undefined if any of these denominators become zero.

Restrictions on Variables

When dealing with equations with variables in denominators, it's crucial to identify any restrictions on the variable before proceeding with solving the equation. This involves determining the values of the variable that would make any denominator zero, as division by zero is not allowed in mathematics.
In our example, both denominators, \( x \) and \( 2x \), become zero when \( x = 0 \). Consequently, the restriction here is that \( x eq 0 \). These restrictions are vital because they help you understand which solutions are potentially invalid or may cause the equation to become undefined.
Always write down these restrictions clearly at the start, so you can refer back to them after solving the equation to ensure that your solution is valid.

Clearing Fractions

Once you've identified the restrictions, the next step is to clear the fractions from the equation, as working with fractions can complicate solving. Clearing fractions involves manipulating the equation to eliminate fractional terms, typically through multiplication.
To clear fractions in an equation like \( \frac{4}{x} = \frac{5}{2x} + 3 \), you aim for a common denominator first, if needed, and then multiply each term by that common denominator. In this case, multiplying through by \( x \) allows you to simplify the equation to \( 4 = 5 + 3x \). Notice how clearing the fractions transforms the equation into a more straightforward arithmetic form.
Proceeding from this point, you're left with an equation that is easier to manage and solve.

Checking Solutions Against Restrictions

After solving the entire equation and finding a potential solution, it's essential to revisit the restrictions you set initially. This ensures that the solutions you've obtained are valid within the original constraints of the equation.
In the equation considered, the solution we reached was \( x = -\frac{1}{3} \). We know from our earlier work that \( x eq 0 \) was the restriction. Since \( -\frac{1}{3} \) does not violate \( x eq 0 \), we can confirm that our solution is indeed valid and lies within acceptable boundaries.
Always remember, if your solution meets the restriction criteria, it's a valid solution for the given equation. If not, the solution would need to be rejected, and you'd need to analyze further to determine valid solutions.