Problem 38 - 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. $$ \frac{4}{x}=\frac{9}{5}-\frac{7 x-4}{5 x} $$

Step-by-Step Solution

Answer

The restriction on the variable x is \(x≠0\). The solution of the equation \(\frac{4}{x}=\frac{9}{5}-\frac{7x-4}{5x}\) is \(x=8\), which does not violate the identified restriction.

Step 1: Identify Denominator Restrictions

Find the value of \(x\) that makes each denominator zero. Setting \(x=0\) and \(5x=0\), we find that the restriction on the variable is \(x≠0\).

Step 2: Finding Equivalent Fractions

Rewrite the equation with a common denominator for all three fractions. A common choice would be \(5x\), so the equation becomes \(\frac{20x}{5x}=\frac{9x}{5}-\frac{7x-4}{5x}\).

Step 3: Simplify the Equation

Remove the common denominator by multiplying all the terms by \(5x\) which gives \(20=9x-7x+4\). Simplify this to get \(20= 2x +4\).

Step 4: Solving for \(x\)

Solving the final equation \(20= 2x +4\) for \(x\) gives us \(x=8\).

Key Concepts

Denominator Restrictions

When working with rational equations, paying attention to denominator restrictions is important. Denominator restrictions arise because if a fraction has a denominator of zero, it becomes undefined. This means that there are specific values for which the variable cannot take on certain numbers.
In the given problem, we have fractions with denominators of \(x\) and \(5x\). To find restrictions, set each denominator equal to zero:

• For \(x=0\), the denominator becomes undefined.
• For \(5x=0\), solving gives us \(x=0\) as well.

After determining where the denominators are zero, it is clear that \(x eq 0\) is our restriction. This means that \(x\) cannot equal zero since it would invalidate the expression by making the denominators zero.

Common Denominator

To solve equations with fractions efficiently, finding a common denominator simplifies the process. A common denominator is essentially the least common multiple of the denominators in the given fractions, allowing you to combine them easily.
In the original equation, we have denominators \(x\) and \(5x\). The least common multiple of these is \(5x\).
By choosing \(5x\) as the common denominator, the equations align neatly:

• Rewrite each term so that it shares this common denominator.
• This involves multiplying both sides of the equation by \(5x\) to eliminate fractions.
• This simplification step is crucial before solving the equation.

Thus, our equation is reformulated with a unified denominator of \(5x\), enabling further simplification and solving in subsequent steps.

Equivalent Fractions

Equivalent fractions are fractions that represent the same value, even if they appear different. Manipulating fractions to have a common denominator is a way of creating equivalent fractions, enabling comparisons or combinations.
In the problem, the task is to adjust each term to have the common denominator \(5x\).
This adjusts the initial expression:

• Convert \(\frac{4}{x}\) to \(\frac{20x}{5x}\).
• \(\frac{9}{5}\) remains \(\frac{9x}{5x}\) given the context.
• \(\frac{7x-4}{5x}\) stays unchanged, maintaining its common denominator form.

By ensuring that the fractions have equivalent forms, you set the stage to eliminate the denominator in bulk, combining terms and effectively moving towards solving the equation.

Solving Equations

Once the equation is simplified to its core form, solving it becomes a straightforward algebra problem. The task is to isolate the variable to find its value that satisfies the equation.
In our modified equation: \( 20 = 9x - 7x + 4 \), follow these steps:

• Simplify by combining like terms on one side to obtain: \( 20 = 2x + 4 \).
• Subtract 4 from both sides to get: \(16 = 2x\).
• Finally, divide both sides by 2 to isolate \(x\): \(x = 8\).

Through these steps, you solve the rational equation successfully, ensuring the solution sits within the confines of the denominator restrictions and maintains equivalent forms throughout the process.