Problem 31 - College Algebra Essentials

Question

Contain rational 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 value for the variable that makes the denominator zero or undefined is \(x = 0\). By solving the equation while keeping this restriction in mind, the value of the variable \(x\) that satisfies the given equation is \(x = 0.5\).

Step 1: Identify Restrictions

The values for the variable that would result in a denominator of zero are calculated by setting \(x = 0\) and \(2x = 0\). For both equations, the variable \(x\) equals 0. Therefore, the only restriction for the variable is \(x ≠ 0\)

Step 2: Normalize the Equation

Since there's a restriction \(≠ 0\), we can multiply every term in the equation by \(2x\), the least common multiple of our denominators \(x\) and \(2x\). This results in: \(2x*\frac{4}{x} = 2x*\frac{5}{2x} + 2x*3\) => \(8 = 5 + 6x\).

Step 3: Solve for x

Solving for x turns the equation into a simple linear equation. First, move \(5\) to the left side by subtracting \(5\) from each side: \(3 = 6x\). Then, divide each side by \(6\) to solve for \(x\): \(x = 0.5\).

Key Concepts

Variable Restrictions

When solving rational equations, it's crucial to identify any restrictions on the variable. This means finding out which values make the denominator zero, as these can cause the equation to be undefined. In our example,

• there are denominators with terms like \(x\) and \(2x\).
• We set each to zero: \(x = 0\) and \(2x = 0\), which both indicate \(x eq 0\).

This tells us that the variable \(x\) cannot take the value 0 in this equation. Making a denominator zero would make the whole term undefined. Keeping these restrictions in mind ensures that the solution is valid.

Denominator Zero

Understanding why a denominator cannot be zero is key in algebra, especially when dealing with rational equations. A rational expression becomes undefined when the denominator is zero, similar to how division by zero doesn't make sense in mathematics. For example, if we have \(\frac{1}{x}\), the expression is undefined when \(x = 0\).
In our exercise, setting the denominator to zero helped us find \(x eq 0\) as a restriction. Avoiding zero ensures equations remain logical and solvable, keeping the math "clean". It's always a first step to check and eliminate any potential values that disrupt the equation before solving further.

Solving Rational Equations

Once variable restrictions are clear, solving rational equations becomes straightforward. A rational equation has variables in the denominator, such as our equation \(\frac{4}{x} = \frac{5}{2x} + 3\). To solve, follow these typical steps:

• First, identify any restrictions and remember them.
• Next, eliminate the denominators by finding a common multiple, like \(2x\), and multiply through the equation to "clear" the fractions.
• Then, simplify the resultant equation. In our problem, multiplying by \(2x\) gives \(8 = 5 + 6x\).

After clearing the fractions, it’s just a matter of solving the simplified equation. Here, we shifted 5 to the other side and divided by 6, which gave us \(x = 0.5\). Just remember to check against the restrictions. Since \(x = 0.5\) isn't a restricted value, it's acceptable.