Problem 32 - 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. $$ \frac{5}{x}=\frac{10}{3 x}+4 $$

Step-by-Step Solution

Answer

The solution to the provided equation is \(x = \frac{5}{12}\) and the restriction on the variable is \(x ≠ 0\).

Step 1: Identify the Restrictions

We need to find the value of \(x\) that makes the denominators \(x\) and \(3x\) zero. This is straightforward since if \(x=0\) then \(x\) and \(3x\) are both zero. So, the restriction for the variable \(x\) in this equation is \(x ≠ 0\).

Step 2: Find a Common Denominator

The common denominator of the fractions \(\frac{5}{x}\) and \(\frac{10}{3x}\) is \(3x\). This is because \(3x\) is the least common multiple of \(x\) and \(3x\).

Step 3: Rewrite the Equation Using the Common Denominator

Now we can rewrite the equation \(\frac{5}{x}=\frac{10}{3x}+4\) as \(\frac{15}{3x}=\frac{10}{3x}+4\).

Step 4: Clear the Fractions

Let's clear fractions by multiplying every term in the equation by \(3x\), which gives \(15 = 10 + 12x\).

Step 5: Simplify and Solve for the Variable

To simplify, move 10 to left side of the equation: \(15 - 10 = 12x\) so \(5 = 12x\). Solving for \(x\), we divide both sides by 12, to get \(x = \frac{5}{12}\).

Key Concepts

Identifying Restrictions in Rational Equations

When working with rational equations that include variables in the denominators, it's essential to determine the values that make the denominators zero—these are called restrictions.

Rational equations are undefined when their denominators are zero, hence the importance of identifying restrictions early in the problem-solving process. In the example equation, we have denominators of \(x\) and \(3x\). It's clear that when \(x=0\), both denominators will be zero, leading to an undefined equation. Thus, the restriction in this case is \(x eq 0\).

It's important not only to identify these restrictions but also to remember them as you solve the equation. Typically, the final answer should exclude any value that violates these restrictions, even if it mathematically satisfies the equation.

Finding Common Denominators

To ease the process of solving equations with fractions, we must find common denominators for all terms involving fractions. A common denominator is the least common multiple of all denominators in the equation.

In our exercise, the fractions are \(\frac{5}{x}\) and \(\frac{10}{3x}\). To combine these fractions, or to eliminate them, one should find a shared denominator, which in this case is \(3x\). Once a common denominator is identified, we can rewrite the fractions as equivalent fractions with the common denominator, laying the groundwork for the next step of clearing fractions from the equation.

Clearing Fractions from Equations

One effective technique for dealing with fractions in an equation is to eliminate them altogether. This can be achieved by multiplying every term in the equation by the common denominator identified in the previous step.

In our example, when we multiply each term by the common denominator \(3x\), the denominators are canceled out, leaving us with a simpler, non-rational equation: \(15 = 10 + 12x\). This is a straightforward linear equation without fractions, thus simplifying the problem significantly and paving the way to find the solution for \(x\).

Solving Rational Equations

After clearing the fractions, the rational equation is transformed into a linear or quadratic equation that's easier to solve. The resulting equation should be in a familiar form, effortlessly maneuverable using basic algebraic principles.

In our provided exercise, the equation simplifies to \(5 = 12x\) after clearing fractions and moving terms around. To find the solution for \(x\), we divide both sides by 12 to isolate the variable, rendering \(x = \frac{5}{12}\). Always check your final answer against the initially identified restrictions. If the answer is not in the restricted set, then it is valid. In our case, \(\frac{5}{12}\) does not violate the restriction \(x eq 0\), so it is indeed the solution to the equation.