Problem 33 - 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. $$ \frac{2}{x}+3=\frac{5}{2 x}+\frac{13}{4} $$

Step-by-Step Solution

Answer

The value of \(x\) that satisfies the given equation and does not violate the restriction is \(x = -2\).

Step 1: Identify Restrictions on the Variable

The variable \(x\) is present in the denominators of the given equation. The value of \(x\) that will make the denominator of any fraction in the equation zero is identified by equating each denominator to zero. The denominator in the given equation are \(x\) and \(2x\). Therefore, equating these values to zero gives: \(x = 0\) for the first and \(x = 0\) for the second fraction, which means \(x\) cannot be \(0\).

Step 2: Clear the Fractions

To simplify the equation, clear the fractions by multiplying each term by the least common multiple (LCM) of the denominators of the fractions. The LCM of \(x\) and \(2x\) is \(2x\). So, the equation becomes: \(2x*2/x + 2x*3 = 2x*5/2x + 2x*13/4 => 4 + 6x = 5 + 13x/2\).

Step 3: Simplify Further and Solve

Combine like terms and start solving for \(x\). This results in: \(6x - 13x/2 = 5 - 4 => (12x - 13x) / 2 = 1 => -x/2 = 1\). Divide both sides by \(-1/2\) to solve for \(x\), which gives \(x = -2\). However, \(x = -2\) does not violate the restriction that \(x\) cannot be 0.

Key Concepts

Restrictions on Variables

When solving rational equations, it's crucial to identify any restrictions on the variable. These restrictions occur when a variable makes any denominator zero, which is undefined in mathematics. In our example, the variable in question is \(x\), and it appears in the denominators of the fractions in the equation.

• The denominators are \(x\) and \(2x\).
• Setting these denominators equal to zero helps find the restrictions: \(x = 0\) and \(2x = 0\), both lead to \(x = 0\).
• Thus, \(x\) cannot be 0 as it would make the denominator undefined.

Therefore, while solving, we must ensure that our solution for \(x\) does not equal 0.

Clearing Fractions

Once we have determined the restrictions, our next step is often to clear fractions in the equation. This procedure simplifies the equation considerably and is executed by multiplying every term by a common multiple.

• Multiplying removes the fractions, thus simplifying the equation.
• It's important to multiply each term of the equation to maintain equality.
• Clearing fractions helps to avoid dealing with complicated fractional arithmetic.

In the given equation, multiplying each term by \(2x\), the least common multiple (LCM), efficiently clears the fractions and results in an equation without denominators.

Least Common Multiple

The Least Common Multiple (LCM) is a useful tool when dealing with rational equations. It's the smallest multiple that two numbers share, and it's critical for clearing fractions. Here's why.

• Finding the LCM involves identifying the smallest number that both denominators divide without leaving a remainder.
• In our example, the LCM of \(x\) and \(2x\) is \(2x\).
• By multiplying every term in the equation by this LCM, we get rid of fractions, simplifying the calculation.

The process of finding and using the LCM streamlines solving rational equations and avoids complications from fractions.