Problem 59 - Intermediate Algebra

Question

Solve equation. If a solution is extraneous, so indicate. \(\frac{x+4}{2 x+14}-\frac{x}{2 x+6}=\frac{3}{16}\)

Step-by-Step Solution

Answer

The solution is \(x = 1\). The extraneous solution is \(x = -11\).

Step 1: Identify a common denominator

The given equation is \(\frac{x+4}{2x+14} - \frac{x}{2x+6} = \frac{3}{16}\). We need a common denominator for the fractions. The denominators are \(2x + 14\), \(2x + 6\), and 16. Notice that \(2x + 14 = 2(x + 7)\) and \(2x + 6 = 2(x + 3)\), indicating they share factors with constants. We choose \((x + 7)(x + 3)\) as the Least Common Denominator (LCD) for the variable parts and 16 for the constant.

Step 2: Express each term with the common denominator

Rewriting each term with the common denominator \((x + 7)(x + 3)\), we have:- \(\frac{x+4}{2(x+7)} \rightarrow \frac{8(x+4)(x+3)}{16(x+7)(x+3)}\)- \(\frac{x}{2(x+3)} \rightarrow \frac{8x(x+7)}{16(x+7)(x+3)}\)- \(\frac{3}{16} \rightarrow \frac{3(x+7)(x+3)}{16(x+7)(x+3)}\)This gives the equation \(\frac{8(x+4)(x+3) - 8x(x+7)}{16(x+7)(x+3)} = \frac{3(x+7)(x+3)}{16(x+7)(x+3)}\).

Step 3: Clear the denominators

Since the denominators are the same, we can multiply through by \(16(x+7)(x+3)\) to clear the fractions. This simplifies to:\[ 8(x + 4)(x + 3) - 8x(x + 7) = 3(x + 7)(x + 3) \]

Step 4: Expand and simplify the numerators

Expand each term:- \(8(x+4)(x+3) = 8(x^2 + 3x + 4x + 12) = 8x^2 + 56x + 96\)- \(-8x(x+7) = -8x^2 - 56x\)- \(3(x+7)(x+3) = 3(x^2 + 3x + 7x + 21) = 3x^2 + 30x + 63\)Combine the left side:\(8x^2 + 56x + 96 - 8x^2 - 56x = 96\).So we have:\(96 = 3x^2 + 30x + 63\).

Step 5: Rearrange and solve the quadratic equation

Rearrange the equation to form \(0 = 3x^2 + 30x - 33\). Factor or use the quadratic formula to solve:\(3x^2 + 30x - 33 = 0\).Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = 30\), and \(c = -33\):\(x = \frac{-30 \pm \sqrt{30^2 - 4 \times 3 \times -33}}{6}\)\\(x = \frac{-30 \pm \sqrt{900 + 396}}{6}\)\\(x = \frac{-30 \pm \sqrt{1296}}{6}\)\\(x = \frac{-30 \pm 36}{6}\).Thus, \(x = 1\) or \(x = -11\).

Step 6: Check for extraneous solutions

We substitute \(x = 1\) and \(x = -11\) back into the original equation to check:- For \(x = 1\), plug into \((2x + 14)\) and \((2x + 6)\) to see they're not zero.- For \(x = -11\), \(2x + 14\) equals 0, making the fraction undefined.Therefore, \(x = -11\) does not satisfy the equation and is extraneous.

Key Concepts

Solving Equations

Solving equations is a crucial skill in mathematics. It involves finding the value of the variable that makes the equation true. When dealing with equations containing fractions, such as \( \frac{x+4}{2x+14} - \frac{x}{2x+6} = \frac{3}{16} \), the process becomes slightly more complicated.
First, it's important to identify the least common denominator (LCD) among the fractions involved. This helps to combine the fractions and simplify the equation. Once you have an LCD, rewrite each fraction so that it has this common denominator.
After rewriting the fractions, you can clear the denominators by multiplying through by the LCD. This results in an equation without fractions, making it much easier to solve. From there, expand and combine like terms, and then rearrange the equation into a standard quadratic form.

• Quadratic equations are typically solved by factoring or using the quadratic formula.
• The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Solving these steps methodically will lead to finding the solution efficiently.

Extraneous Solutions

Extraneous solutions are potential solutions that, when substituted back into the original equation, do not satisfy it. This typically happens when solving equations that were originally more complex and involved forms like fractions.
For instance, when you solve the equation \( \frac{x+4}{2x+14} - \frac{x}{2x+6} = \frac{3}{16} \) and find solutions like \( x = 1 \) and \( x = -11 \), it is crucial to verify them.
- Substituting \( x = 1 \) into the original equation results in defined terms, so it remains a valid solution.- However, substituting \( x = -11 \) causes \( 2x + 14 \) to equal zero, which makes the fraction undefined. Therefore, \( x = -11 \) is an extraneous solution.

• Always check solutions by substituting them back into the original equation.
• Watch out for signs that could lead to division by zero, as they signal extraneous roots.

By validating your solutions, you can confirm their correctness and ensure accuracy in your work.

Least Common Denominator (LCD)

The concept of the least common denominator (LCD) is critical when working with equations involving fractions. The LCD refers to the smallest common multiple of the denominators.
Consider the equation \( \frac{x+4}{2x+14} - \frac{x}{2x+6} = \frac{3}{16} \). To solve it, you must find a common denominator that allows you to combine the fractions.
The denominators in this problem can be factorized as follows:

• \( 2x + 14 = 2(x + 7) \)
• \( 2x + 6 = 2(x + 3) \)

The LCD is then formed by multiplying the distinct factors \((x + 7)\) and \((x + 3)\), which deals with variable parts, and the constant 16.

• Finding the LCD ensures you can rewrite each fraction with a common denominator, simplifying the process of solving.
• This leads directly into clearing the fractions and solving the simplified equation.

Effectively finding and using the LCD is a potent tool in handling complex fraction-heavy equations.