Problem 31 - Intermediate Algebra

Question

Solve equation. \(\frac{2}{5 x-5}+\frac{x-2}{15}=\frac{4}{5 x-5}\)

Step-by-Step Solution

Answer

The solutions are \(x = 4\) and \(x = -1\).

Step 1: Simplify the Equation

Notice that \(\frac{2}{5x-5}\) and \(\frac{4}{5x-5}\) have the same denominator. Let's subtract \(\frac{4}{5x-5}\) from both sides to simplify:\[\frac{2}{5x-5} - \frac{4}{5x-5} + \frac{x-2}{15} = 0\]This simplifies to:\[\frac{-2}{5x-5} + \frac{x-2}{15} = 0\]

Step 2: Find a Common Denominator

The common denominator for \(\frac{-2}{5x-5}\) and \(\frac{x-2}{15}\) is \(15(5x-5)\). Rewrite each fraction with this common denominator:The first term:\[\frac{-2 \times 15}{(5x-5) \times 15} = \frac{-30}{15(5x-5)}\]The second term:\[\frac{(x-2)(5x-5)}{15(5x-5)} = \frac{5x^2-10x-5x+10}{15(5x-5)} = \frac{5x^2 - 15x + 10}{15(5x-5)}\]Combine the terms:\[\frac{-30 + 5x^2 - 15x + 10}{15(5x-5)} = 0\]

Step 3: Simplify the Numerator and Solve for x

Combine the terms in the numerator:\[\frac{5x^2 - 15x + 10 - 30}{15(5x-5)} = \frac{5x^2 - 15x - 20}{15(5x-5)} = 0\]A fraction is zero only when the numerator is zero, solve:\[5x^2 - 15x - 20 = 0\]

Step 4: Solve the Quadratic Equation

Use the quadratic formula \(x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\) for \(ax^2 + bx + c = 0\). Here: \(a=5, b=-15, c=-20\).Calculate the discriminant:\[b^2 - 4ac = (-15)^2 - 4 \times 5 \times (-20) = 225 + 400 = 625\]Substitute into quadratic formula:\[x = \frac{-(-15) \pm \sqrt{625}}{2 \times 5} = \frac{15 \pm 25}{10}\]

Step 5: Calculate Roots

Solve for the values of \(x\):1. \(x = \frac{15 + 25}{10} = \frac{40}{10} = 4\)2. \(x = \frac{15 - 25}{10} = \frac{-10}{10} = -1\)So, \(x=4\) or \(x=-1\)

Step 6: Verify the Solutions

Check if the solutions make the original equation undefined:For \(x=4\): denominator is \(5(4) - 5 = 15\), valid.For \(x=-1\): denominator is \(5(-1) - 5 = -10\), valid.Both \(x=4\) and \(x=-1\) are valid solutions.

Key Concepts

Common Denominator

When working with fractions, a common denominator is crucial for combining or comparing different fractions. A common denominator is a shared multiple of the denominators of all fractions involved. It is important because it allows us to perform addition, subtraction, or comparison of fractions effectively.

In the given problem, we have two fractions: \( \frac{-2}{5x-5} \) and \( \frac{x-2}{15} \). To solve the equation, we need a common denominator so that we can combine these fractions. The common denominator here is the product of the individual denominators \( (5x-5) \) and \( 15 \), resulting in \( 15(5x-5) \).

• This common denominator allows you to rewrite each fraction in a form that shares the same denominator.
• This step simplifies the operation and enables the equation to be combined and further simplified.

Finding a common denominator may seem complicated, but it's a systematic approach. Once you have it, operations become much more manageable.

Fraction Operations

Fraction operations can be intricate but straightforward to manage with practice. In this exercise, the subtraction and addition of fractions required mastering fraction operations.
When you have fractions with a common denominator, combining them involves straightforward arithmetic:

• Subtract (or add) the numerators.
• Keep the common denominator unchanged.

For instance, \( \frac{-2}{5x-5} \) and \( \frac{4}{5x-5} \) were simplified first because they shared the same denominator. Thus, \( \frac{2}{5x-5} - \frac{4}{5x-5} = \frac{-2}{5x-5} \).

Then, for \( \frac{-30}{15(5x-5)} \) and \( \frac{(x-2)(5x-5)}{15(5x-5)} \), the process required distributing and combining like terms in the numerators:

• Expand and simplify \( \frac{(x-2)(5x-5)}{15(5x-5)} \).
• Add the simplified forms to create a single fraction.

Mastering these operations is essential for dealing with algebraic terms in fractions.

Quadratic Formula

The quadratic formula is a reliable and robust tool for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). When we approach such equations, the quadratic formula is our go-to solution:
\[ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
In our exercise, the equation \( 5x^2 - 15x - 20 = 0 \) needed solving. Here, \( a=5 \), \( b=-15 \), and \( c=-20 \).

• The discriminant, \( b^2-4ac \), calculates whether the roots are real numbers. A positive discriminant implies real solutions.
• Our discriminant was \( 625 \), indicating two distinct real solutions.
• Substitute \( a \), \( b \), and \( c \) into the quadratic formula to find \( x \).

Using the quadratic formula led to calculated results, yielding roots of \( x=4 \) and \( x=-1 \). Each solution was verified by substituting back into the original equation to ensure no undefined behavior with the denominators.