Problem 35 - Elementary Algebra

Question

\(-1-\frac{5}{x-2}=\frac{3}{x-2}\)

Step-by-Step Solution

Answer

The solution to the equation is \(x = -6\).

Step 1: Recognize Common Denominators

In the given equation, \(-1-\frac{5}{x-2}=\frac{3}{x-2}\), the fractions \( \frac{5}{x-2}\) and \( \frac{3}{x-2}\) share the same denominator, \(x-2\). This allows us to work on simplifying the equation by getting rid of the denominators.

Step 2: Combine Like Terms

Since the fractions on both sides have the same denominator, combine the equations to remove the fractions: \[-1 - \frac{5}{x-2} = \frac{3}{x-2}\] becomes \[-1 = \frac{3}{x-2} + \frac{5}{x-2}\].This results in the same equation \(-1 = \frac{3 + 5}{x-2}\), where \(-1 = \frac{8}{x-2}\).

Step 3: Clear the Denominator

To eliminate the fraction, multiply both sides of the equation by \(x-2\):\[ -1 \cdot (x-2) = 8 \] leads to \[-(x-2)=8\].

Step 4: Simplify the Equation

Distribute the \(-1\) on the left side:\[-x + 2 = 8\].

Step 5: Solve for x

Subtract 2 from both sides to isolate \(-x\):\[-x = 8 - 2 \] results in \[-x = 6\].Then multiply both sides by \(-1\) to solve for \(x\):\[x = -6\].

Key Concepts

Solving Rational Equations

Rational equations are equations that include fractions with polynomials in their numerators, denominators, or both. Solving them involves finding a value for the variable that makes the equation true.
The solution process typically involves a few steps:

• First, ensure the fractions are clear or can be manipulated easily by identifying common denominators.
• Next, manipulate the equation to combine terms and remove fractions if necessary.
• Solve the simplified equation for the variable.

In the exercise \(-1-\frac{5}{x-2}=\frac{3}{x-2}\), we solve the rational equation by eliminating fractions through strategic operations. The goal is to transform the equation into a more straightforward form that can be solved as a simple algebraic equation.

Common Denominators

Identifying common denominators in rational equations simplifies the process of solving them. When fractions have the same denominator, you can combine or compare them more easily.
In the example \(-1-\frac{5}{x-2}=\frac{3}{x-2}\), both fractions share the denominator \(x-2\).
Here's how common denominators help:

• You can directly subtract or add fractions on either side of an equation when they share a denominator.
• This step can lead to the elimination of fractions, simplifying the equation significantly.

The common denominator allows for direct actions, like combining numerators and thus advancing to simpler equation forms.

Combining Like Terms

After establishing a common denominator, combining like terms is essential to streamline the equation. "Like terms" are terms that have identical variable parts and can be combined through addition or subtraction.
In this algebra problem, we focus on simplifying the equation to make problem-solving easier.In the equation\(-1= \frac{3}{x-2} + \frac{5}{x-2}\),
we combine the numerators:\ \(\frac{3+5}{x-2}\).

• Add the like terms (numerators) directly when denominators are the same: resulting in \(\frac{8}{x-2}\).
• This yields a simpler expression that retains variable position through numeric simplification.

Combining like terms helps streamline expressions, making them easier to work with, especially when working towards isolating the variable.

Clearing Fractions in Equations

Clearing fractions is often a necessary step in algebra for simplifying and solving equations. When you clear fractions, you're aiming to make the equation easier to understand and solve by eliminating the fractional parts.
In our problem\(-1= \frac{8}{x-2}\),
the goal is to remove the fraction by multiplying both sides by the denominator, which is \(x-2\).Here’s the process:

• Multiply every term by the common denominator to eliminate fractions: \(-1 \times (x-2) = 8\).
• After this, you're left with a standard equation to solve: \(-x + 2 = 8\).

This procedure leads to a straightforward algebraic expression where the main fraction component has been cleared, simplifying the task of solving for the variable \(x\).