Problem 33 - Algebra A Combined Function

Question

Solve each equation. $$ \frac{x+1}{3}-\frac{x-1}{6}=\frac{1}{6} $$

Step-by-Step Solution

Answer

The solution to the equation is \( x = -2 \).

Step 1: Eliminate Fractions

The given equation is \( \frac{x+1}{3} - \frac{x-1}{6} = \frac{1}{6} \). To simplify this, let's eliminate the fractions by finding the least common denominator (LCD), which is 6 in this case. Multiply every term by 6: \[ 6 imes \frac{x+1}{3} - 6 \times \frac{x-1}{6} = 6 \times \frac{1}{6} \] This simplifies to: \[ 2(x+1) - (x-1) = 1 \].

Step 2: Simplify the Equation

Distribute the numbers in front of the parentheses: \[ 2(x+1) - (x-1) = 1 \] becomes:\[ 2x + 2 - x + 1 = 1 \].

Step 3: Combine Like Terms

Combine the \(x\) terms and the constants:\[ (2x - x) + (2 + 1) = 1 \] simplifies to:\[ x + 3 = 1 \].

Step 4: Solve for x

To isolate \(x\), subtract 3 from both sides of the equation:\[ x + 3 - 3 = 1 - 3 \]This results in:\[ x = -2 \].

Key Concepts

Fractions

Fractions can be a bit daunting, but they are simply numbers written in the form of one number divided by another. In the exercise you tackled, you encounter fractions represented as \( \frac{x+1}{3} \) and \( \frac{x-1}{6} \). The number on the top is the numerator, and the number on the bottom is the denominator. Here, the denominators are 3 and 6, which indicate the divisions by these numbers.

• The goal is to simplify these fractions whenever dealing with equations, as they can make calculations more straightforward.
• Eliminating fractions transforms the equation into a simpler form that is easier to solve.

Understanding fractions and being comfortable manipulating them is a key part of mastering algebraic equations. You often start by identifying them in the equation and then work towards a way to simplify or eliminate them. Working with fractions involves ensuring you've got the same denominator across all parts of the equation, which leads us to the next important concept.

Least Common Denominator

When working with fractions in equations, finding the Least Common Denominator (LCD) is vital. Think of the LCD as a common ground for all the different pieces in the fractions to meet. In the given problem, the denominators are 3 and 6. - A common denominator is useful because it allows you to eliminate fractions by converting each term to an equivalent fraction with this shared denominator.- The smallest common multiple of the denominators becomes the LCD that you use to simplify the fractions.In this exercise, the LCD is 6, since 6 is the smallest number that both 3 and 6 divide evenly into. Multiplying every term by this LCD (6) helps eliminate the fractions:\[ 6 \times \frac{x+1}{3} - 6 \times \frac{x-1}{6} = 6 \times \frac{1}{6} \] This action effectively gets rid of the denominators, leading you to a neater equation: \( 2(x+1) - (x-1) = 1 \). Now, you're working with a simpler equation with whole numbers.

Isolating Variables

Once your equation is free of fractions, the next step is to isolate the variable. In many algebra problems, isolating the variable helps you solve the equation and find your answer. The equation \( 2(x+1) - (x-1) = 1 \) already sees some simplification but still contains the variable \( x \) on both sides.To isolate \( x \):- First, distribute the numbers outside of the parentheses across the terms inside, leading to: \[ 2x + 2 - x + 1 = 1 \]- Next, combine like terms: - Combine \( x \) terms: \( 2x - x = x \) - Combine the constant terms: \( 2 + 1 = 3 \)After these steps, your equation simplifies to \( x + 3 = 1 \). Now, your task is to get \( x \) by itself:- Subtract 3 from both sides of the equation: \[ x + 3 - 3 = 1 - 3 \]This action results in \( x = -2 \). Now, you've successfully isolated \( x \) and found its value. Isolating variables by moving other elements of the equation to the opposite side is how you solve and interpret solutions in algebraic equations.