Problem 32 - Algebra A Combined Function

Question

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

Step-by-Step Solution

Answer

The solution is \(x = 2\).

Step 1: Simplify the Equation

First, isolate the fractional terms in the equation:\[1 + \frac{3}{x+1} = \frac{x}{x-1}\]Subtract 1 from both sides to get:\[\frac{3}{x+1} = \frac{x}{x-1} - 1\]

Step 2: Combine the Right Side Fractions

To subtract 1 from \(\frac{x}{x-1}\), express 1 as \(\frac{x-1}{x-1}\):\[\frac{x}{x-1} - 1 = \frac{x}{x-1} - \frac{x-1}{x-1}\]Combine the fractions:\[\frac{x - (x-1)}{x-1} = \frac{x - x + 1}{x-1} = \frac{1}{x-1}\]The equation becomes:\[\frac{3}{x+1} = \frac{1}{x-1}\]

Step 3: Cross-Multiply

To solve the equation, cross-multiply the two fractions:\[3(x-1) = 1(x+1)\]This simplifies to:\[3x - 3 = x + 1\]

Step 4: Solve for x

Rearrange the equation to find \(x\):\[3x - x = 1 + 3\]Simplify the terms:\[2x = 4\]Divide by 2:\[x = 2\]

Step 5: Verify the Solution

Substitute \(x = 2\) back into the original equation to verify:\[1 + \frac{3}{2+1} = \frac{2}{2-1}\]Calculate each side:Left side: \(1 + \frac{3}{3} = 1 + 1 = 2\)Right side: \(\frac{2}{1} = 2\)Both sides are equal, so \(x = 2\) is a solution.

Key Concepts

Fractional Equations

Fractional equations are a type of algebraic equation where the variable is located in the denominator of one or more fractions. These equations can seem intimidating at first, but they are manageable with a systematic approach. When dealing with fractional equations, our main goal is to eliminate the fractions to simplify the solving process.

Here are some steps to consider when approaching fractional equations:

• Identify all the fractional terms in the equation.
• Focus on eliminating these fractions — often, this can be done by finding a common denominator.
• Keep in mind that any solution that makes a denominator zero is invalid, as division by zero is undefined.

In our example, the equation initially involves two fractions: \(\frac{3}{x+1}\) and \(\frac{x}{x-1}\). The solution begins by isolating and simplifying these fractions before moving to solve for the variable.

Cross-Multiplication

Cross-multiplication is a vital technique in solving fractional equations. It provides a way to eliminate fractions by creating a proportional equation that is much easier to solve. The method involves multiplying the numerator of each fraction by the denominator of the other fraction, effectively 'crossing' the products.

Consider using cross-multiplication if you encounter an equation of the form \(\frac{a}{b} = \frac{c}{d}\). Cross-multiply to get:

• \(a \times d = b \times c\)

This step results in a simpler linear equation. In the problem, after simplifying the fractions, we apply cross-multiplication to \(\frac{3}{x+1} = \frac{1}{x-1}\), resulting in \(3(x-1) = 1(x+1)\).

This equivalence removes the denominators entirely, leaving us with a straightforward linear equation to solve.

Solving Linear Equations

Solving linear equations is an essential skill in algebra. Once the fractions are eliminated from a fractional equation using cross-multiplication, you'll often be left with a linear equation. Linear equations have a straightforward structure and typically involve finding the value of a variable that makes the equation true.

Consider these essential steps in solving a linear equation:

• Combine like terms on both sides of the equation, if necessary.
• Move all terms containing the variable to one side of the equation and constants to the other.
• Simplify by combining like terms and isolating the variable.
• Solve for the variable using inverse operations (addition/subtraction, multiplication/division).

In our problem, after cross-multiplying, we simplified \(3x - 3 = x + 1\) to \(2x = 4\), eventually finding \(x = 2\).

When you've found a solution, it's crucial to check your work. Substitute the solution back into the original equation to verify that both sides are equal.