Question

Find the real solution \((s),\) if any, of each equation. $$ \frac{x}{x+3}+\frac{1}{x-3}=1 $$

Short answer

The solution is \( x = 6 \).

Step-by-step solution

- Identify a common denominator

The common denominator for the fractions \( \frac{x}{x+3} \) and \( \frac{1}{x-3} \) is \((x+3)(x-3)\).

- Rewrite each term with the common denominator

Rewrite each fraction so they have the same denominator: \[ \frac{x(x-3)}{(x+3)(x-3)} + \frac{1(x+3)}{(x-3)(x+3)} = \frac{(x^2 - 3x) + (x + 3)}{(x+3)(x-3)} \]

- Simplify the fractions

Combine the numerators: \[ \frac{x^2 - 3x + x + 3}{(x+3)(x-3)} = \frac{x^2 - 2x + 3}{(x+3)(x-3)} \]

- Set the simplified expression equal to 1

Set the fraction equal to 1: \[ \frac{x^2 - 2x + 3}{(x+3)(x-3)} = 1 \]

- Eliminate the denominator

Multiply both sides by \((x+3)(x-3)\) to eliminate the denominator: \[ x^2 - 2x + 3 = (x+3)(x-3) \]

- Expand the equation on the right

Expand the right-hand side: \[ x^2 - 2x + 3 = x^2 - 9 \]

- Solve for x

Subtract \( x^2 \) from both sides to simplify: \[ -2x + 3 = -9 \]. Then isolate \( x \) by subtracting 3 from both sides: \[ -2x = -12 \]. Finally, divide by -2: \[ x = 6 \]

- Validate the solution

Check if \( x = 6 \) satisfies the original equation: \[ \frac{6}{6+3} + \frac{1}{6-3} = \frac{6}{9} + \frac{1}{3} = \frac{2}{3} + \frac{1}{3} = 1 \]. Since the equation is satisfied, \( x = 6 \) is a valid solution.

Key concepts

Common Denominator

The first key step in solving rational equations is to identify a common denominator. This allows us to combine fractions with different denominators easily.
In this problem, the fractions are \( \frac{x}{x+3} \) and \( \frac{1}{x-3} \). The common denominator for these fractions is \((x+3)(x-3)\).
Using a common denominator converts the equation into a more manageable form, allowing for straightforward simplification and solving.

Rewriting Fractions with the Common Denominator

After finding the common denominator, we rewrite each fraction to share this common denominator.
Our fractions become:
\[ \frac{x(x-3)}{(x+3)(x-3)} + \frac{1(x+3)}{(x-3)(x+3)} \]

This adjustment lets us combine the fractions into a single expression. Simplifying the numerators and denominators in this way allows us to focus on solving the resulting equation.

Eliminating the Denominator

With the fractions presented under a common denominator, the next step is to eliminate this denominator.
This is accomplished by multiplying both sides of the equation by the common denominator \((x+3)(x-3)\).
For example,
\[ \frac{x^2 - 2x + 3}{(x+3)(x-3)} = 1 \]
Multiplying both sides by \((x+3)(x-3)\), we get rid of the denominator:
\[ x^2 - 2x + 3 = x^2 - 9 \].
Removing the denominator reduces the equation to a simpler polynomial form, making it easier to solve for the variable.

Multiplying Both Sides

To clear a rational equation of its fractions, we multiply both sides by the common denominator.
This action allows us to deal with polynomials rather than complex fractions. After multiplying:
\[ x^2 - 2x + 3 = x^2 - 9 \]
we simplify the resulting polynomial (if necessary) and solve for the variable. Multiplication is a standard method for handling rational expressions, making complex equations simpler and more understandable.

Validating Solutions

After finding a solution, we must verify it by plugging it back into the original equation.
This ensures the solution is accurate and does not create undefined expressions.
For instance, when \( x = 6 \), substituting back, we see:
\[ \frac{6}{6+3} + \frac{1}{6-3} = \frac{6}{9} + \frac{1}{3} = \frac{2}{3} + \frac{1}{3} = 1 \]
Since both sides equal, \( x = 6 \) is a valid solution. During validation, always check if any values create zero denominators to avoid invalid solutions.