Question

Solve each equation, if possible. $$ \frac{2 x}{x^{2}-4}=\frac{4}{x^{2}-4}-\frac{3}{x+2} $$

Short answer

No solutions.

Step-by-step solution

Identify Common Denominators

Recognize that the denominators are all based on the same terms: \(x^2 - 4\) and \(x + 2\). Notice that \(x^2 - 4\) can be factored as \((x - 2)(x + 2)\).

Rewrite the Equation with Factored Denominators

Rewrite the original equation to use the factored form of the denominator: \[ \frac{2x}{(x-2)(x+2)} = \frac{4}{(x-2)(x+2)} - \frac{3}{x+2} \]

Combine the Fractions on the Right

Combine the fractions on the right side of the equation into a single fraction over the common denominator \((x-2)(x+2)\). To do this, multiply the numerator and the denominator of the second fraction by \(x-2\): \[ \frac{2x}{(x-2)(x+2)} = \frac{4 - 3(x-2)}{(x-2)(x+2)} \]

Simplify the Numerator

Distribute and combine like terms in the numerator on the right side: \[ 4 - 3(x-2) = 4 - 3x + 6 = 10 - 3x \]. Now the equation is: \[ \frac{2x}{(x-2)(x+2)} = \frac{10 - 3x}{(x-2)(x+2)} \]

Set the Numerators Equal

Because the denominators are the same, the numerators must be equal: \[ 2x = 10 - 3x \]

Solve for x

Combine like terms to solve for \(x\): \[ 2x + 3x = 10 \] \[ 5x = 10 \] \[ x = 2 \]

Check for Extraneous Solutions

Substitute \(x = 2\) back into the original equation to check for extraneous solutions: Since there would be a division by zero in the original equation, \(x = 2\) is not a valid solution. Therefore, there are no solutions to the equation.

Key concepts

Factoring Denominators

Factoring denominators is an important first step in solving rational equations. By breaking down a polynomial into simpler expressions, you can more easily manage and simplify the equation.

In our given equation, the denominator is initially represented as \(x^2 - 4\). Notice that \(x^2 - 4\) can be factored into \((x - 2)(x + 2)\). This helps us recognize common factors across different terms:

\[ \frac{2x}{(x-2)(x+2)} = \frac{4}{(x-2)(x+2)} - \frac{3}{x+2} \]
Factoring simplifies the problem and aligns all terms to a common denominator, which allows further simplification and comparison.

Combining Fractions

Combining fractions means putting them under a common denominator to form a single fraction. This process simplifies the equation and helps move towards solving for the variable.

After factoring the denominators, the common denominator here is \((x-2)(x+2)\). Let’s put the fractions on the right-hand side of the equation under this common denominator:

First, rewrite the right-hand side:

\[ \frac{4}{(x-2)(x+2)} - \frac{3}{x+2} \]

We notice that \(\frac{3}{x+2}\) needs to be written over a common denominator \((x-2)(x+2)\). By multiplying the numerator and the denominator by \(x-2\), we get:

\[ \frac{3(x-2)}{(x-2)(x+2)} \]
Simplifying and combining, we obtain:

\[ \frac{4 - 3(x-2)}{(x-2)(x+2)} = \frac{4 - 3x + 6}{(x-2)(x+2)} = \frac{10 - 3x}{(x-2)(x+2)} \]
So, the updated equation is:

\[ \frac{2x}{(x-2)(x+2)} = \frac{10 - 3x}{(x-2)(x+2)} \]

Checking Extraneous Solutions

Checking for extraneous solutions is a crucial step. Extraneous solutions are results that arise from the solving process but do not actually satisfy the original equation.

In our worked example, we ultimately solve \(2x = 10 - 3x\), yielding \(x = 2\). However, it’s essential to verify this potential solution in the context of the initial equation:

\[\frac{2x}{x^{2}-4}=\frac{4}{x^{2}-4}-\frac{3}{x+2} \]
Substituting \(x = 2\) into the original equation, we get:

\[x^{2}-4 = 0\]
This leads to division by zero, which is undefined. Therefore, \(x = 2\) is an extraneous solution and cannot be accepted as a valid answer.

Properly checking for extraneous solutions ensures that only the correct, viable solutions are considered. In this case, it means acknowledging that no valid solutions exist.