Question

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

Short answer

No solution

Step-by-step solution

- Identify a common denominator

The common denominator in the equation \[\frac{x}{x-2}+3=\frac{2}{x-2}\] is \(x-2\).

- Combine like terms

To eliminate the denominator, multiply every term by \(x-2\): \[x + 3(x-2) = 2\]

- Simplify the equation

Expand and simplify: \[x + 3x - 6 = 2\], which simplifies to \[4x - 6 = 2\]

- Solve for x

Isolate \(x\) by adding 6 to both sides: \[4x = 8\] then divide by 4: \[x = 2\]

- Check for extraneous solutions

Since \(x = 2\) results in division by zero in the original equation, it is an extraneous solution. The equation has no solution.

Key concepts

Common Denominator

In rational equations, it's crucial to find a common denominator to simplify the equation. A common denominator is a shared multiple of the denominators in each rational expression. In our exercise, the denominators are both \(x-2\). Once identified, you can use this common denominator to eliminate fractions by multiplying each term by it. This will make your equation easier to manipulate and solve.

Combining Like Terms

After eliminating the denominators, you might find similar terms that can be combined. Like terms have identical variable parts, making them easy to add or subtract. In our example, after multiplying by the common denominator \(x-2\): \(x + 3(x-2) = 2\) becomes \[ x + 3x - 6 = 2 \]. Combining \(x\) and \(3x\) yields \(4x - 6 = 2\). This step simplifies the equation, making it easier to solve for the variable.

Simplifying Equations

Simplifying equations involves reducing them to their simplest form. This includes expanding expressions, combining like terms, and performing basic arithmetic operations. From our exercise, once we have \[4x - 6 = 2\], we add 6 to both sides and get \[4x = 8\]. Dividing both sides by 4 simplifies it further to \[x = 2\]. Simplifying is key in progressing to find the actual solution.

Extraneous Solutions

Extraneous solutions are results that, despite emerging from the solving process, do not satisfy the original equation. These often arise when dealing with rational equations that involve variable denominators. In our example, \(x = 2\) creates a denominator of zero in the original equation \(\frac{x}{x-2} + 3 = \frac{2}{x-2}\). Since division by zero is undefined, \(x = 2\) is classified as an extraneous solution. Therefore, the original equation actually has no valid solution.