Question

Equations with Unknown in Denominator. \(\frac{x+5}{x-2}=5\)

Short answer

The solution to the equation \(\frac{x+5}{x-2}=5\) is \(x = \frac{15}{4}\).

Step-by-step solution

Isolate the fraction

Given the equation \(\frac{x+5}{x-2}=5\), note that the fraction is already isolated on one side of the equation.

Eliminate the denominator

To eliminate the denominator, multiply both sides of the equation by \(x-2\). This gives us \(x+5 = 5(x-2)\).

Distribute and simplify

Distribute the 5 on the right side of the equation to get \(x + 5 = 5x - 10\). Then, simplify the equation by combining like terms.

Isolate the variable

Subtract x from both sides to get \(5 = 4x - 10\). Then, add 10 to both sides to isolate \(4x\), resulting in \(4x = 15\).

Solve for the variable

Divide both sides by 4 to solve for \(x\), obtaining \(x = \frac{15}{4}\).

Check for extraneous solutions

Substitute \(x = \frac{15}{4}\) back into the original equation to ensure that the denominator does not equal zero and the equation holds true. There are no restrictions, so the solution is valid.

Key concepts

Unknown in Denominator

Rational equations often pose a unique challenge due to the presence of variables in the denominator. In the equation \(\frac{x+5}{x-2}=5\), the variable \(x\) in the denominator introduces the possibility of division by zero, which is undefined in mathematics. Therefore, whenever you encounter an unknown in the denominator, it is crucial to remember that the values which make the denominator zero are not part of the solution set. This caution is especially important when considering the domain of the function before proceeding with solving the equation.

When solving such an equation, one typically starts by identifying values for the variable that would invalidate the equation (like \(x = 2\) in this case, which would make the denominator zero) and then excluding those values from the solution set. This pre-solving analysis ensures the solutions are meaningful and valid within the domain of rational functions.

Eliminate the Denominator

To simplify the process of solving rational equations, the first step is often to eliminate the denominator. This allows the equation to be converted into a simpler form that is more straightforward to solve. To do this, you multiply both sides of the equation by the least common denominator (LCD) of all the fractions involved.

In the given example \(\frac{x+5}{x-2}=5\), the denominator \(x-2\) is multiplied across both sides of the equation to cancel out the fraction: \(x+5=5(x-2)\). This step transforms the rational equation into a linear equation without fractions, which is generally easier to solve. It's important to note that when you multiply by the denominator, you are assuming that it is not zero since you cannot multiply by zero, which further echoes the importance of considering the domain beforehand.

Equation Simplification

After clearing the fractions from the equation, simplification is an essential process. It involves expanding expressions, combining like terms, and rearranging the equation to isolate the variable. For the equation \(x+5=5(x-2)\), distributing the 5 across \(x-2\) yields \(5x-10\) and simplifying further by combining like terms gives \(5=4x-10\).

Equation simplification may include various arithmetic operations and should be performed with careful attention to maintaining the balance on both sides of the equation. Mistakes in simplification can lead to incorrect solutions, thus it is essential to do each step methodically, checking work as you go. As equations become simpler, the path to finding the value of the variable becomes clearer.

Extraneous Solutions

Extraneous solutions are non-valid solutions that emerge during the process of solving equations, most often as a result of multiplying both sides of an equation by an expression that contains a variable. They are called 'extraneous' because they do not satisfy the original equation even though they may emerge through the algebraic process.

For example, if we multiply both sides of an equation by \(x-2\), and later find \(x=2\) as one of the solutions, this solution would be extraneous because it creates a division by zero in the original equation. To remedy this, it's critical to always check potential solutions in the original equation as performed in Step 6 of the problem. If a solution makes any denominator zero or does not satisfy the original equation, it must be discarded. This final verification is an important step in validating the correctness of the solutions and ensuring they are not extraneous.