- The equation has a domain constraint of \(x
eq -4\).
- Simplifying the rational expression leads to the quadratic equation \(x^2 + 4x - 12 = 0\).
- Factoring gives the roots \(x = -6\) and \(x = 2\), both of which are valid as they satisfy the domain constraint.
- Therefore, there are two valid solutions and no extraneous solutions.
- Final answer: The equation has two valid solutions and no extraneous solutions.
Explanation
- Analyzing the Equation and Domain Constraints
We are given the rational equation \(\frac{x}{3} + \frac{x-1}{x+4} = \frac{x+3}{x+4}\). Our objective is to determine the number of valid solutions and extraneous solutions. First, we identify the domain of the equation. Since the denominator cannot be zero, we must have \(x + 4
eq 0\), which implies \(x
eq -4\). Any solution we find must satisfy this constraint. If a potential solution results in \(x = -4\), it is considered extraneous.
- Simplifying the Equation
To solve the equation, we isolate the terms with the common denominator. Subtract \(\frac{x-1}{x+4}\) from both sides:
\[\frac{x}{3} = \frac{x+3}{x+4} - \frac{x-1}{x+4}\]
Since the denominators on the right side are the same, we can combine the numerators:
\[\frac{x}{3} = \frac{(x+3) - (x-1)}{x+4}\]
\[\frac{x}{3} = \frac{x + 3 - x + 1}{x+4}\]
\[\frac{x}{3} = \frac{4}{x+4}\]
Now, we have a simple proportion that we can solve by cross-multiplying.
- Solving the Quadratic Equation
Cross-multiplying gives us:
\[x(x+4) = 3 \times 4\]
\[x^2 + 4x = 12\]
Subtracting 12 from both sides results in a quadratic equation:
\[x^2 + 4x - 12 = 0\]
We can factor this quadratic expression by looking for two numbers that multiply to -12 and add to 4. These numbers are 6 and -2. Thus, the equation factors as:
\[(x + 6)(x - 2) = 0\]
This gives us two potential solutions: \(x = -6\) and \(x = 2\).
- Verifying Solutions and Conclusion
Now, we check our potential solutions against the original domain constraint \(x
eq -4\). Both \(x = -6\) and \(x = 2\) satisfy this condition. Therefore, both values are valid solutions to the original equation. Since neither potential solution causes a division by zero, there are no extraneous solutions. We conclude that the equation has two valid solutions and no extraneous solutions.
Examples
In physics or engineering, rational equations often arise when modeling balance points or rate-based problems. For example, when calculating the effective resistance of a complex circuit or determining the time taken for two machines to finish a job, you may end up with rational expressions. Identifying extraneous solutions is critical because a mathematical result might imply a physically impossible situation, such as a negative time interval or a configuration that violates circuit components, allowing engineers to filter out invalid model predictions.