Solve the equation, and enter solutions from least to greatest. If there is only one solution, enter "n.a." for the second solution.
[tex]
\frac{1}{x}+\frac{1}{x-3}=\frac{x-2}{x-3}
x=1 \quad \text { or } x=\square
[/tex]
Answer (2)
Multiply both sides of the equation by x ( x − 3 ) to eliminate the fractions.
Simplify the equation to obtain a quadratic equation: x 2 − 4 x + 3 = 0 .
Factor the quadratic equation to find the possible solutions: x = 1 and x = 3 .
Check for extraneous solutions; x = 3 is extraneous, so the only solution is x = 1 . The second solution is n.a. Therefore, the final answer is n . a .
Explanation
Understanding the Problem We are given the equation x 1 + x − 3 1 = x − 3 x − 2 and we need to solve for x . We are also given that x = 1 is one of the solutions. We need to find the other solution, if it exists, and enter the solutions from least to greatest. If there is only one solution, we enter 'n.a.' for the second solution.
Eliminating Fractions To solve the equation, we first multiply both sides by x ( x − 3 ) to eliminate the fractions. This gives us ( x − 3 ) + x = x ( x − 2 ) Simplifying, we get 2 x − 3 = x 2 − 2 x
Forming a Quadratic Equation Now, we rearrange the equation to form a quadratic equation: x 2 − 4 x + 3 = 0
Factoring the Quadratic We can factor the quadratic equation as follows: ( x − 1 ) ( x − 3 ) = 0
Checking for Extraneous Solutions This gives us two possible solutions: x = 1 and x = 3 . However, we must check if these solutions are valid by plugging them back into the original equation. If x = 3 , the denominator x − 3 becomes zero, which is not allowed. Therefore, x = 3 is an extraneous solution.
Final Answer The only valid solution is x = 1 . Since we are asked to provide two solutions, and we only have one valid solution, we enter 'n.a.' for the second solution.
Examples
When designing a bridge, engineers often use rational equations to model the distribution of weight and stress. Solving these equations helps them determine the optimal placement of supports to ensure the bridge's stability. Similarly, in electrical engineering, rational equations are used to analyze circuits and determine the values of components needed to achieve a desired performance. By understanding how to solve these equations, engineers can design safer and more efficient structures and systems.
The valid solution to the equation is x = 1 . The other potential solution, x = 3 , is extraneous. Therefore, the final answer is x = 1 and x = n . a .
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