[tex]\frac{2 a-2}{a-3}-\frac{a-3}{a-2}=\frac{a^2-8 a+27}{a^2-5 a+6}[/tex]
Answer (2)
Multiply both sides of the equation by ( a − 2 ) ( a − 3 ) to eliminate the denominators.
Expand and simplify the equation to get a 2 − 5 = a 2 − 8 a + 27 .
Solve for a : 8 a = 32 , which gives a = 4 .
Verify that a = 4 is not an extraneous solution. The final answer is 4 .
Explanation
Analyze the problem We are given the equation a − 3 2 a − 2 − a − 2 a − 3 = a 2 − 5 a + 6 a 2 − 8 a + 27 We need to solve for a . First, note that a 2 − 5 a + 6 = ( a − 2 ) ( a − 3 ) . Thus, we must have a = 2 and a = 3 .
Clear denominators and expand To solve the equation, we multiply both sides by ( a − 2 ) ( a − 3 ) to clear the denominators: ( a − 2 ) ( a − 3 ) ( a − 3 2 a − 2 − a − 2 a − 3 ) = ( a − 2 ) ( a − 3 ) ( ( a − 2 ) ( a − 3 ) a 2 − 8 a + 27 ) ( 2 a − 2 ) ( a − 2 ) − ( a − 3 ) ( a − 3 ) = a 2 − 8 a + 27 Expanding the terms, we get: ( 2 a 2 − 4 a − 2 a + 4 ) − ( a 2 − 6 a + 9 ) = a 2 − 8 a + 27 2 a 2 − 6 a + 4 − a 2 + 6 a − 9 = a 2 − 8 a + 27 a 2 − 5 = a 2 − 8 a + 27
Simplify and solve for a Now, we simplify the equation: a 2 − 5 = a 2 − 8 a + 27 Subtract a 2 from both sides: − 5 = − 8 a + 27 Add 8 a to both sides and add 5 to both sides: 8 a = 32 Divide by 8: a = 4
Check the solution We need to check if the solution satisfies the conditions a = 2 and a = 3 . Since a = 4 , the conditions are satisfied. Therefore, the solution is a = 4 .
Examples
Understanding how to solve rational equations is crucial in many real-world applications, such as calculating the flow rate in pipes or determining the optimal concentration of a chemical solution. For instance, if you're designing a water distribution system, you might use rational equations to model the relationship between pipe diameter, water pressure, and flow rate. By solving these equations, you can ensure that the system delivers the required amount of water to different locations efficiently and safely. Similarly, in chemical engineering, rational equations can help you optimize reaction rates and yields by adjusting the concentrations of reactants.
To solve the equation, we eliminate the denominators, expand and simplify the expression, and find that a = 4 is the solution. This value meets all the necessary restrictions. Hence, the solution is confirmed to be valid.
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