The solution to $\frac{15}{a^2-1}=\frac{5}{2 a-2}$ is
$a=$
$\square$ .
The extraneous solution is $a=$ $\square$ .
Answer (2)
Factor the denominators of the rational expressions.
Multiply both sides of the equation by the least common multiple of the denominators to eliminate fractions.
Solve the resulting equation for a , finding a = 5 .
Identify extraneous solutions by finding values of a that make the original denominators zero, which is a = 1 . The solution is 5 and the extraneous solution is 1 .
Explanation
Understanding the Problem We are given the equation a 2 − 1 15 = 2 a − 2 5 . Our goal is to find the value(s) of a that satisfy this equation and to identify any extraneous solutions that may arise during the solving process.
Factoring the Denominators First, let's factor the denominators of both fractions. We have a 2 − 1 = ( a − 1 ) ( a + 1 ) and 2 a − 2 = 2 ( a − 1 ) . Substituting these into the original equation, we get ( a − 1 ) ( a + 1 ) 15 = 2 ( a − 1 ) 5 .
Clearing the Denominators To eliminate the fractions, we multiply both sides of the equation by the least common multiple of the denominators, which is 2 ( a − 1 ) ( a + 1 ) . This gives us 15 ⋅ 2 = 5 ( a + 1 ) .
Solving for a Now, we simplify the equation: 30 = 5 a + 5 . Subtracting 5 from both sides, we get 25 = 5 a . Dividing both sides by 5, we find a = 5 .
Checking for Extraneous Solutions Next, we need to check for extraneous solutions. Extraneous solutions occur when a value of a makes the original denominators equal to zero. The original equation is undefined when a 2 − 1 = 0 or 2 a − 2 = 0 . Solving a 2 − 1 = 0 , we get a = 1 or a = − 1 . Solving 2 a − 2 = 0 , we get a = 1 . Thus, a = 1 and a = − 1 are the values that make the denominators zero.
Identifying Extraneous Solutions Since our solution is a = 5 , and 5 is not equal to 1 or − 1 , it is a valid solution. However, a = 1 is an extraneous solution because it makes the denominators in the original equation equal to zero, which is not allowed.
Final Answer Therefore, the solution to the equation is a = 5 , and the extraneous solution is a = 1 .
Examples
When solving equations involving rational expressions, it's crucial to identify and exclude extraneous solutions. These are values that satisfy the transformed equation but not the original one, often due to making a denominator zero. For instance, in circuit analysis, determining current flow through different branches may lead to rational equations. Extraneous solutions would represent physically impossible scenarios, such as infinite current or zero resistance, ensuring the model accurately reflects real-world constraints. Similarly, in chemical reaction kinetics, extraneous solutions could arise when modeling reaction rates, where negative concentrations or infinite reaction speeds are not physically viable.
The solution to the equation a 2 − 1 15 = 2 a − 2 5 is a = 5 , and the extraneous solution is a = 1 . Both values were evaluated to ensure they do not make the original denominators zero. The valid solution is 5 as it does not cause any division by zero in the original equation.
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