Answers26-27 CFISD Algebra II B CRAdding and Subtracting Rational Expressions

Rational Equations Answers

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⇒ -4

Answer:

-8, -4

3

Name the extraneous solution(s) to the equation.✔ x = 0x = 1x = 8x = 0, x = 1, and x = 8x = 1 and x = 8There are no extraneous

A
✔ x = 0
B
x = 1
C
x = 8
D
x = 0, x = 1, and x = 8
E
x = 1 and x = 8
F
There are no extraneous
4

x = 1 is a solution.x = 0 is a solution.x = –1 is a solution.x = –6 is a solution.

A
x = 1 is a solution.
B
x = 0 is a solution.
C
x = –1 is a solution.
D
x = –6 is a solution.
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Option A
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Answer:

The student is not correct in claiming that 𝑎 = − 2 a=−2 is a solution. The original equation is 3 𝑎 + 2 − 6 𝑎 𝑎 2 − 4 = 1 𝑎 − 2 a 2 −4 3a+2−6a ​ = a−2 1 ​ . Simplifying the numerator on the left gives − 3 𝑎 + 2 −3a+2, and factoring the denominator gives ( 𝑎 − 2 ) ( 𝑎 + 2 ) (a−2)(a+2). Multiplying both sides by 𝑎 − 2 a−2 (assuming 𝑎 ≠ 2 a  =2) leads to the equation − 3 𝑎 + 2 𝑎 + 2 = 1 a+2 −3a+2 ​ =1, which simplifies to 𝑎 = 0 a=0. The value 𝑎 = − 2 a=−2 is not valid because it makes the denominators 𝑎 2 − 4 a 2 −4 and 𝑎 + 2 a+2 equal to zero, resulting in division by zero. Therefore, 𝑎 = − 2 a=−2 is an extraneous solution, and the correct solution to the original equation is 𝑎 = 0 a=0.

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Option A
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8

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Answers:
Selection 1:x = 1 and x = 8
Selection 2:x = 0
10

Which of the following did you include in your answer?

A
No, the student did not check the solution to the derived equation in the original equation.
B
No, a possible solution is a = –2, but it does not check in the original equation because it makes two of the denominators equal to zero.
C
No, a = –2 is an extraneous solution, and there is no solution to the original rational equation.

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