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Papessa [141]
2 years ago
11

Find all solutions to the following equation. If there are any extraneous solutions, identify them and explain why they are extr

aneous.(7/b+3)+(5/b-3)=(10b/b²-9)
Mathematics
1 answer:
vagabundo [1.1K]2 years ago
7 0

Answer:

b = 3, which is an extraneous solution because if we replace b = 3 in the equation, we must be 5/0 and 30/0, and there is division by 0.

Step-by-step explanation:

\frac{7}{b+3} + \frac{5}{b-3} = \frac{10b}{b^2-9}

The minus common multiply must be (b-3)*(b+3) = b² - 9

\frac{7*(b-3)}{b^2-9} + \frac{5*(b+3)}{b^2 -9}  = \frac{10b}{b^2-9}

7b - 21 + 5b + 15 = 10b

2b = 6

b = 3

Which is an extraneous solution because if we replace b = 3 in the equation, we must be 5/0 and 30/0, and there is division by 0. So this solution is invalid.

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Somebody please help me on this?! What is the correct postulate or theorem for this one..
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Answer:

Segment Addition Postulate

Step-by-step explanation:

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Step-by-step explanation:

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Read 2 more answers
Help I need it done now
Paha777 [63]
You need to submit the question you would like for us to answer.
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CAN SOMEONE JUST PLEASE ANSWER THIS ASAP FOR BRAINLIEST!!!
Rom4ik [11]
A=lw
42xy^3=(14xy)l
isolate l
divide both sides by 14xy to isolate l
(42xy^3)/ (14xy)=14xy / 14xy *l
the 14xy in the right cancel out
now you have:
42xy^3/14xy=l
42/14= 3, now just do division of powers for xy
x-x =0x, y^3-y= y^2
therefore the answer is 3y^2
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