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If c ≠ −5, then (5ab+abc)/(abc^2+10abc+25ab) is equivalent to which of

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Sajjad1994
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If c ≠ −5, then (5ab+abc)/(abc^2+10abc+25ab) is equivalent to which of [#permalink] New post   24 Nov 2020, 02:19
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If \(c ≠ −5,\) then \(\frac{5ab+abc}{abc^2+10abc+25ab}\) is equivalent to which of the following?

(A) \(\frac{1}{(c+5)^2}\)

(B) \(c + 5\)

(C) \(\frac{1}{c+5}\)

(D) \(c-5\)

(E) \(\frac{1}{c-5}\)
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C
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Re: If c ≠ −5, then (5ab+abc)/(abc^2+10abc+25ab) is equivalent to which of [#permalink] New post   24 Nov 2020, 04:52
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Sajjad1994 wrote:
If \(c ≠ −5,\) then \(\frac{5ab+abc}{abc^2+10abc+25ab}\) is equivalent to which of the following?

(A) \(\frac{1}{(c+5)^2}\)

(B) \(c + 5\)

(C) \(\frac{1}{c+5}\)

(D) \(c-5\)

(E) \(\frac{1}{c-5}\)




\(\frac{5ab + abc}{abc^2 + 10abc + 25ab}\)

Taking ab common in the numerator and denominator


\(\frac{(5 + c)}{ab(c^2 + 10c + 25)} = \frac{c + 5}{(c + 5)^2} = \frac{1}{c + 5}\)


Option C

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Re: If c ≠ −5, then (5ab+abc)/(abc^2+10abc+25ab) is equivalent to which of [#permalink] New post   28 Nov 2020, 03:27
Expert Reply

Solution



Given
In this question, we are given that
    • The number c is not equal to -5

To find
We need to determine
    • The value of the given expression \(\frac{5ab+abc}{abc^2+10abc+25ab}\)

Approach and Working out
Simplifying the numerator and the denominator individually,
    • 5ab + abc = ab (5 + c)
    • \(abc^2 + 10abc + 25ab = ab (c^2 + 10c + 25) = ab (c^2 + 2 * c * 5 + 5^2) = ab (c + 5)^2\)

Hence, the value = \(\frac{ab (5 + c)}{ab(c + 5)^2} = \frac{1}{c + 5}\)
[here, we can cancel (c + 5) as we already know c is not equal to -5]

Thus, option C is the correct answer.

Correct Answer: Option C
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If c ≠ −5, then (5ab+abc)/(abc^2+10abc+25ab) is equivalent to which of [#permalink] New post   02 Dec 2020, 04:14
Hi,

Could someone please help?

I struggle with the factorization of the denominator

I understand that 5ab+abc becomes ab(5+c) and I also understand that abc^2 + 10 abc + 25 ab becomes ab(c^2+10c+5^2)

However, when the result ab(c+5)^2 is given, I don't understand where did the 10c go?

Thank you!
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giovannisumano
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Re: If c ≠ −5, then (5ab+abc)/(abc^2+10abc+25ab) is equivalent to which of [#permalink] New post   20 Jan 2021, 16:42
Sabby wrote:
Hi,

Could someone please help?

I struggle with the factorization of the denominator

I understand that 5ab+abc becomes ab(5+c) and I also understand that abc^2 + 10 abc + 25 ab becomes ab(c^2+10c+5^2)

However, when the result ab(c+5)^2 is given, I don't understand where did the 10c go?

Thank you!


\((c+5)^2 = (c+5)(c+5) = c^2+c5+5c+5^2 = c^2+\)10c\(+5^2\)
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Re: If c ≠ −5, then (5ab+abc)/(abc^2+10abc+25ab) is equivalent to which of [#permalink] New post   21 Jan 2021, 03:48
giovannisumano wrote:
Sabby wrote:
Hi,

Could someone please help?

I struggle with the factorization of the denominator

I understand that 5ab+abc becomes ab(5+c) and I also understand that abc^2 + 10 abc + 25 ab becomes ab(c^2+10c+5^2)

However, when the result ab(c+5)^2 is given, I don't understand where did the 10c go?

Thank you!


\((c+5)^2 = (c+5)(c+5) = c^2+c5+5c+5^2 = c^2+\)10c\(+5^2\)


I got it after some practice, thank you for your reply anyway!

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Re: If c ≠ −5, then (5ab+abc)/(abc^2+10abc+25ab) is equivalent to which of [#permalink]
21 Jan 2021, 03:48
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