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Bunuel
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If (5x^2 + 65x + 60)/(x^2 +10x - 24) = (5x + 5)/(x - 2), then which of
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28 May 2019, 02:10
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95% (hard)

Question Stats:

41% (02:18) correct 59% (02:33) wrong based on 122 sessions

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If \(\frac{(5x^2 + 65x + 60)}{(x^2 +10x - 24)} = \frac{(5x + 5)}{(x - 2)}\), then which of the following are possible values of x?


A. −60
B. −12
E. 2

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III
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If (5x^2 + 65x + 60)/(x^2 +10x - 24) = (5x + 5)/(x - 2), then which of
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 Updated on: 30 May 2019, 01:22
Originally posted by AvidDreamer09 on 28 May 2019, 02:44.
Last edited by AvidDreamer09 on 30 May 2019, 01:22, edited 2 times in total.
Bunuel
If \(\frac{(5x^2 + 65x + 60)}{(x^2 +10x - 24)} = \frac{(5x + 5)}{(x - 2)}\), then which of the following are possible values of x?


A. −60
B. −12
E. 2

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III
Show more


\(\frac{5(x^2 + 13x + 12)}{(x^2 +10x - 24)} = \frac{5(x + 1)}{(x - 2)}\)

\(\frac{(x^2 + 12x+x + 12)}{(x^2 +12x-2x - 24)} = \frac{(x + 1)}{(x - 2)}\)

\(\frac{(x(x+12)+1(x + 12))}{(x(x+12)-2(x + 12)} = \frac{(x + 1)}{(x - 2)}\)

\(\frac{(x+1)(x + 12)}{(x-2)(x+12)} = \frac{(x + 1)}{(x - 2)}\)

You cannot divide by (x-2) when (x-2)=0, or when x=2
You cannot divide by (x+12) when (x+12)=0, or when x=-12

Answer A
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If (5x^2 + 65x + 60)/(x^2 +10x - 24) = (5x + 5)/(x - 2), then which of
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28 May 2019, 03:11
AvidDreamer09
Bunuel
If \(\frac{(5x^2 + 65x + 60)}{(x^2 +10x - 24)} = \frac{(5x + 5)}{(x - 2)}\), then which of the following are possible values of x?


A. −60
B. −12
E. 2

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III


\(\frac{5(x^2 + 13x + 12)}{(x^2 +10x - 24)} = \frac{5(x + 1)}{(x - 2)}\)

\(\frac{(x^2 + 12x+x + 12)}{(x^2 +12x-2x - 24)} = \frac{(x + 1)}{(x - 2)}\)

\(\frac{(x(x+12)+1(x + 12))}{(x(x+12)-2(x + 12)} = \frac{(x + 1)}{(x - 2)}\)

\(\frac{(x+1)(x + 12)}{(x-2)(x+12)} = \frac{(x + 1)}{(x - 2)}\)

You cannot divide by (x-2) when (x-2)=0, or when x=2

Answer D
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By the same reasoning, x can't be -12
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If (5x^2 + 65x + 60)/(x^2 +10x - 24) = (5x + 5)/(x - 2), then which of
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28 May 2019, 10:21
x cannot be 2 or -12 since (x-2) and (x+12) appear in the factorization, therefore answer is A (I only)
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If (5x^2 + 65x + 60)/(x^2 +10x - 24) = (5x + 5)/(x - 2), then which of
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29 May 2019, 02:16
Mo2men
AvidDreamer09
Bunuel
If \(\frac{(5x^2 + 65x + 60)}{(x^2 +10x - 24)} = \frac{(5x + 5)}{(x - 2)}\), then which of the following are possible values of x?


A. −60
B. −12
E. 2

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III


\(\frac{5(x^2 + 13x + 12)}{(x^2 +10x - 24)} = \frac{5(x + 1)}{(x - 2)}\)

\(\frac{(x^2 + 12x+x + 12)}{(x^2 +12x-2x - 24)} = \frac{(x + 1)}{(x - 2)}\)

\(\frac{(x(x+12)+1(x + 12))}{(x(x+12)-2(x + 12)} = \frac{(x + 1)}{(x - 2)}\)

\(\frac{(x+1)(x + 12)}{(x-2)(x+12)} = \frac{(x + 1)}{(x - 2)}\)

You cannot divide by (x-2) when (x-2)=0, or when x=2

Answer D

By the same reasoning, x can't be -12
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I am new here. Since (x+12) appears both at the numerator and denominator, I thought it just cancel. Hence the answer should be D
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If (5x^2 + 65x + 60)/(x^2 +10x - 24) = (5x + 5)/(x - 2), then which of
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30 May 2019, 01:19
KONE456
If \(\frac{(5x^2 + 65x + 60)}{(x^2 +10x - 24)} = \frac{(5x + 5)}{(x - 2)}\), then which of the following are possible values of x?


A. −60
B. −12
E. 2

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III


\(\frac{5(x^2 + 13x + 12)}{(x^2 +10x - 24)} = \frac{5(x + 1)}{(x - 2)}\)

\(\frac{(x^2 + 12x+x + 12)}{(x^2 +12x-2x - 24)} = \frac{(x + 1)}{(x - 2)}\)

\(\frac{(x(x+12)+1(x + 12))}{(x(x+12)-2(x + 12)} = \frac{(x + 1)}{(x - 2)}\)

\(\frac{(x+1)(x + 12)}{(x-2)(x+12)} = \frac{(x + 1)}{(x - 2)}\)

You cannot divide by (x-2) when (x-2)=0, or when x=2

Answer D

By the same reasoning, x can't be -12

I am new here. Since (x+12) appears both at the numerator and denominator, I thought it just cancel. Hence the answer should be D
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i made the same mistake. You cannot cancel out 0. When x=-12, then x+12 is equal to 0
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If (5x^2 + 65x + 60)/(x^2 +10x - 24) = (5x + 5)/(x - 2), then which of
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16 Jan 2022, 10:41
Asked: If \(\frac{(5x^2 + 65x + 60)}{(x^2 +10x - 24)} = \frac{(5x + 5)}{(x - 2)}\), then which of the following are possible values of x?

5x^2 + 65x + 60 = 5x^2 + 5x + 60x + 60 = 5(x+12)(x+1)
x^2 +10x - 24 = x^2 + 12x - 2x - 24 = (x-2)(x+12)

\(\frac{(5x^2 + 65x + 60)}{(x^2 +10x - 24)} = \frac{5(x+12)(x+1) }{(x-2)(x+12)} = \frac{5(x+1)}{(x-2)}\) except for x = - 12
x = 2 is also not possible since denominator becomes 0.

I. −60: Possible
II. −12: Not Possible
III. 2: Not Possible

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

IMO A
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