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If y >2, and (x−2)^2+6/y=70y+70/y+y2 which of the following could be
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10 Nov 2021, 01:26
10 Nov 2021, 01:26
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Question Stats:
75% (01:10) correct
25% (05:42) wrong based on 4 sessions
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If y >2, and \(\frac{(x−2)^2+6}{y}\) = \(\frac{70y+70}{y+y^2}\) which of the following could be the value of 2x + 3?
A) -9
B) -6
C) 0
D) 10
E) 17
A) -9
B) -6
C) 0
D) 10
E) 17
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Re: If y >2, and (x−2)^2+6/y=70y+70/y+y2 which of the following could be
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10 Nov 2021, 06:05
10 Nov 2021, 06:05
Carcass wrote:
If y >2, and \(\frac{(x−2)^2+6}{y}\) = \(\frac{70y+70}{y+y^2}\) which of the following could be the value of 2x + 3?
A) -9
B) -6
C) 0
D) 10
E) 17
A) -9
B) -6
C) 0
D) 10
E) 17
Given: \(\frac{(x−2)^2+6}{y}=\frac{70y+70}{y+y^2}\)
Factor the right side: \(\frac{(x−2)^2+6}{y}=\frac{70(y+1)}{y(1+y)}\)
Since \(y > 2\), we know that \(y + 1 \neq 0\), which means we can safely divide numerator and denominator by \(y + 1\) to get: \(\frac{(x−2)^2+6}{y}=\frac{70}{y}\)
Since both sides of the equation have the same nonzero denominator (\(y\)), we can conclude that: \((x−2)^2+6 = 70\)
Expand and simplify the left side: \(x^2-4x+10 = 70\)
Set the equation equal to \(0\) to get: \(x^2-4x-60 = 0\)
Factor: \((x-10)(x+6) = 0\)
So, either \(x = 10\) or \(x = -6\)
If \(x = 10\), then \(2x + 3 = 2(10) + 3 = 20 + 3 = 23\)
\(23\) is not among the other choices, so we'll try the other possible x-value
If \(x = -6\), then \(2x + 3 = 2(-6) + 3 = (-12) + 3 = -9\)
Answer: A
Re: If y >2, and (x−2)^2+6/y=70y+70/y+y2 which of the following could be
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10 Nov 2021, 21:56
10 Nov 2021, 21:56
\(\frac{(x-2)^2 + 6}{y} = \frac{70 + 70y}{y + y^2}\)
\(\frac{(x-2)^2 + 6}{y} = \frac{70}{y}\frac{1 + y}{y + 1}\)
\(\frac{(x-2)^2 + 6}{y} = \frac{70}{y}\)
\((x-2)^2 + 6 = 70\)
\((x-2)^2 = 64 = (8)^2\)
Which means, \(|x-2| = 8\)
Therefore \(x - 2 = 8\) and \(-(x - 2) = 8\)
So, \(x = \) 10 and -6
Now, \(2x + 3 = 2(10) + 3\) and \(2(-6) + 3 = \)23 and -9
Hence, Answer is A
\(\frac{(x-2)^2 + 6}{y} = \frac{70}{y}\frac{1 + y}{y + 1}\)
\(\frac{(x-2)^2 + 6}{y} = \frac{70}{y}\)
\((x-2)^2 + 6 = 70\)
\((x-2)^2 = 64 = (8)^2\)
Which means, \(|x-2| = 8\)
Therefore \(x - 2 = 8\) and \(-(x - 2) = 8\)
So, \(x = \) 10 and -6
Now, \(2x + 3 = 2(10) + 3\) and \(2(-6) + 3 = \)23 and -9
Hence, Answer is A
