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505-555 (Easy)|   Algebra|      
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Russ19
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If t = 1/x − 1, then in terms of t, x + 2/x − 1 is equal to A) t + 3/t 30 May 2021, 07:56
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82% (01:34) correct 18% (02:08) wrong based on 83 sessions

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If \(t = \frac{1}{x − 1}\), then in terms of \(t, \frac{x + 2}{x − 1}\) is equal to

A) \(\frac{t + 3}{t}\)

B) \(\frac{t}{t + 3}\)

C) \(\frac{t}{3t + 1}\)

D) \(\frac{3t + 1}{t}\)

E) \(3t + 1\)
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E
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Bunuel
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If t = 1/x − 1, then in terms of t, x + 2/x − 1 is equal to A) t + 3/t 30 May 2021, 08:30
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sjuniv32
If \(t = \frac{1}{x − 1}\), then in terms of \(t, \frac{x + 2}{x − 1}\) is equal to

A) \(\frac{t + 3}{t}\)

B) \(\frac{t}{t + 3}\)

C) \(\frac{t}{3t + 1}\)

D) \(\frac{3t + 1}{t}\)

E) \(3t + 1\)
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\(\frac{x + 2}{x − 1}=\frac{x }{x − 1}+\frac{2}{x − 1}=x*\frac{1}{x − 1}+2*\frac{1}{x − 1}=xt+2t\).

Next, from \(t = \frac{1}{x − 1}\) we get that \(x =\frac{1}{t} + 1\).

Thus, \(xt+2t=(\frac{1}{t} + 1)t+2t=1+t+2t=3t+1\).

Answer: E.

Hope it helps.
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PTibrewal
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If t = 1/x − 1, then in terms of t, x + 2/x − 1 is equal to A) t + 3/t 30 May 2021, 08:40
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t = 1/x-1; x=(1+t)/t
x+2 = (1+t)/t+2 = (3t+1)/t
x-1 = (1+t)/t - 1 = (1+t-t)/t = 1/t

Therefore, (x+2)/(x-1) = (3t+1)/t * t = 3t+1

Answer : E

Hope thats clear :)
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If t = 1/x − 1, then in terms of t, x + 2/x − 1 is equal to A) t + 3/t 30 May 2021, 08:42
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sjuniv32
If \(t = \frac{1}{x − 1}\), then in terms of \(t, \frac{x + 2}{x − 1}\) is equal to

A) \(\frac{t + 3}{t}\)

B) \(\frac{t}{t + 3}\)

C) \(\frac{t}{3t + 1}\)

D) \(\frac{3t + 1}{t}\)

E) \(3t + 1\)
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\(t = \frac{1}{x − 1}\)
\(x - 1 = \frac{1}{t}\) -(1)
\(x = \frac{1}{t} + 1\) -(2)

So, \(\frac{x + 2}{x − 1}\) (From 1 & 2)
= (1/t + 3)/(1/t) = 1 + 3t -(E)
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If t = 1/x − 1, then in terms of t, x + 2/x − 1 is equal to A) t + 3/t 02 Jun 2021, 18:56
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Showmeyaa
sjuniv32
If \(t = \frac{1}{x − 1}\), then in terms of \(t, \frac{x + 2}{x − 1}\) is equal to

A) \(\frac{t + 3}{t}\)

B) \(\frac{t}{t + 3}\)

C) \(\frac{t}{3t + 1}\)

D) \(\frac{3t + 1}{t}\)

E) \(3t + 1\)

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

So, \(\frac{x + 2}{x − 1}\) (From 1 & 2)
= (1/t + 3)/(1/t) = 1 + 3t -(E)
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Showmeyaa I have a question regarding the solving. 1/t + 3 x t = 3t + 1 x t/1 right? How to eliminate the t in the denominator? I would have selected answer D. Because 3t + 1 x t/1 results in 3t + 1 x t.
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GMAT 1: 760 Q51 V42
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If t = 1/x − 1, then in terms of t, x + 2/x − 1 is equal to A) t + 3/t 02 Jun 2021, 20:34
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\(t = \frac{1}{x - 1}\) or \(x - 1 = \frac{1}{ t}\)

=> Adding '3' on both the sides: \(x - 1 + 3 = \frac{1}{ t} + 3\)

=> \(x + 2 = \frac{1 + 3t}{ t}\)

=> \(\frac{(x + 2)}{(x - 1)} = (x + 2) * \frac{1}{ x - 1} = (x + 2) * t\)

=> \(\frac{(x + 2)}{(x - 1)} = \frac{1 + 3t}{ t} * t = 3t + 1\)

Answer E
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If t = 1/x − 1, then in terms of t, x + 2/x − 1 is equal to A) t + 3/t 02 Jun 2021, 22:40
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AbhaGanu
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sjuniv32
If \(t = \frac{1}{x − 1}\), then in terms of \(t, \frac{x + 2}{x − 1}\) is equal to

A) \(\frac{t + 3}{t}\)

B) \(\frac{t}{t + 3}\)

C) \(\frac{t}{3t + 1}\)

D) \(\frac{3t + 1}{t}\)

E) \(3t + 1\)

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

So, \(\frac{x + 2}{x − 1}\) (From 1 & 2)
= (1/t + 3)/(1/t) = 1 + 3t -(E)

Showmeyaa I have a question regarding the solving. 1/t + 3 x t = 3t + 1 x t/1 right? How to eliminate the t in the denominator? I would have selected answer D. Because 3t + 1 x t/1 results in 3t + 1 x t.
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I'm not sure if I understood your question correctly but 2 things to note here:
\(x = \frac{1}{t}+1\)
So, \(x + 2 = \frac{1}{t}+1 + 2\)
\(x + 2 = \frac{1}{t} + 3\)
Similarly, \(x - 1 = \frac{1}{t}\)
So, we are asked to find \(\frac{x + 2}{x - 1}\)

\((1/t + 3)/(1/t)\)

\((\frac{1}{t} + 3)*(t)\)

\(\frac{3t + 1}{t}*(t)\)

\(3t + 1\)

AbhaGanu, I hope I have answered your question correctly, if not please rephrase the question and let me know which step you are getting stuck on :)
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