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If \(x ≠ 1\), is y equal to \(x + 1\)?


(1) \(\frac{y - 2}{x - 1} = 1\) --> suff: y-2 = x-1 => y = x+1

(2) \(y^2 = (x + 1)^2\) --> insuff: y = x+1 or y = -(x+1)

Answer: A
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If \(x ≠ 1\), is y equal to \(x + 1\)?


(1) \(\frac{y - 2}{x - 1} = 1\)

(2) \(y^2 = (x + 1)^2\)


DS67602.01

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Condition 1)

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

Since condition 1) is equivalent to the question.
Condition 1) is sufficient.

Condition 2)
If \(x = 0, y = 1\), then \(y = x + 1 \)and the answer is 'yes'.
If \(x = -2, y = 1\), then \(y = x + 3\) and the answer is 'not.

Since condition 2) does not yield a unique solution, it is not sufficient.

Therefore, A is the answer.
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in statement two why can't we sq rt both the sides as y^2=(x+1)^2 can also be written as

y*y=(x+1)*(X+1)
which will give
y = x+1.
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If x≠1, is y equal to x+1?

(1) (y−2)/(x−1)=1
(2) y^2=(x+1)^2

1: y-2 = x-1 thus y = x+1

2:
now y = +-(x+1)
as (-x)^2 = (x)^2

thus only A
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