If x ≠ 1, is y equal to x + 1? (1) (y-2)/(x-1)=1 (2) y^2=(x+1)^2

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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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.

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.

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