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My logic is following: (1) not sufficient equation is virtually a circle around (0,0) point with the radius 1. If y ≠ 1, than x can be almost anything within following limits [-1;0) & (0;1] (2) not sufficient: consider y = 0 => x=1; y = 3 => x=-2

(1)+(2) Substitute Y from (2) in (1): \(X^2 + (1-X)^2=1\) ... \(2X (X-1) = 0\)

My logic is following: (1) not sufficient equation is virtually a circle around (0,0) point with the radius 1. If y ≠ 1, than x can be almost anything within following limits [-1;0) & (0;1] (2) not sufficient: consider y = 0 => x=1; y = 3 => x=-2

(1)+(2) Substitute Y from (2) in (1): \(X^2 + (1-X)^2=1\) ... \(2X (X-1) = 0\)

Hense X= 0 or X = 1.

Not sufficient?

PLS Help!

The point is that x=0 is not a valid solution. For x = 0, from y = 1 - x it follows that y = 1 but we are told in the stem that y ≠ 1. Thus x can only be 1. Sufficient.

My logic is following: (1) not sufficient equation is virtually a circle around (0,0) point with the radius 1. If y ≠ 1, than x can be almost anything within following limits [-1;0) & (0;1] (2) not sufficient: consider y = 0 => x=1; y = 3 => x=-2

(1)+(2) Substitute Y from (2) in (1): \(X^2 + (1-X)^2=1\) ... \(2X (X-1) = 0\)

Hense X= 0 or X = 1.

Not sufficient?

PLS Help!

The point is that x=0 is not a valid solution. For x=1, from y=1-x it follows that y=1 but we are told in the stem that y≠1. Thus x can only be 1. Sufficient.

Answer: C.

Does this make sense?

bunuel, i have a doubt here... x^2 + y^2 = 1 and y = 1-x => x+y = 1 squaring both sides => x^2 + y^2 + 2xy = 1 substituting x^2 + y^2 = 1 => 1 + 2xy = 1 => 2xy = 0 => xy = 0 now x can be anything on the basis of the above equatn hence getting E. what am i missing

secondly, what might be the level of this question??

Re: If y is not equal to 1, is x = 1? [#permalink]

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26 Nov 2014, 02:16

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My quick thought. Look at the picture, there are only two points satisfy the condition (1,0) and (0,1) but y is not equal 1 then only x = 1, y = 0 satisfy the conditions. The answer is C, and you can come up with the answer in 30 seconds.

Re: If y is not equal to 1, is x = 1? [#permalink]

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vasili wrote:

If y is not equal to 1, is x = 1?

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

Quote:

Please help.

In the question it is given that y can take ANY VALUE except 1.

If we look at the first statement \(x^2 + y^2 = 1\), we are provided with a equation of a circle with center at (0,0) and radius 1. This circle will intersect x axis at points (1,0) and (-1,0) and y axis at (0,1) and (0,-1). Over the entire circumference of the circle there are infinite number of points that will satisfy statement 1 even if we exclude one specific point where value of y is 1. Therefore statement ONE alone is not sufficient.

Moving on to second statement \(y = 1 - x\), we are provided with a equation of a line passing through points (0,1) and (1,0). Yet again there are infinite number of points that will satisfy statement 2 even if we exclude one specific point where value of y is 1. Therefore statement TWO alone is not sufficient.

Now when we combine both first and second statement we get two specific points [i.e. (0,1) and (1,0)] where these curves meet. We are given that y is not equal to 1. Hence we are left with one unique value (1,0). Therefore answer option (c) is correct.

I have also uploaded the image of containing both the figures.

----------------------------------------------------------- +1 Kudos if you find the reply useful.

Attachments

File comment: Figure of circle and line

Figure circle and line.JPG [ 10.85 KiB | Viewed 23756 times ]

My logic is following: (1) not sufficient equation is virtually a circle around (0,0) point with the radius 1. If y ≠ 1, than x can be almost anything within following limits [-1;0) & (0;1] (2) not sufficient: consider y = 0 => x=1; y = 3 => x=-2

(1)+(2) Substitute Y from (2) in (1): \(X^2 + (1-X)^2=1\) ... \(2X (X-1) = 0\)

Hense X= 0 or X = 1.

Not sufficient?

PLS Help!

The point is that x=0 is not a valid solution. For x=1, from y=1-x it follows that y=1 but we are told in the stem that y≠1. Thus x can only be 1. Sufficient.

Answer: C.

Does this make sense?

bunuel, i have a doubt here... x^2 + y^2 = 1 and y = 1-x => x+y = 1 squaring both sides => x^2 + y^2 + 2xy = 1 substituting x^2 + y^2 = 1 => 1 + 2xy = 1 => 2xy = 0 => xy = 0 now x can be anything on the basis of the above equatn hence getting E. what am i missing

secondly, what might be the level of this question??

would really appreciate your help. thanks.

The very post you are quoting answers the question.

xy = 0 means that x = 0 or y = 0. But x = 0 is not a valid solution. If x = 0, from y = 1 - x it follows that y=1 but we are told in the stem that y≠1. Thus y = 0 and from y = 1 - x it follows that x = 1.

As for the difficulty level of the question: stats in the original post say that it's a 700 level question.
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Hi I am not sure if this is correct solution, someone kindly help

If y is not equal to 1, is x = 1? 1) x^2 + y^2 = 1 2) y = 1 - x

------------------------------------------------ Statement1 x^2 + y^2 = 1 y=0 , x=1 --> x=1 Yes but y and x need not be integers so y = x = sqrt(0.5) --> x=1 No ------------------------------------------------ Statement2 y = 1 - x x=1, Y=0 --> x=1 Yes x=0.5 , Y = 0.5 --> x=1 No -------------------------------------------------- 1+2 Here I checked all ranges

Case1 If x=0 then Y=1 which can not be true So x=1 and Y=0 x=1 Yes

Case2 x between 0 and 1 x=0.1 y=0.9 x=0.5 y=0.5 x=0.9 y=0.1 As we know number between 0 and 1 , then x2<x so in this case x^2 + y^2 <> 1 so we cant consider this range

Case3 x between 0 and -1 this will be same as case2 , because square will be same for + and -ve

Case4 x >=1 here again x^2 + y^2 <> 1 so we cant consider this range

x<=-1 here again x^2 + y^2 <> 1 so we cant consider this range

st.1 here different values of x and y can satisfy the equation \(x^2 + y^2 =1\). for example

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

as x is not equal to 1. hence answer to the original question is no.

also, x=1 and y=0 will satisfy this equation. as x=1, thus answer to the original question is yes.

x=-1 and y=0 will satisfy this equation. as x is not equal to 1, thus answer to the original question is no.

st. 2 y= 1-x again different values are possible. hence not sufficient

combining st.1 and st.2

put y=1-x in \(x^2+y^2=1\)

\(x^2 +1+x^2-2x=1\) \(2(x^2-2x)=0\) \(x(x-1)=0\)

x=0 or x=1

but x=0 is not possible, as y is not equal to 1. hence x=1. thus answer should be C

Hi manpreetsingh86! The correct version of the bold part is 2(x^2-x)=0. In your solution 2 in the brackets is wrong since you factor out 2 already. Please correct
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Statement 1: Circle with infinite values for x when y is not 1. reject.

Statement 2: Line with infinite values for x when y is not 1. reject

Combines statement: Only two point of intersection (0,1) and (1,0). Thus solvable.

As pointed in the above posts the easiest way is the graphical way in most of these type of questions. I think we should not even think on the track of solving equations.
_________________

y is "not 1" is an unusual information. There is no reason for which it needs to be "not 1" such as (y - 1) in denominator.

(1) x^2 + y^2 = 1 x can take many values since this is an equation in two variables. y will take corresponding value(s). What we need to check is whether x can be 1. The only time the statement may be sufficient is if x is 1 only when y = 1. Put x = 1, you get y = 0. So x may or may not be 1. Insufficient

(2) y = 1 - x Similarly, x and y may take infinite different value. When x = 1, y = 0 here. So again, x may or may not be 1. Not sufficient.

Using both, x^2 + (1-x)^2 = 1 x(x - 1) = 0 Either x is 0 or 1. But if x is 0, y is 1 which is not allowed. So x must be 1.

It is obvious that neither statement alone is sufficient. Now, let's look two statements together.

Think about it x+y = 1 but x^2 + y^2 = 1. Can you think of a pair of two numbers whose sum is 1. But when you square those numbers and add together the results, you again get 1? The only pair 1 and 0

Got this question in GMATPrep EP1 and below is the approach I followed: 1. x^2 + y^2 = 1 Not given that x and y are integers. Hence, two cases possible: a. If y = 0, x = 1 or x = -1 b. If y =1/sqrt2 , x = 1/sqrt2. Not sufficient

2. y = 1-x x can take any value except for zero. Hence not sufficient.

(1) + (2), x^2 + y^2 = 1 and y = 1-x We know that (x+y)^2 = x^2 + y^2 + 2xy 1^2 = 1 + 2xy 2xy = 0 Given that y ≠ 1, for xy to be zero x has to be 1. Sufficient