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What is the value of x? (1) x+y=2 (2) xy=z2+1

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What is the value of x? (1) x+y=2 (2) xy=z2+1 [#permalink]

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New post 31 Oct 2014, 13:12
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What is the value of x?

(1) \(x+y=2\)
(2) \(xy\)\(=\)\(z^2\)\(+1\)
[Reveal] Spoiler: OA

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Re: What is the value of x? (1) x+y=2 (2) xy=z2+1 [#permalink]

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New post 31 Oct 2014, 13:23
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You should quickly see that neither statement alone is sufficient, which should signify to you that you'll have to do some work on the C vs. E decision. And here, algebra is the way to roll up your sleeves and get to work. If you take the first equation, x+y=2, and set it up to substitute in the second, you'll have:

x=2−y
and

xy=z2+1
Substituting, that's:

y(2−y)=z2+1
Which works out to:

2y−y2=z2+1
If you then set this up as a quadratic in terms of y, you have:

−z2=y2−2y+1
And then factor the quadratic:

−z2=(y−1)2
Now it's time to take a look logically. Since the left hand side cannot be positive (z2 can't be negative, so once you apply that negative sign to whatever the result is it's either 0 or something negative) and the right hand side cannot be negative (whatever (y−1) is, when you square it it's either positive or 0), you can determine that y−1 must be 0, and therefore that y=1. Given that, plug back into the first equation to find that x=1, proving the two statements together to be sufficient.
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Re: What is the value of x? (1) x+y=2 (2) xy=z2+1 [#permalink]

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S1:

x + Y = 2....what are the options for X & Y...X,Y can be combinations of [1,1] [0,2] [3/2, 1/2] [1/3, 5/3] etc....hence insufficient.


S2:

XY = z^2 + 1 .....with this we can not say what is x....hence insufficient.

Combining S1 & S2 together lets us plug-in those options of S1 and see...

First lets take the fractions....e.g. 3/2, 1/2......so 3/4 = Z^2+ 1...now Z^2 = -1/4 But Z^2 can't be negative...so 3/2,1/2 is ruled out. Similarly all fractions options can be ruled out.

Now lets take 0,2 or 2,0......so now the eqn becomes 0=Z^2+1....Z^2 can't be -1 so...this option is also ruled out.

So finally we are left out with only option [1,1]. Hence X=1. So Answer C.

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Re: What is the value of x? (1) x+y=2 (2) xy=z2+1 [#permalink]

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Re: What is the value of x? (1) x+y=2 (2) xy=z2+1 [#permalink]

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New post 16 Apr 2016, 08:18
JusTLucK04 wrote:
What is the value of x?

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



st1> Many values of x and y to get value 2. So clearly insuff

St2> xy= z^2+1

so z^2= positive number and adding one will be positive number
Z can be z= -1 or z= 1/8 and z= 1 but their sq will always be positive number
xy = +ve
so x= +ve and y= +ve
or x=-ve and y = -ve

combining both:

x can be 1 and y can be 1


am I right?

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Re: What is the value of x? (1) x+y=2 (2) xy=z2+1 [#permalink]

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New post 31 Jul 2017, 02:41
JusTLucK04 wrote:
You should quickly see that neither statement alone is sufficient, which should signify to you that you'll have to do some work on the C vs. E decision. And here, algebra is the way to roll up your sleeves and get to work. If you take the first equation, x+y=2, and set it up to substitute in the second, you'll have:

x=2−y
and

xy=z2+1
Substituting, that's:

y(2−y)=z2+1
Which works out to:

2y−y2=z2+1
If you then set this up as a quadratic in terms of y, you have:

−z2=y2−2y+1
And then factor the quadratic:

−z2=(y−1)2
Now it's time to take a look logically. Since the left hand side cannot be positive (z2 can't be negative, so once you apply that negative sign to whatever the result is it's either 0 or something negative) and the right hand side cannot be negative (whatever (y−1) is, when you square it it's either positive or 0), you can determine that y−1 must be 0, and therefore that y=1. Given that, plug back into the first equation to find that x=1, proving the two statements together to be sufficient.


Using the knowledge that both RHS and LHS are not negative in this equation derived by combining both statements , how do we get to know that y=1?

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Re: What is the value of x? (1) x+y=2 (2) xy=z2+1 [#permalink]

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New post 01 Aug 2017, 09:15
email2vm wrote:
JusTLucK04 wrote:
What is the value of x?

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



st1> Many values of x and y to get value 2. So clearly insuff

St2> xy= z^2+1

so z^2= positive number and adding one will be positive number
Z can be z= -1 or z= 1/8 and z= 1 but their sq will always be positive number
xy = +ve
so x= +ve and y= +ve
or x=-ve and y = -ve

combining both:

x can be 1 and y can be 1


am I right?


Who said anything about the two being integers?
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Re: What is the value of x? (1) x+y=2 (2) xy=z2+1 [#permalink]

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New post 03 Aug 2017, 05:50
TheMastermind wrote:

Using the knowledge that both RHS and LHS are not negative in this equation derived by combining both statements , how do we get to know that y=1?


Hi TheMastermind ,

Good question.

I know that you are already aware of the fact that square of any integer is always positive.

Now, you got a^2 = - b^2

This is only possible when we have both a and b = 0. Any non zero values of a and b will not make this expression hold true.

Hence, I can conclude that y-1=0.

Hence, y = 1 or x = 1. Thus, C is correct.
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Re: What is the value of x? (1) x+y=2 (2) xy=z2+1 [#permalink]

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New post 01 Sep 2017, 07:54
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JusTLucK04 wrote:
What is the value of x?

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


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

-(a^2) = (b^2) IF AND ONLY IF a=b=0

x = 1
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Re: What is the value of x? (1) x+y=2 (2) xy=z2+1   [#permalink] 01 Sep 2017, 07:54
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