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

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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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20 Nov 2014, 10:11

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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.

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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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

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

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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?

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

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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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