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

Question

What is the value of x?

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

BrentGMATPrepNow · Accepted Answer

Target question :    What is the value of x?

Statement 1 : x + y = 2  
We can see that x can have infinitely many values. 
Statement 1 is NOT SUFFICIENT
For example, we could have  x = 1  and y = 1 OR we could have  x = 0  and y = 2

Statement 2 : xy = z² + 1 
Statement 2 is clearly NOT SUFFICIENT
For example, we could have  x = 1 , y = 2 and z = 1 OR we could have  x = 2 , y = 0.5 and z = 0

Statements 1 and 2 combined    
Take statement 1, x + y = 2, and rewrite as  y = 2 - x 
Take statement 2, xy = z² + 1, and replace  y  with  2 - x  to get: x( 2 - x ) = z² + 1
Expand to get: 2x - x² = z² + 1
Multiply both sides by -1 to get: -2x + x² = -(z²) - 1
Rearrange to get: x² - 2x = -(z²) - 1
Add 1 to both sides to get: x² - 2x + 1 = -(z²)
Factor left side to get: (x - 1)² = -(z²)

IMPORTANT: 
(x - 1)² is GREATER than  or equal to 0  for ALL values of x
-(z²) is LESS than  or equal to 0  for ALL values of z
So, the only way that the equation, (x - 1)² = -(z²), can be true is when (x - 1)² and -(z²)  both equal 0 
If (x - 1)² = 0, then it MUST be the case that  x = 1  
Since we can answer the  target question  with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent

JusTLucK04 · Answer

Veritas Official Explanation

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.

Vetrik · Answer

S1:

x + Y = 2....what are the options for X & Y...X,Y can be combinations of         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  . Hence X=1. So Answer C.

--------

Please give Kudos if this helps .

email2vm · Answer

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?

profileusername · Answer

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?

rekhabishop · Answer

Who said anything about the two being integers?

abhimahna · Answer

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.

BillyZ · Answer

(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

Kinshook · Answer

Asked: What is the value of x?

(1)  x+y=2 
x = 2-y and y is unknown
NOT SUFFICIENT

(2)  xy = z^2  +1 
x = \frac{( z^2 + 1)}{y}  and value of z and y are unknown
NOT SUFFICIENT

Combining (1) & (2) 
(1)  x+y=2 
y = 2-x
(2)  xy = z^2  +1 
 y = \frac{(z^2 + 1 )}{x} 
 2-x = \frac{(z^2 + 1 )}{x} 
 2x - x^2 = z^2 + 1 
 x^2 - 2x + 1 + z^2 = 0 
 (x-1)^2 + z^2 = 0 
Since  (x-1)^2   \geq  0 &  z^2   \geq  0
x-1=0 & z =0
x = 1
SUFFICIENT

IMO C

nick1816 · Answer

Statement 1- x can be any real number
Insufficient

Statement 2- xy≥1
Again x can be any real number
Insufficient

Combining both statement
Sum and product of x and y are positive; hence, x,y>0

From statement 1
 \frac{x+y}{2} ≥ \sqrt{xy} 
1≥xy

From statement 2
xy≥1

so xy=1

AM=GM; so x=y=1

Sufficient

BhaveshGMAT · Answer

Great Question and I was trapped too. The Veritas explanation does not go down well with me. So, here is my attempt-

Statement 1 : X and Y can take any real values. Clearly insufficient.

Statement 2 : Again, clearly insufficient. But square of a variable is always a point of inference.

So what can we infer from \(xy = z^2 + 1?\\
\)
Since min value of \(Z^2=0\) So, \(xy\geq1\)

Therefore, x and y will have same sign.

Combining 1 & 2,

x+y=2 and from statement 2, x & y having same sign.. so it follows that both x and y are of +sign.

Let's now plug in some values of (x,y) to satisfy both x+y=2 and \(xy\geq1\) >>
(A) (0.1, 1.9) - Incorrect
(B) (0.5, 1.5) - Incorrect
(C) (1,1) - Correct. Voila, that's a specific value and I select Answer Choice C.

Kudos if you liked!!

ProfChaos · Answer

Great question, almost stumped me in my mock test
Looking at S1 alone and S2 alone, it is clear that both alone are not sufficient

So it is between C and E now

Lets start with S2:
xy=z^2+1
z^2 is non negative (z^2 can be zero or positive), that means xy has to be either 1 or more than 1 (xy will be 1 if z^2 is zero or xy will be more than 1 if z^2 is positive)
Boiling this down to xy>=1
for xy to be greater than or equal to 1 ---> x and y can be (+,+) OR (-,-)

But we are told in S1 that x+y=2
If both x and y are negative, we cannot get 1 as an addition of two negative numbers
Thus, x and y both have to be positive and the only possible combination of x and y is 1 and 1 which satisfies x+y=2

Hence value of x is 1

Option C

Palladin · Answer

ext{Obviously, neither of the statements alone is sufficient.}\
	ext{So, take the statements together.}

ext{Statement (1): } x + y = 2 \implies x = 2-y

ext{Statement (2):} 
  \begin {alignat} {2}
&&xy &= z^2 + 1\
&\implies &(2-y)y &= z^2 + 1\
&\implies &-y^2+2y-1 &= z^2\
&\implies &(y-1)^2 &= -z^2\
&\implies &(y-1)^2 &\le 0
\end {alignat}

ext{However, }  
 \begin {alignat}{2}
&&(y-1)^2 &\ge 0 	ext{ (Since the square of any real number is greater than/equal to zero.)}\
&\implies &(y-1)^2 &= 0\
&\implies &y &= 1\
&\implies &x &= 2-y = 1
\end {alignat}

ext{Hence, the statements taken together are sufficient. Answer is C.}

Basshead · Answer

Each statement alone is clearly insufficient.

Combined:

x = 2 - y

xy ≥ 1

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

2y - y^2 = z^2 + 1

-z^2 = y^2 - 2y + 1

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

Now lets take a second to analyze. z^2 must be non-negative. If we add a negative sign, then  -z^2  can't be positive either.

Therefore  z = 0  and  y - 1 = 0 .

y = 1

SUFFICIENT. Answer is C.

Hoozan · Answer

When we see such equations, it gets very easy to think of (E) as the answer since we have 2 equations and 3 variables.

How do we not fall for such a trap?

Victor19 · Answer

Hello all,

All the answers in this thread would give you the solution. I have a slightly different approach to solve this problem.

Statement 1 - "X" could be any value. So options A & D are ruled out.

Statement 2 - Again "X" could be any value. But only thing we can learn here is, Since (Z^2 + 1) should be positive, both X and Y are of the same signs. We can rule out option B as well.

Remaining is only options C and E.

Combining both the statements :
Sub x = 2-y in statement 2 and rearrange it. Finally we'll get equation Z^2 = -((y-1)^2).

Sub y = 2-x in statement 2 and rearrange it. We get equation Z^2 = -((x-1)^2).

From above to equations we can equate RHS since LHS is the same.

So (x-1)^2 = (y-1)^2. When we take sq. root and solve this we will have 4 cases. When you solve all 4 cases, 2 of the cases will give back Statement 1 and 2 of the cases will give you the equation " x-y=0 ". When we solve this with statement 1 we will obviously get an unique solution.

So you can choose option C without actually solving these 2 equations. The advantage I find here is, we don't have to logically guess anything and can blindly go with the option.

Feel free to let me know if you think my reasoning is flawed or if someone has already posted this ( I read only the top 3 solutions and went about typing this).

rvgmat12 · Answer

OE:

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.

GMATBusters · Answer

St 1 :  x+y=2

We can't find x as there can be an infinite combination of x and y

St2:  xy  =  z^2  +1 
Here also we can't find x and y as there can be many combinations of x and y. 
However, we can say that xy > or = 1
as z^2 is non-negative and hence, its min value = 0

While combining St 1 and St 2, we can say that as x+y = 2, the max value of xy will be 1 ( when x=y=1)
So xy< or = 1 from St 1
xy > or = 1 from St2
combining xy = 1

Sufficient, C

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards