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

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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Please give Kudos if this helps.

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!!

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?

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?

Who said anything about the two being integers?

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

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.

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

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

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

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

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

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

\( \text{Hence, the statements taken together are sufficient. Answer is C.}\)

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.

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?

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).

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.

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