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(x-1) (y + 3) (xy + 3) = 0

one of the factors need to be 0 for this to work

xy = -3

(-3,1) (3, -1) (1,-3) (-1, 3)

A. If x = -1 |y| is prime, thats true.

B. If |y| is not prime, x is not prime, not true, if |y| = 1, x is 3. False.

C. If x is a prime, y/x is not an integer = False.
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Taking statement 1:
for all 3 multiplication to be qual to 0, atleast one bracket should be 0
So, either x=1 or y = -3 or xy= -3
Now, if x= -1
solving with y= -1, we will get either y= -3 or when solving with xy= -3 , y = 3
in either case |y| = 3 which is prime
statement 1 correct

Statement 2:
take y= 1
we get x= 3 --it is prime
statement 2 may not always be true

statement 3:
take x= 3 and y = -3
we will get x/y = -1
which is a negative integer
so statement 3 not always true..

Hence A, only I
Bunuel
If x and y are integers and (x - 1)(y + 3)(xy + 3) = 0, which of the following must be true?

I. If x = -1, |y| is a prime number
II. If |y| is not a prime number, x is not a prime number
III. If x is a prime number, y/x is not an integer

A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III


 


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i am going with option a.
if x = -1, |y| is prime, x=-1, the first case is not valid, 2nd case y=-3, |y| = 3, 3rd case -1y = -3, |y| = 3, S1 correct MBT
S2 incorrect, x = 3, 3rd case 3(xy=-3): 3y=-3, |y|=-1 = 1, not a prime, false
S3 incorrect, x = 3, 2nd case 2(y=-3) = 2x0x-6=0, y/x = -3/3 = -1, as y/x can be an integer when x is prime, false.
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Answer: A) I only

(x-1)(y+3)(xy+3)=0
x = 1, y = -3, xy = -3

Statement I.
When x = 1, y = -3, and abs(y)=3 -> CORRECT

Statement II.
From xy = -3 we know that x could also be 3 and y could be -1. Here, y is not a prime number but x is -> INCORRECT

Statement III.
If x = -3 and y = 1, then x/y = -3/1 = -3 = an integer -> INCORRECT
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Bunuel
If x and y are integers and (x - 1)(y + 3)(xy + 3) = 0, which of the following must be true?

I. If x = -1, |y| is a prime number
II. If |y| is not a prime number, x is not a prime number
III. If x is a prime number, y/x is not an integer

A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III


 


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x = 1 or y = -3 or xy = -3

I. If x = -1, then |y| must be 3. Hence this is always true.

2. Not always true. Say x = 3, and y = -1, in this case |y| is not prime, but x is prime.

3. x = 3, y = -3. In this case, y/x is an integer.

Only I

Option A
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x=1 or
y=-3 or
xy=-3

I if x=-1, then y=+-3---!y!=3 (prime) correct
II counter example: (x,y)= (3,-1) !y!=1 not prime, but x=3(prime) incorrect
III counter example:(x,y)=(3,-3), Then y/x=-1 ( integar)
I only
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If x and y are integers and (x - 1)(y + 3)(xy + 3) = 0, which of the following must be true?

I. If x = -1, |y| is a prime number
(-1-1)(y+3)(-1*y+3) = (-2)(y+3)(3-y) = 2(y+3)(y-3) = 0
y = 3 or -3; |y| = 3; a prime number
MUST BE TRUE

II. If |y| is not a prime number, x is not a prime number
Let y = k ; where k is NOT a prime number

y = k
(x-1)(k+3)(kx + 3) = 0
x = 1 or -3/k
-3/k is NOT an integer unless k = 1, -1 or 3, - 3; -3/k = -3,3,-1,1 respectively in these cases
k = 1; x = 1 or -3; x is NOT a prime number
k = -1; x = 1 or 3; x is a prime number when x = 3

MAY OR MAY NOT BE TRUE / COULD BE TRUE

III. If x is a prime number, y/x is not an integer

x = 2;
(2-1)(y+3)(2y+3) = 0
y = -3 or -3/2; y/x = -3/2 or -3/4; Not integers

In general, x = k where k is a prime number
(k-1)(y+3)(ky+3) = 0

y = -3 or -3/k; y/x = -3/k or -3/k^2
x = 3; y = -3 or -1; y/x = -1 or -1/3; y/x may or may not be an integer

MAY OR MAY NOT BE TRUE / COULD BE TRUE


A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III

IMO A
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Given

(x-1)(y+3)(xy+3) = 0

in this case either-
x-1 can be zero => x = 1
y+3 can be zero => y = -3
xy+3 can be zero => y = -3/x

lets look at 1

given x = -1, sub => -2 (y+3) (-y+3) = 0
as -2 is not zero, 9 - y^2 = 0
9 = y^2
|y|=3 which is prime
satisfies

statement 2
it talks about a condition where |y| is not 3, so y+3 can never be zero in this case
so either x-1 is zero, xy = -3
let us check examples of xy being -3
x = 1 and y = -3, which cant be the case as Y cannot be Prime
x = 3, and y = -1. x turned out to be prime again false
x = -3. y = 1 again false
x = -1, y= 3, again false
so statement 2 always false, one has to be prime

Statement 3
let us consider 2 examples again, lets pick x as 7 a prime , and y+3 = 0, meaning y as -3
so y/x = -3/7 which is not an integer satisfies
but for the above examples of Statement 2, where both meet the conditions, y/x will always be an integer
so this statement not always holds true

so ans = A
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Given: x and y are integers

(x-1)(y+3)(xy+3) = 0
- This means at least one or all of the brackets should be equal to zero

Statement 1. If x = -1, |y| is prime

If x = -1, then absolute value of y should be prime
x-1 can not be zero

so either y+3 = 0 or xy+3 =0 or both
which gives us a prime value of y (either -3 or 3)

This is true

Statement II. If |y| is not prime, then x is not prime- either x should be less than 0 or if it is greater than 0, then it should be a composite number

If |y| is not prime, then y+3 can not be zero
Either xy+3 = 0 or x-1 = 0 or both

If xy + 3 = 0
Then, xy = -3
Now, for x to be an integer, absolute value of y should be less than 3- so the only choices are 0, 1, 2 (positive or negative)

Now, y is not 0, because then xy+3 can not be 0
y is not 2 because |y| is not prime
So, y has to be 1 or -1, which makes for a case where x = 3

So, even when y is not prime, x can be prime

Won't go to statement III because there is no option with I and III- So Option (A) Only I
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If x and y are integers and (x - 1)(y + 3)(xy + 3) = 0, which of the following must be true?

I. If x = -1, |y| is a prime number.
(-1-1)(y+3)(-y+3)=0
y= -3 or +3
|y|= 3.
It must be true.

II. If |y| is not a prime number, x is not a prime number.
Let, x=3
(3-1)(y+3)(3y+3)=0
6(y+3)(y+1)=0
y= -3 or -1
|y|= 1 or 3. If x is prime, y could be prime or not.
Must not be true

III. If x is a prime number, y/x is not an integer.
Let, x=3 . As in II. y=-3 or -1
y/x= -3/3 = Integer
Must not be true

A
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(x-1)(y+3)(xy +3) = 0

there one of the terms needs to be zero

x = 1
y = -3
or xy = -3

the possibilities for xy = -3 are (-1,3) (1, -3) (3,-1) and (-3, 1)

1. if x is -1 than absolute value of y is 3 which is prime, true

2. if absolute value of y is not a prime number, x is not a prime number. y = -1, absolute value is 1, not prime
Therefor not true

3. If x is a prime number , y/x is not an integer
if y/x = -1/3. this is not an integer.

If x=3 (a prime number) and y =-3. Testing the original equation we get

(3-1)(-3+3)(3x-3 +3) = 2 x 0 x -6 = 0. works

y / x = -3/3 = -1. which is an integer. Therefore statement is false.


ANSWER. A Only statement on is true.
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The correct answer is A. I only

Lets breakdown the equation in the question.
1. x-1=0
x=1
2. y+3=0
y=-3
3. xy+3=0
xy=-3

Now, evaluating the statements,
Statement 1,
if x=-1, this means either y=-3 or xy=-3 must be true.
If y=-3
mod y=3, and 3 is a prime number

If xy=-3
Substituting x=-1, we get -y=-3
y=3, which is a prime number

In both scenarios, mod y=3, which is a prime number.
Statement 1 is always true.

Statement 2
Lets assume x=3
xy+3=0
3y+3=0
3y=-3
y=-1
mod y=1
This is not a prime number.
This proves the statement is wrong because here when mod y is not a prime number x is a prime number. Hence, the statement is false.

Statement 3
Lets assume x=3
y+3=0
y=-3
now putting the values of x and y
-3/3=-1
-1 is an integer. This contradicts the rule, and thus the statement is false.

Only 1 is true.
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Better to solve this using the inequalities principle itself:

(x-1)(y+3)(xy+3) = 0
So x = 1 or y = -3 or xy = -3

I) If x = -1, |y| is a prime:

(-1-1)(y-3)(-1y + 3) = 0
(-2y - 6)(-y+3) = 0
-2y-6 = 0 OR -y+3 = 0
2y = 6
y=3

OR
y=-3

Hence |y| = 3 -> Hence Statement 1 is true

II) |y| is not a prime, then x is not a prime

Let us substitute x = 2
We have (3-1)(y+3)(3y+3) = 0
So 2y + 6 = 0
OR
3y+3 = 0

Hence y=-1 or y=-3
Since we are getting two different values for y when x is a prime, then Statement 2 must not be true. Hence false.

III) As regards Statement 3: x is a prime, y/x not an integer

Take x = 3
After substituting into the formula, we are left with the two different values of 'x'
Then the two values for x are -1 or -1/3
Here -1/3 is not an integer but -1 is an integer out of the two values. Hence Statement 3 is also not true.

Only Statement 1 is true.
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In must be true question I always try to find a case in which I could reject the given statement.

First, I observed the given equation carefully and realized one of the brackets have to be zero.

S1: put x = -1
then the brackets would be: (-2) (y+3) (-y+3)

in order to equate it to zero y should be +3 or -3. Hence it must be true

S2: put x = 3, which is prime
now the brackets would be: (2) (y+3) (3x + 3)
if we put y = -1 then the brackets product would be equal to zero as the third bracket would be zero - HENCE NOT MUST BE TRUE

S3: Put x=3 and y=-3,
the brackets would be: (2)(0)(-6) = 0
And y/x = -3/3 = 1 which is integer, HENCE NOT MUST BE TRUE

So, Only Statement 1 must be true

IMO A
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Given: (x-1)(y+3)(xy+3) = 0
so one of them must be true
* x =1
* y=-3
* xy =-3, giving (-1,3),(1,-3),(-3,1),(3.-1)

Now check each statement:
I. If x =-1, then xy = -3
=> y =3. Hence |y| = 3
true

II. if |y| is not prime:
* x = 1 (not prime), or
* x,y = (-3,1) (here x =-3, y =1 not prime)
No counterexample
true

III. counterexample (x,y) = (3,-3)
* equation holds since y+3 = 0
* y/x = -3/3 = -1, an integer
False

So, I & II
Option: D
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Taking the critical points: x = 1, y = -3, xy = -3
For each statement:
for x = -1 is only possible if y = + - 3 = |y| so statement 1 is true.

Statement 2 example: (x,y) = (3, -1) satisfies the equation and |y| is 1 which is not prime but 3 is prime so statement 22 cannot be true.

Statement 3: in cases if both the numbers are same like (3,-3) it is an integer so statement not true.

IMO ans is A.
Bunuel
If x and y are integers and (x - 1)(y + 3)(xy + 3) = 0, which of the following must be true?

I. If x = -1, |y| is a prime number
II. If |y| is not a prime number, x is not a prime number
III. If x is a prime number, y/x is not an integer

A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III


 


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solutions can be when: i: x = 1 or ii: y = -3 or iii: xy = -3, (1,-3)(-1,3)(3,-1)(-3,1)
S1:
if x = -1, either y = -3 or 3 so |y| = 3 = prime number yes
S2:
|y| = not prime, then x = 1 i, x = 3, -3 iii, x can be prime or not no
S3:
if x = prime: y = -3 ii; y = -1 iii
in case x = 3, y = -3
y/x = -3/3 = -1 which is integer, so no
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