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15 Aug 2022, 18:11
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Thank you! this is explanation is great and much easy
TusharViv
What values of x have a corresponding value of y that satisfies both xy > 0 and xy = x + y ?
A. x≤1
B. −1<x≤0
C. 0<x≤1
D. x>1
E. All real numbers
xy > 0 this means either:
1. x>0 AND y>0
2. x<0 AND y<0
xy = x + y
Now the other part of the question. Therefore x+ y > 0 since xy > 0
x + y > 0 can only be true if x or y are positive or both. And since we have the 2 statements above. x<0 AND y<0 can be eliminated.
Therefore eliminate A, B, E
Now between C and D. Get y on one side
xy = x + y
xy - y = x
y(x-1) = x
y = x / (x-1)
x cannot be 1. Eliminate C Hence answer is D
Also if x=1 was not part of C. The denominator would be negative if x was a fraction between 0 and 1, hence giving a negative y value but we know this cannot be true since BOTH x AND y must be positive
A. x≤1
B. −1<x≤0
C. 0<x≤1
D. x>1
E. All real numbers
xy > 0 this means either:
1. x>0 AND y>0
2. x<0 AND y<0
xy = x + y
Now the other part of the question. Therefore x+ y > 0 since xy > 0
x + y > 0 can only be true if x or y are positive or both. And since we have the 2 statements above. x<0 AND y<0 can be eliminated.
Therefore eliminate A, B, E
Now between C and D. Get y on one side
xy = x + y
xy - y = x
y(x-1) = x
y = x / (x-1)
x cannot be 1. Eliminate C Hence answer is D
Also if x=1 was not part of C. The denominator would be negative if x was a fraction between 0 and 1, hence giving a negative y value but we know this cannot be true since BOTH x AND y must be positive
29 Nov 2024, 10:07
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We know that xy>0 so we also know that x/y>0. Keep also in mind that x and y cannot be 0.
Now just rewrite the second equation as x-(x/y)=1 and it becomes clear that x must be >1
Now just rewrite the second equation as x-(x/y)=1 and it becomes clear that x must be >1
parkhydel
What values of x have a corresponding value of y that satisfies both xy > 0 and xy = x + y ?
A. \(x \leq 1\)
B. \(-1 < x \leq 0\)
C. \(0 < x \leq 1\)
D. \(x > 1\)
E. All real numbers
PS22680.02
A. \(x \leq 1\)
B. \(-1 < x \leq 0\)
C. \(0 < x \leq 1\)
D. \(x > 1\)
E. All real numbers
PS22680.02
30 Dec 2024, 02:23
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XY>0
Both have same signs - Either both positive or both negative. None 0
XY=X+Y
Both cant be -ve as RHS will be -ve but LHS will be +ve
X,Y must be>0
Eliminates A,B,E
Now between C and E
Lets take a value greater than 1
If both x and y are 2 each the equation is held. So D
Both have same signs - Either both positive or both negative. None 0
XY=X+Y
Both cant be -ve as RHS will be -ve but LHS will be +ve
X,Y must be>0
Eliminates A,B,E
Now between C and E
Lets take a value greater than 1
If both x and y are 2 each the equation is held. So D
parkhydel
What values of x have a corresponding value of y that satisfies both xy > 0 and xy = x + y ?
A. \(x \leq 1\)
B. \(-1 < x \leq 0\)
C. \(0 < x \leq 1\)
D. \(x > 1\)
E. All real numbers
PS22680.02
A. \(x \leq 1\)
B. \(-1 < x \leq 0\)
C. \(0 < x \leq 1\)
D. \(x > 1\)
E. All real numbers
PS22680.02
