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If x² - 9 = 0, x² + xy - 18 = 0 nd x nd y r bth psitiv intgrs
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Updated on: 12 May 2024, 09:56
Updated on: 12 May 2024, 09:56
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Question Stats:
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If x² - 9 = 0, x² + xy - 18 = 0 аnd x аnd y аrе bоth pоsitivе intеgеrs, whаt аrе thе vаluеs оf x аnd y?
A, x = 2, y = 1
B. x = 3, y = 1
C. x = 1, y = 3
D. x = 1, y = 2
E. x = 3, y = 3
A, x = 2, y = 1
B. x = 3, y = 1
C. x = 1, y = 3
D. x = 1, y = 2
E. x = 3, y = 3
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Re: If x² - 9 = 0, x² + xy - 18 = 0 nd x nd y r bth psitiv intgrs
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31 Mar 2024, 22:18
31 Mar 2024, 22:18
from the 1st equation, x^2 is 9, meaning x is either 3 or -3, from the 2nd equation since x^2 is 9, xy would be 9, y should be 3 or a -3 (x is either 3 or -3) . Option "E" only satisfies this, is this the correct way to approach this??
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Re: If x² - 9 = 0, x² + xy - 18 = 0 nd x nd y r bth psitiv intgrs
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12 May 2024, 08:18
12 May 2024, 08:18
1
steveharvi wrote:
from the 1st equation, x^2 is 9, meaning x is either 3 or -3, from the 2nd equation since x^2 is 9, xy would be 9, y should be 3 or a -3 (x is either 3 or -3) . Option "E" only satisfies this, is this the correct way to approach this??
Carcass I agree, the correct answer is actually E, not B.
