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If x + y ≠ 0, which of the following is a solution to the inequality
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Updated on: 31 Mar 2021, 11:13
Updated on: 31 Mar 2021, 11:13
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39% (01:39) correct
60% (01:50) wrong based on 63 sessions
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If x + y ≠ 0, which of the following is a solution to the inequality \(\frac{(x^2 - y^2 - 1)}{(x + y)} > \frac{-1}{(x + y)}\) ?
Indicate all solutions.
1) x=3 and y=7
2) x=-3 and y=7
3) x=-11 and y=-9
4) x=9 and y=-6
5) x=-20 and y=-24
6) x=12 and y=9
7) x=-2 and y=16
Indicate all solutions.
1) x=3 and y=7
2) x=-3 and y=7
3) x=-11 and y=-9
4) x=9 and y=-6
5) x=-20 and y=-24
6) x=12 and y=9
7) x=-2 and y=16
Originally posted by ayeshaakhtar209 on 31 Mar 2021, 06:31.
Last edited by Carcass on 31 Mar 2021, 11:13, edited 1 time in total.
Last edited by Carcass on 31 Mar 2021, 11:13, edited 1 time in total.
Formatted
Retired Moderator
Joined: 16 Apr 2020
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Posts: 1546
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If x + y ≠ 0, which of the following is a solution to the inequality
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31 Mar 2021, 11:11
31 Mar 2021, 11:11
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Carcass wrote:
KarunMendiratta wrote:
ayeshaakhtar209 wrote:
If x + y ≠ 0, which of the following is a solution to the inequality [x^2 - y^2 - 1][/x + y] > [-1][/x + y] ?
Indicate all solutions.
1) x=3 and y=7
2) x=-3 and y=7
3) x=-11 and y=-9
4) x=9 and y=-6
5) x=-20 and y=-24
6) x=12 and y=9
7) x=-2 and y=16
Indicate all solutions.
1) x=3 and y=7
2) x=-3 and y=7
3) x=-11 and y=-9
4) x=9 and y=-6
5) x=-20 and y=-24
6) x=12 and y=9
7) x=-2 and y=16
Is it \(\frac{(x^2 - y^2 - 1)}{(x + y)} > \frac{-1}{(x + y)}\) or \((x^2 - y^2 - 1)(x + y) > -1(x + y)\)??
I think the first elaboration is correct
Waiting response
regards
Then;
\(\frac{(x^2 - y^2 - 1)}{(x + y)} > \frac{-1}{(x + y)}\)
\(\frac{(x^2 - y^2)}{(x + y)} - \frac{1}{(x + y)} > \frac{-1}{(x + y)}\)
Adding \(\frac{1}{(x + y)}\) to both sides;
\(\frac{(x^2 - y^2)}{(x + y)} > 0\)
\(\frac{(x + y)(x - y)}{(x + y)} > 0\)
\((x - y) > 0\)
\(x > y\)
A. \(3 > 7\)
B. \(-3 > 7\)
C. \(-11 > -9\)
D. \(9 > -6\)
E. \(-20> -24\)
F. \(12 > 9\)
G. \(-2 > 16\)
Only option D, E, and F satisfy this
General Discussion
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Re: If x + y ≠ 0, which of the following is a solution to the inequality
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31 Mar 2021, 09:55
31 Mar 2021, 09:55
1
ayeshaakhtar209 wrote:
If x + y ≠ 0, which of the following is a solution to the inequality [x^2 - y^2 - 1][/x + y] > [-1][/x + y] ?
Indicate all solutions.
1) x=3 and y=7
2) x=-3 and y=7
3) x=-11 and y=-9
4) x=9 and y=-6
5) x=-20 and y=-24
6) x=12 and y=9
7) x=-2 and y=16
Indicate all solutions.
1) x=3 and y=7
2) x=-3 and y=7
3) x=-11 and y=-9
4) x=9 and y=-6
5) x=-20 and y=-24
6) x=12 and y=9
7) x=-2 and y=16
ayeshaakhtar209
Is it \(\frac{(x^2 - y^2 - 1)}{(x + y)} > \frac{-1}{(x + y)}\) or \((x^2 - y^2 - 1)(x + y) > -1(x + y)\)??
