Answer:(-3, 4).
Step-by-step explanation:
Let's check each option to see if the given solution, (-3, 4), satisfies the equation:
A.
οΏ½
οΏ½
β
2
(
οΏ½
β
οΏ½
)
=
1
β
2
(
β
7
)
xyβ2(xβy)=1β2(β7)
(
β
3
)
(
4
)
β
2
(
(
β
3
)
β
4
)
=
1
β
2
(
β
7
)
(β3)(4)β2((β3)β4)=1β2(β7)
(
β
12
)
β
2
(
β
7
)
=
1
β
14
(β12)β2(β7)=1β14
β
12
+
14
=
1
β
14
β12+14=1β14
2
=
β
13
2=β13 (False)
B.
3
(
οΏ½
οΏ½
)
=
οΏ½
β
οΏ½
3(xy)=xβy
3
(
(
β
3
)
(
4
)
)
=
(
β
3
)
β
4
3((β3)(4))=(β3)β4
3
(
β
12
)
=
β
3
β
4
3(β12)=β3β4
β
36
=
β
7
β36=β7 (False)
C.
οΏ½
οΏ½
β
5
(
οΏ½
β
οΏ½
)
=
5
(
1
)
β
7
xyβ5(xβy)=5(1)β7
(
β
3
)
(
4
)
β
5
(
(
β
3
)
β
4
)
=
5
β
7
(β3)(4)β5((β3)β4)=5β7
β
12
β
5
(
β
7
)
=
β
2
β12β5(β7)=β2
β
12
+
35
=
β
2
β12+35=β2
23
=
β
2
23=β2 (False)
D.
β
4
(
οΏ½
οΏ½
)
=
οΏ½
β
οΏ½
β4(xy)=xβy
β
4
(
(
β
3
)
(
4
)
)
=
(
β
3
)
β
4
β4((β3)(4))=(β3)β4
β
4
(
β
12
)
=
β
3
β
4
β4(β12)=β3β4
48
=
β
7
48=β7 (False)
E.
οΏ½
οΏ½
β
οΏ½
οΏ½
=
1
β
7
xyβxy=1β7
(
β
3
)
(
4
)
β
(
β
3
)
(
4
)
=
1
β
7
(β3)(4)β(β3)(4)=1β7
β
12
+
12
=
β
6
β12+12=β6
0
=
β
6
0=β6 (False)
