Not quadratic!
⎩ ⎨ ⎧ x 2 + x y − 1 0 x + 2 y = − 4 y 2 + x y − 1 0 y + 2 x = − 1 2
Let x and y be numbers satisfying the system of equations above. Find the value of x − y .
The answer is -1.
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2 solutions
Almost the same way!!
After getting
(
x
+
y
)
=
4
;
I subtracted the second equation from the first to get:
(
x
+
y
)
(
x
−
y
)
−
1
0
(
x
−
y
)
−
2
(
x
−
y
)
=
8
=
=
>
(
4
−
1
0
−
2
)
(
x
−
y
)
=
8
=
=
>
x
−
y
=
−
8
8
=
=
>
x
−
y
=
−
1
SAME WAY!!!!
Nice method @Yatin Khanna 👍👍👍
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Thank you very much!!
First I added the two equation and got x+y=4 then I subtracted the two and got (x-y)(x+y)-12(x-y)=8 then substitute value of x+y and I got x-y = -1
Good idea but you must prove that there are solutions to this, as Calvin has told me quite a few times.
Adding both equations:
x 2 + y 2 + 2 x y − 8 x − 8 y = − 1 6
⟹ x 2 + 2 x y + y 2 − 8 x − 8 y + 1 6 = 0
⟹ ( x + y ) 2 − 8 ( x + y ) + 1 6 = 0
⟹ ( ( x + y ) − 4 ) 2 = 0
⟹ ( x + y ) = 4
⟹ x = 4 − y
Substituting this in the second equation we get:
y 2 + ( 4 − y ) y − 1 0 y + 2 ( 4 − y ) = − 1 2
⟹ − 8 y + 8 = − 1 2
⟹ y = 2 5
⟹ x = 4 − 2 5 = 2 3
So x − y = 2 3 − 2 5 = − 1