I have to solve this system
Is solution to solve 4 cases where:
1. x+y>0 and xy>0
2. x+y>0 and xy<0
3. x+y<0 and xy>0
4. x+y<0 and xy<0
and accordingly to that write 4 systems of equations and solve each of them?
Four equations? Maybe.Originally Posted by DenMac21
But here is one way.
Just by looking, [x=2 and y=3] or [x = -2 and y = -3] or [x=3, y=2] or [x= -3, y= -2] are the answers. Four sets of them.
By computations, let us see.
|x +y| = 5 means [x+y = 5] or [x+y = -5]
|xy| = 6 means [xy = 6] or [xy = -6]
-----------------------------------
When x+y = 5, and xy = 6:
From the 2nd equation, y = 6/x
So, in the 1st equation,
x +6/x = 5
x^2 +6 = 5x
x^2 -5x +6 = 0
(x-2)(x-3) = 0
x = 2 or 3
And so, y = 6/2 or 6/3 = 3 or 2
Meaning, [x=2, y=3] or [x=3, y=2] ----------answer.
-----------------------------------
When x+y = -5, and xy = 6:
From the 2nd equation, y = 6/x
So, in the 1st equation,
x +6/x = -5
x^2 +6 = -5x
x^2 +5x +6 = 0
(x+2)(x+3) = 0
x = -2 or -3
And so, y = 6/(-2) or 6/(-3) = -3 or -2
Meaning, [x= -2, y= -3] or [x= -3, y= -2] ----------answer.
-----------------------------------
When x+y = 5, and xy = -6:
From the 2nd equation, y = -6/x
So, in the 1st equation,
x -6/x = 5
x^2 -6 = 5x
x^2 -5x -6 = 0
(x-6)(x+1) = 0
x = 6 or -1
And so, y = -6/6 or -6/(-1) = -1 or 6
Meaning, [x=6, y= -1] or [x= -1, y=6] ----------answer.
---------------------------------------
When x+y = -5, and xy = -6:
From the 2nd equation, y = -6/x
So, in the 1st equation,
x -6/x = -5
x^2 -6 = -5x
x^2 +5x -6 = 0
(x+6)(x-1) = 0
x = -6 or 1
And so, y = -6/(-6) or -6/1 = 1 or -6
Meaning, [x= -6, y=1] or [x=1, y= -6] ----------answer.
Originally Posted by DenMac21
We may write the system we want to solve as::
Substituting the second of these into the first (remembering that
the ambiguous signs are independent).
Multiply through by :
So solving this using the quadratic formula::
,
with all the signs independent. So:
So if we plug each of the 's above into we
will get the complete set of solutions. So the solution set is (I hope):
.
RonL