Thread: System of equations with absolute value

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?

Originally Posted by DenMac21

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.

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.

Here is a graphical approach,


Note, that ticbol's solutions is much better than mine. Because this graph is not a proof that these are all the solutions and the only solutions. But it makes things easier.

Perhaps you can do this,
since
Thus, solve these two systems,


Which becomes,



Or if you want you may do this,


But then you have to go through each possibility

Originally Posted by DenMac21

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?

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

Thanks!!!!!!!!!!!!!!!