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?

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- February 1st 2006, 05:59 AMDenMac21System 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? - February 1st 2006, 11:42 AMticbolQuote:

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. - February 1st 2006, 12:24 PMThePerfectHacker
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. - February 1st 2006, 01:13 PMThePerfectHacker
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 - February 1st 2006, 01:41 PMCaptainBlackQuote:

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 - February 1st 2006, 03:57 PMDenMac21
Thanks!!!!!!!!!!!!!!!