# System of equations with absolute value

• Feb 1st 2006, 04:59 AM
DenMac21
System of equations with absolute value
I have to solve this system

$\displaystyle $\left\{ \begin{array}{l} \left| {x + y} \right| = 5 \\ \left| {xy} \right| = 6 \\ \end{array} \right.$$

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?
• Feb 1st 2006, 10:42 AM
ticbol
Quote:

Originally Posted by DenMac21
I have to solve this system

$\displaystyle $\left\{ \begin{array}{l} \left| {x + y} \right| = 5 \\ \left| {xy} \right| = 6 \\ \end{array} \right.$$

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.
• Feb 1st 2006, 11:24 AM
ThePerfectHacker
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.
• Feb 1st 2006, 12:13 PM
ThePerfectHacker
Perhaps you can do this,
since $\displaystyle |x|=\sqrt{x^2}$
Thus, solve these two systems,
$\displaystyle \sqrt{(x+y)^2}=5$
$\displaystyle \sqrt{(xy)^2}=6$
Which becomes,
$\displaystyle (x+y)^2=25$
$\displaystyle (xy)^2=36$

Or if you want you may do this,
$\displaystyle x+y=\pm5$
$\displaystyle xy=\pm6$
But then you have to go through each possibility
• Feb 1st 2006, 12:41 PM
CaptainBlack
Quote:

Originally Posted by DenMac21
I have to solve this system

$\displaystyle $\left\{ \begin{array}{l} \left| {x + y} \right| = 5 \\ \left| {xy} \right| = 6 \\ \end{array} \right.$$

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::

$\displaystyle x+y=\pm 5$
$\displaystyle x=\pm6/y$

Substituting the second of these into the first (remembering that
the ambiguous signs are independent).

$\displaystyle \pm6/y+y= \pm 5$

Multiply through by $\displaystyle y$:

$\displaystyle \pm 6+y^2 \pm 5y=0$

So solving this using the quadratic formula::

$\displaystyle y=\frac{\pm5\pm \sqrt{25\pm24}}{2}$,

$\displaystyle y=\frac{\pm5\pm (1\ \vee \ 7)}{2}$

with all the $\displaystyle \pm$ signs independent. So:

$\displaystyle y=\pm3\ \vee\ \pm 2\ \vee\ \pm6\ \vee\ \pm1$

So if we plug each of the $\displaystyle y$'s above into $\displaystyle x+y=\pm 5$ we
will get the complete set of solutions. So the solution set is (I hope):

$\displaystyle \ (x=2, y=3) \vee (x=-2, y=-3)$
$\displaystyle \vee(x=3, y=2) \vee (x=-3, y=-2)$
$\displaystyle \vee (x=-1, y=6) \vee (x=1, y=-6)$
$\displaystyle \vee (x=-6, y=1) \vee (x=6, y=-1)$.

RonL
• Feb 1st 2006, 02:57 PM
DenMac21
Thanks!!!!!!!!!!!!!!!