1. ## Simultaneous equations

Determine the number of solutions of the following set, depending on the value of a.

2. Hello, atreyyu!

Did you sketch the graphs?

Determine the number of solutions of the following system, depending on the value of $\displaystyle a$.

. . $\displaystyle \begin{array}{ccccc}|x| + |y| &=& 1 & {\color{blue}[1]} \\ |x| + a &=& y & {\color{blue}[2]} \end{array}$

The graph of [1] looks like this:
Code:
                  |
1*
* | *
*   |   *
*     |     *
- - * - - - + - - - * - -
-1 *     |     * 1
*   |   *
* | *
-1*
|

The graph of [2] looks like this:
Code:
                  |
*       |       *
*     |     *
*   |   *
* | *
a*
|
- - - - - - + - - - - - -
|

Superimpose the two graphs and "slide" $\displaystyle a$ up and down the y-axis
. . and you will observe the following . . .

. . $\displaystyle \begin{array}{cc}\text{Domain} & \text{Solutions} \\ \hline \\[-3mm] a < -1 & 0 \\ a = -1 & \text{infinite} \\ -1 < a < 1 & 2 \\ a = 1 & 1 \\ a > 1 & 0 \end{array}$

3. Originally Posted by atreyyu
Determine the number of solutions of the following set, depending on the value of a.

|x| means x can either be negative or positive....but its absolute value is always positive or zero.
Or, the x in the |x| is +,-x.
If we square +,-x, it will result to x^2 always.
So (|x|)^2 = x^2.

|x| +|y| = 1 -------------(1)
|x| +a = y ---------------(2)

Eq.(1) minus Eq.(2),
|y| -a = 1 -y
|y| = (a+1) -y
Square both sides,
y^2 = (a+1)^2 -2(a+1)y +y^2
The y^2 cancels out,
0 = (a+1)[(a+1) -2y]

So,
(a+1) = 0
a = -1 -----------***

And,
[(a+1) -2y] = 0
a = 2y -1 ------------***

--------------------------------------------
When a = -1,

In Eq.(2),
|x| -1 = y
Or, y = |x| -1 -----------(3)
The graph of that is a V opening upwards whose vertex is at (0,-1) and whose x-intercepts are (-1,0) and (1,0).

In Eq.(2), |x| +|y| = 1,
x cannot be less than -1, nor more than 1.
Therefore, in Eq.(3), y cannot be more than zero.
So, whn a = -1, the solutions for the set....Eq.(1) and Eq.(2)... are only those represented in the graph of
y = |x| -1
from y = -1 up to y = 0 only..

Hence, when a = -1, the solutions for the set are infinitely many, which include, in (x,y):
(0,-1), (+,-0.2, -0.8), (+,-0.5, -0.5), (+,-6.4, -3.6), (+,-0.9, -0.1), (+,-1, 0).

--------------------------------------------------------------------------
When a = 2y -1,

In Eq.(2),
|x| +(2y -1) = y
|x| = y -2y +1
|x| = 1 -y
y = 1 -|x| -----------(4)
The graph of that is a V opening downwards whose vertex is at (0,1) and whose x-intercepts are (-1,0) and (1,0).

Again, in Eq.(2), |x| +|y| = 1,
x cannot be less than -1, nor more than 1.
Therefore, in Eq.(4), y cannot be less than zero.
So, whn a = 2y -1, the solutions for the set are only those represented in the graph of
y = |x| -1
from y = 1 down to y = 0 only..

Hence, when a = 2y -1, the solutions for the set are infinitely many again, which inclue:
(0,1), (+,-0.2, 0.8), (+,-0.5, 0.5), (+,-6.4, 3.6), (+,-0.9, 0.1), (+,-1, 0).

4. Holy God, thanks, I thought I'd struggle with that to death! ;P