# Simultaneous equations

• Aug 2nd 2008, 12:46 AM
atreyyu
Simultaneous equations
Determine the number of solutions of the following set, depending on the value of a.

• Aug 2nd 2008, 05:32 AM
Soroban
Hello, atreyyu!

Did you sketch the graphs?

Quote:

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 . . .

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

• Aug 2nd 2008, 06:14 AM
ticbol
Quote:

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).
• Aug 2nd 2008, 09:40 AM
atreyyu
Holy God, thanks, I thought I'd struggle with that to death! ;P