Determine the number of solutions of the following set, depending on the value of a.
http://www.forummatematyka.pl/latexr...1c9dad5e15.png
Please explain! (Crying)
Printable View
Determine the number of solutions of the following set, depending on the value of a.
http://www.forummatematyka.pl/latexr...1c9dad5e15.png
Please explain! (Crying)
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 yaxis
. . and you will observe the following . . .
. . $\displaystyle \begin{array}{cc}\text{Domain} & \text{Solutions} \\ \hline \\[3mm]
a < 1 & 0 \\
a = 1 & \text{infinite} \\
1 < a < 1 & 2 \\
a = 1 & 1 \\
a > 1 & 0 \end{array}$
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 xintercepts 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 xintercepts 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).
Holy God, thanks, I thought I'd struggle with that to death! ;P