# Thread: System of equations

1. ## System of equations

Dear Forum I am having issues with the following question any feedback would be appreciated.

"Describe all possible outcomes of solving a system of two equations. Give examples"

Thanks -AC-

2. Hello, AlgebraicallyChallenged!

I would have assumed that you've run into these cases by now.

Describe all possible outcomes of solving a system of two equations.
Give examples.
There are three possible outcomes.

[1] The system has a unique solution.

Example: .$\displaystyle \begin{array}{ccc}x + y &=& 8 \\ x - y &=& 2\end{array}$

The solution is: .$\displaystyle (x,y) \,=\,(5,3)$

The problem asks for two numbers whose sum is 8 and whose difference is 2.
. . The numbers are 5 and 3.

If we graph the two lines, they intersect at (5,3).

[2] The system has no solutions.

Example: .$\displaystyle \begin{array}{ccc}x + y &=& 4 \\ x + y &=& 3\end{array}$

If we try to solve the system, we get a false statement, like: .$\displaystyle 0 = 1$
. . This indicates that the system has no solution.

The problem asks for two numbers whose sum is 4 and whose sum is 3.
. . Of course, there are no such numbers.

If we graph the two line, we find that they are parallel.
. . They do not intersect.

[3] The system has infinite soluions.

Example: .$\displaystyle \begin{array}{ccc}x + y &=& 4 \\ 2x + 2y &=&8\end{array}$

If we try to solve the system, we get a true statement, like: .$\displaystyle 0 = 0$
. . This indicates that the system has infinitely many solutions.

The problem says: find two numbers whose sum is 4
. . and if we double the numbers and add them, we get 8.

We see that $\displaystyle (3,1)$ is a solution . . . but so is $\displaystyle (4,0),\;(5,-1),\;\hdots$

If we graph the lines, we get the same line twice.
. . So, of course, they "intersect" each other a zillion times.

3. examples ( 2 variables ) :

1. case

x + y = 2

x - y = 1

this system has only one solution x = 3/2 , y = 1/2 ( 3/2, 1/2 )

2. case

x + y = 2

2x + 2y = 4

this system has infinite set of solutions

( x = t => y = 2 - t ) => solution is ( t , 2 - t ) = t*( 1, -1 ) + ( 0, 2 )

where t is element of F ( F = R or C )

3. case

x + y = 2

x + y = 4

this system obvious has no solutions