Determine if the solution set for the system of equations shown is the empty set, contains one point or is infinite. x + y = 6 x - y = 0

  [B]:  "contains one point" .

Explanation:

Given:

x + y  = 6  ;
x - y = 0 ;

 
To solve for "x" ;

Consider the first equation:

x + y = 6 ;

subtract "y" from each side of the equation ; to isolate "x" on one side of the equation; and to solve for "x" ;

x + y - y = 6 - y ;

x = 6 - y ;

Take the second equation:

 x - y = 0 ;

Solve for "x" ;

Add "y" to EACH SIDE of the equation; to isolate "x" on one side of the equation; and to solve for "x" ;

x - y + y = 0 + y ;

    x = y

x = 6 - y

Substitute "x" for "y" ;

x = 6 - x ;

Add "x" to Each side of the equation:

x + x = 6 - x + x ;

2x = 6 ;

Now, divide EACH SIDE of the equation by "2" ; to isolate "x" on one side of the equation; and to solve for "x" ;

2x/2 = 6/2 ;

x = 3 .

Now, since "x = 3" ;  substitute "3" for "x" in both original equations; to see if we get the same value for "y" ;

x + y  = 6  ;
x - y = 0

Start with the first equation:

x + y = 6 ;

3 + y = 6 ;

Subtract  "3" from each side of the equation; to isolate "y" on one side of the equation; and to solve for "y" ;

3 + y - 3 = 6 - 3 ;

y = 3 .

Now, continue with the second equation; {Substitute "3" for "x" to see the value we get for "y"} ;

The second equation given is:

x - y = 0 ;

Substitute "3" for "x" to solve for "y" ;

3 - y = 0 ;

Subtract "3" from EACH side of the equation:

3 - y - 3 = 0 - 3 ;

      -1y = -3  ;

Divide EACH side of the equation by "-1" ;  to isolate "y" on one side of the equation; and to solve for "y" ;

-1y/-1 = -3/-1 ;

    y = 3 .

So, for both equations, we have one value:  x = 3, y = 3;  or:  write as: 
"(3, 3)" ;  { which is:  "one single point" ;  which is:  Answer choice:  [B] } .