identify the system of equality which has no solution

Question

1. x-2y=-6

   2x=4y-12

2. x+2y-1

   2x+4y

3. 5x+2y=4

    2x-2y+10

4. -x+3y+1

   x+3y-1

Answer 1

I.  x - 2y = -6   .....(1)
   2x = 4y - 12 ....(2)
   Let's (2) by "2" and move "y" to the right side of an equation
   x - 2y = -6 , so you can see, that (1) and (2) are identical.
The system has an infinite number of solutions.

II. x = 2y - 1  ......(1)
    2x =  4y    ......(2)
    Let's divide (2) by "2" we will get
    x = 2y      ......(2)
This system has no solutions. There is contradiction, from one side x = 2y - 1 and  from another side x = 2y

III. For this system let's use method of illumination:
    5x + 2y = 4
 +
    2x - 2y = 10
‾‾‾‾‾‾‾‾‾‾‾‾‾‾
   7x        = 14
     x = 2
Let's plug in value of "x" into first equation
5 • (2) + 2y = 4
10 + 2y = 4
2y = - 6
y = - 3
The solution of the given system is pair of numbers (2, - 3)

IV.  - x = 3y + 1  .....(1)
         x = 3y - 1   ....(2)
Let's live "3y" by itself for both equations
      - x - 1 = 3y
    —
        x + 1 = 3y
    ‾‾‾‾‾‾‾‾‾‾‾‾‾
     - 2x - 2 = 0

    - 2x = 2 
       x = - 1
Let's plug in value of "x" into second equation
   - 1 = 3y - 1
    3y = 0
     y = 0

The solution of the given system is pair of numbers (- 1, 0)

Answer 2

(2)

This system will not have a unique solution.  The slopes of the two lines are identical - so either they are parallel (and never intersect) or are identical lines (and an infinite number of identical points).

Its not quite possible to tell - as the second equation in the (2), (3) and (4) answers are all missing the equal sign.  But in the (3) and (4) case - one can see that the slopes are different - so they must intersect somewhere.