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
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
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)
(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.
Comments
i wrote it wrong its
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