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

Tutors, please sign in to answer this question.

* 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.

II.

2x = 4y ......(2)

Let's divide (2) by "2" we will get

x = 2y ......(2)

III. For this system let's use method of illumination:

5x + 2y = 4

+

2x - 2y = 10

‾‾‾‾‾‾‾‾‾‾‾‾‾‾

7x = 14

Let's plug in value of "x" into first equation

5 • (2) + 2y = 4

10 + 2y = 4

2y = - 6

x = 3y - 1 ....(2)

Let's live "3y" by itself for both equations

- x - 1 = 3y

—

x + 1 = 3y

‾‾‾‾‾‾‾‾‾‾‾‾‾

- 2x - 2 = 0

- 2x = 2

Let's plug in value of "x" into second equation

- 1 = 3y - 1

3y = 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