For b and c, both lines are parrallel to each other so theres no solution as they never meet

For b) the lines are parallel.

y = 2x ①
4x - 2y = 5 ② → 2y = 4x - 5 → y = 2x - 5/2

(Same gradient)

Sub ① into ②,

4x - 2(2x) = 5

4x - 4x = 5

0 = 5

Realise that this leads to a contradiction since 0 ≠ 5, so the two equations do not have any real solutions.

For c), they are essentially talking about the same line because :

3x + 4y = 12 ①

Multiply both sides by 2,

2(3x + 4y) = 2(12)
6x + 8y = 24 ②

For such pairs, we use the term 'coincident'

There are infinitely many solutions since on the line, there are many points of x and y that fit the equation.


If you try to do any substitution,

Eg. From ①,

3x = 12 - 4y
6x = 24 - 8y

Sub 6x = 24 - 8y into ②,

24 - 8y + 8y = 24

24 = 24


You'll realise there's no unique solution.

The third one is certainly collinear (I take the word “parallel” to not include the collinear case)

The correct term is 'coincident' .

'collinear' refers to a set of points, i.e lie on the same line. It doesn't refer to a group of lines.

The two terms should not be conflated with each other.

https://mathworld.wolfram.com/CoincidentLines.html

https://mathworld.wolfram.com/Collinear.html