Question -
Answer -
i) x + y = 5, 2x + 2y = 10
(ii) x – y = 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
Answer
(i)Given, x + y = 5 and 2x + 2y = 10
(a1/a2) = 1/2
(b1/b2) = 1/2
(c1/c2) = 1/2
Since (a1/a2) = (b1/b2) = (c1/c2)
∴The equations are coincident and they have infinite number of possible solutions.
So, the equations are consistent.
For, x + y = 5 or x = 5 – y
For 2x + 2y = 10 or x = (10-2y)/2
So, the equations are represented in graphs as follows:
From the figure, we can see, that the lines are overlapping each other.
Therefore, the equations have infinite possible solutions.
(ii) Given, x – y = 8 and 3x – 3y = 16
(a1/a2) = 1/3
(b1/b2) = -1/-3 = 1/3
(c1/c2) = 8/16 = 1/2
Since, (a1/a2) = (b1/b2) ≠ (c1/c2)
The equations are parallel to each other and have no solutions. Hence, the pair of linear equations is inconsistent.
(iii) Given, 2x + y – 6 = 0 and 4x – 2y – 4 = 0
(a1/a2) = 2/4 = ½
(b1/b2) = 1/-2
(c1/c2) = -6/-4 = 3/2
Since, (a1/a2) ≠ (b1/b2)
The given linear equations are intersecting each other at one point and have only one solution. Hence, the pair of linear equations is consistent.
Now, for 2x + y – 6 = 0 or y = 6 – 2x
And for 4x – 2y – 4 = 0 or y = (4x-4)/2
So, the equations are represented in graphs as follows
:
From the graph, it can be seen that these lines are intersecting each other at only one point,(2,2).
(iv) Given, 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0
(a1/a2) = 2/4 = ½
(b1/b2) = -2/-4 = 1/2
(c1/c2) = 2/5
Since, a1/a2 = b1/b2 ≠ c1/c2
Thus, these linear equations have parallel and have no possible solutions. Hence, the pair of linear equations are inconsistent.