Question -
Answer -
(i) 3x + 2y = 5 ; 2x – 3y = 7
(ii) 2x – 3y = 8 ; 4x – 6y = 9
(iii)(3/2)x+(5/3)y = 7; 9x – 10y = 14
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
(v)(4/3)x+2y = 8 ; 2x + 3y = 12
Answer
(i) Given : 3x + 2y = 5 or 3x + 2y -5 = 0
and 2x – 3y = 7 or 2x – 3y -7 = 0
Comparing these equations with a1x+b1y+c1 = 0
And a2x+b2y+c2 = 0
We get,
a1 = 3, b1 = 2, c1 = -5
a2 = 2, b2 = -3, c2 = -7
(a1/a2) = 3/2
(b1/b2) = 2/-3
(c1/c2) = -5/-7 = 5/7
Since, (a1/a2) ≠ (b1/b2)
So, the given equations intersect each other at one point and they have only one possible solution. The equations are consistent.
(ii) Given 2x – 3y = 8 and 4x – 6y = 9
Therefore,
a1 = 2, b1 = -3, c1 = -8
a2 = 4, b2 = -6, c2 = -9
(a1/a2) = 2/4 = 1/2
(b1/b2) = -3/-6 = 1/2
(c1/c2) = -8/-9 = 8/9
Since , (a1/a2) = (b1/b2) ≠ (c1/c2)
So, the equations are parallel to each other and they have no possible solution. Hence, the equations are inconsistent.
(iii)Given (3/2)x + (5/3)y = 7 and 9x – 10y = 14
Therefore,
a1 = 3/2, b1 = 5/3, c1 = -7
a2 = 9, b2 = -10, c2 = -14
(a1/a2) = 3/(2×9) = 1/6
(b1/b2) = 5/(3× -10)= -1/6
(c1/c2) = -7/-14 = 1/2
Since, (a1/a2) ≠ (b1/b2)
So, the equations are intersecting each other at one point and they have only one possible solution. Hence, the equations are consistent.
(iv) Given, 5x – 3y = 11 and – 10x + 6y = –22
Therefore,
a1 = 5, b1 = -3, c1 = -11
a2 = -10, b2 = 6, c2 = 22
(a1/a2) = 5/(-10) = -5/10 = -1/2
(b1/b2) = -3/6 = -1/2
(c1/c2) = -11/22 = -1/2
Since (a1/a2) = (b1/b2) = (c1/c2)
These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.
(v)Given, (4/3)x +2y = 8 and 2x + 3y = 12
a1 = 4/3 , b1= 2 , c1 = -8
a2 = 2, b2 = 3 , c2 = -12
(a1/a2) = 4/(3×2)= 4/6 = 2/3
(b1/b2) = 2/3
(c1/c2) = -8/-12 = 2/3
Since (a1/a2) = (b1/b2) = (c1/c2)
These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.