Pa.Linear Eq Ex 3.4 Solutions
Question - 11 : - Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
Answer - 11 : -
(i) x – 3y – 3 = 0 and 3x – 9y – 2 = 0 (ii) 2x + y = 5 and 3x + 2y = 8
(iii) 3x – 5y = 20 and 6x – 10y = 40 (iv) x – 3y – 7 = 0 and 3x – 3y – 15 = 0
Solutions:
(i) Given, x – 3y – 3 =0 and 3x – 9y -2 =0
a1/a2=1/3 , b1/b2= -3/-9 =1/3, c1/c2=-3/-2 = 3/2
(a1/a2) = (b1/b2) ≠ (c1/c2)
Since, the given set of lines are parallel to each other they will not intersect each other and therefore there is no solution for these equations.
(ii) Given, 2x + y = 5 and 3x +2y = 8
a1/a2 = 2/3 , b1/b2 = 1/2 , c1/c2 = -5/-8
(a1/a2) ≠ (b1/b2)
Since they intersect at a unique point these equations will have a unique solution by cross multiplication method:
x/(b1c2-c1b2) = y/(c1a2 – c2a=) = 1/(a1b2-a2b1)
x/(-8-(-10)) = y/(15+16) = 1/(4-3)
x/2 = y/1 = 1
∴ x = 2 and y =1
(iii) Given, 3x – 5y = 20 and 6x – 10y = 40
(a1/a2) = 3/6 = 1/2
(b1/b2) = -5/-10 = 1/2
(c1/c2) = 20/40 = 1/2
a1/a2 = b1/b2 = c1/c2
Since the given sets of lines are overlapping each other there will be infinite number of solutions for this pair of equation.
(iv) Given, x – 3y – 7 = 0 and 3x – 3y – 15 = 0
(a1/a2) = 1/3
(b1/b2) = -3/-3 = 1
(c1/c2) = -7/-15
a1/a2 ≠ b1/b2
Since this pair of lines are intersecting each other at a unique point, there will be a unique solution.
By cross multiplication,
x/(45-21) = y/(-21+15) = 1/(-3+9)
x/24 = y/ -6 = 1/6
x/24 = 1/6 and y/-6 = 1/6
∴ x = 4 and y = 1.
Question - 12 : - (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?
Answer - 12 : -
2x + 3y = 7
(a – b) x + (a + b) y = 3a + b – 2
(ii) For which value of k will the following pair of linear equations have no solution?
3x + y = 1
(2k – 1) x + (k – 1) y = 2k + 1
Solution:
(i) 3y + 2x -7 =0
(a + b)y + (a-b)y – (3a + b -2) = 0
a1/a2 = 2/(a-b) , b1/b2 = 3/(a+b) , c1/c2 = -7/-(3a + b -2)
For infinitely many solutions,
a1/a2 = b1/b2 = c1/c2
Thus 2/(a-b) = 7/(3a+b– 2)
6a + 2b – 4 = 7a – 7b
a – 9b = -4 ……………………………….(i)
2/(a-b) = 3/(a+b)
2a + 2b = 3a – 3b
a – 5b = 0 ……………………………….….(ii)
Subtracting (i) from (ii), we get
4b = 4
b =1
Substituting this eq. in (ii), we get
a -5 x 1= 0
a = 5
Thus at a = 5 and b = 1 the given equations will have infinite solutions.
(ii) 3x + y -1 = 0
(2k -1)x + (k-1)y – 2k -1 = 0
a1/a2 = 3/(2k -1) , b1/b2 = 1/(k-1), c1/c2 = -1/(-2k -1) = 1/( 2k +1)
For no solutions
a1/a2 = b1/b2 ≠ c1/c2
3/(2k-1) = 1/(k -1) ≠ 1/(2k +1)
3/(2k –1) = 1/(k -1)
3k -3 = 2k -1
k =2
Therefore, for k = 2 the given pair of linear equations will have no solution.