Question 1 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:
(i) x-3y-3=0 (ii) 2x+y=5
3x-9y-2=0 3x+2y=8
(iii) 3x-5y=20 (iv) x-3y-7=0
6x-10y=40 3x-3y-15=0
Class 10 - Math - Pair of Linear Equations in Two Variables Page 62
Answers
a1x + b1y + c1 = 0 ,
a2x + b2y + c2 = 0 ,
Condition 1: Intersecting Lines
If a 1 / a 2 ≠ b 1 / b 2 , then the pair of linear equations has a unique solution.
Condition 2: Coincident Lines
If a 1 / a 2 = b 1 / b 2 = c 1 / c 2 ,then the pair of linear equations has infinite solutions.
A pair of linear equations, which has a unique or infinite solutions are said to be a consistent pair of linear equations.
A pair of linear equations, which has infinite many distinct common solutions are said to be a consistent pair or dependent pair of linear equations.
Condition 3: Parallel Lines
If a 1/ a 2 = b 1/ b 2 ≠ c 1 / c 2 , then a pair of linear equations has no solution.
A pair of linear equations which has no solution is said to be an inconsistent pair of linear equations.
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(i) x – 3y – 3 = 0
3x – 9y – 2 =0
on comparing with ax+by+c+0
a1= 1 , b1=-3, c1= -3
a2=3, b2=-9, c2= - 2
a1/a2 = 1/3b1/b2 = -3/-9 = 1/3
c1/c2 = -3/-2 = 3/2
a1/a2 = b1/b2 ≠ c1/c2
the given sets of lines are parallel to each other.
Therefore, they will not intersect each other and thus, there will not be any solution for these equations.
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(ii) 2x + y = 5 , 2x + y- 5 =0
3x +2y = 8 , 3x +2y -8 =0
on comparing with ax+by+c+0
a1= 2 , b1=1, c1= -5
a2=3, b2=2, c2= -8
a1/a2 = 2/3
b1/b2 = 1/2
c1/c2 = -5/-8 = 5/8
a1/a2 ≠ b1/b2
Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.
formula for cross multiplication method
x/ b 1 c 2 - b 2 c 1 = y/ c 1 a 2 - c 2 a 1 = 1/a 1 b 2 - a 2 b 1
x / (1)(-8) - (2)(-5) = y / (-5)(3) - (-8) (2) = 1 / (2) (2) - (3) (1)
x/-8-(-10) = y/-15+16 = 1/4-3
x/2 = y/1 = 1/1
Now take x/2 =1/1 & y/1 = 1/1
∴ x = 2, y = 1.
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(iii) 3x – 5y = 20 , 3x – 5y - 20 =0
6x – 10y = 40 , 6x -10y -40 =0
on comparing with ax+by+c+0
a1= 3, b1=-5, c1= -20
a2=6, b2=10, c2= -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
Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite many solutions .
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(iv) x – 3y – 7 = 0
3x – 3y – 15= 0
on comparing with ax+by+c+0
a1= 1 , b1=-3, c1= -7
a2=3, b2=-3, c2= -15
a1/a2 = 1/3
b1/b2 = -3/-3 = 1
c1/c2 = -7/-15 = 7/15
a1/a2 ≠ b1/b2
Therefore, they will intersect each other at a unique point and thus, there will be a unique solution
formula for cross multiplication method
x/ b 1 c 2 - b 2 c 1 = y/ c 1 a 2 - c 2 a 1 = 1/a 1 b 2 - a 2 b 1
x / (-3) (-15) - (3)(-7) = y / (-7)(3) - (-15)(1) = 1 / (1)(-3) - (3)(-3)
x/45-(21) = y/-21-(-15) = 1/-3-(-9)
x/24 = y/-6 = 1/6
x/24 = 1/6 & y/-6 = 1/6
x = 4 and y = -1
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Hope this will help you....
Answer:
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