Practice Test: Pair of Linear Equations in Two Variables - Class 10 MCQ

Homogeneous system of  linear equations are of the form ax+by=0 and cx+dy=0. These lines pass through the origin. Thus, there is always at least one solution, the point (0,0).So, Homogeneous system of linear is always consistent, meaning that it has at least one solution which is (0,0). So B is the exact explanation.

x = 6 km/h  and  y = 4km/h 

Given :

2 k x + 5 y – 7 = 0  ...( i )
6 x – 5 y – 1 = 0   ... ( ii )
Pair of linear equations has a unique solution.
We know for unique solution.

Comparing from ( i ) and ( ii ) we have

Put these values in formula.

Thus we get answer many values of k but leaving k ≠ -3.

An equation has infinitely many solutions when the lines are coincident.
The lines are coincident when 
So 3x + 4y = k, 9x + 12y = 6 are coincident when

An equation has infinitely many solutions when the lines are coincident.
The lines are coincident when 
So 2x + 5y = k, kx + 15y = 18 are coincident when

3/12 = 7/2k  [ applying a1/a2=b1/b2 ]
3 x 2k = 7 x 12
k=14

Work done by 1 girl and 1 boy in x and y days respectively.

work done by 1 girl and 1 boy in 1 day is (1/x) and (1/y).

so, work done by 8 girls and 12 boys in 1 day is (8/x) + (12/y) = 1/10

let (1/x) = a and (1/y) = b

so, 8a + 12b = 1/10

→ 80a + 120b = 1 ---- (1)

work done by 6 girls and 8 boys in 1 day is (6/x) + (8/y) = 1/14

6a + 80 = 1/14

→ 84a + 112b = 1 ---- (2)

By elimination method, Multiple equation 1 by 21 on both sides, we get

1680a + 2520b = 21 ---- (3)

Multiply equation 2 by 20 on both sides, we get

1680a + 2240b = 20 ---- (4)

On solving equation 3 and 4, we get

2520 b - 2240b = 21 - 20

→ 280 b = 1

→ b = 1/280

b =1/y

→ 1/280 = 1/y

→ y = 280

80a + 120 x (1/280) = 1 (From 1)

→ 80a + (3/7) = 1

→ 80a = 1 - (3/7)

→ 80a = (7 - 3)/7

→ 80a = 4/7

→ a = 4/(7 × 80)

→ a = 1/140

→ a = 1/x

→ 1/140 = 1/x

→ x = 140

1 girl and 1 boy alone take 140 days and 280 days to complete a work.

kx + 4y = 5, 3x + 2y = 5
Here, a1=k,b1=4,c1​=−5
and a2=3,b2=2,c2=-5

So , The equation is consistent when 

k ≠ 6

Given equations are:
x+2y=5 ....(1)
7x+3y=13 .....(2)
Multiplying equation (1) by 7
7x+14y=35 .....(3)
Subtracting equation (2) from equation (3), we have
11y=22
y=2
Now putting y value in equation (1)
⇒x+2(2)=5
⇒x=5−4
⇒x=1
x=1,y=2

Let the present age of man is x and of son is y.
Six years hence,
Man’s age =x+6
Son’s age=y+6
Man’s age is 3 times son’s age
x+6=3(y+6)
x+6=3y+18
x=3y+12    …...1
Three years ago,
Man’s age =x-3
Son’s age=y-3
Man’s age was 9 times as of son
x-3=9(y-3)
x-3=9y-27
x=9y-24   ….2
From 1 and 2
3y+12=9y-24
6y=36
y=6
x=3*6+12=18+12=30 years

x - y  = 2
x = 2 + y
Substituting the value of x in the second equation we get;
2 + y + y = 4
2 + 2y = 4
2y = 2
Y = 1
Now putting the value of y, we get;
x = 2 + 1 = 3
Hence, the solutions are x = 3 and y = 1.

Given, 9x + 3y + 12 = 0 and 18x + 6y + 26 = 0 
a₁/a₂ = 9/18 = 1/2 
b₁/b₂ = 3/6 = 1/2 
c₁/c₂ = 12/26 = 6/13 
Since, a₁/a₂ = b₁/b₂ ≠ c₁/c₂ 
So, the pairs of equations are parallel and the lines never intersect each other at any point, therefore there is no possible solution.

a₁/a₂ = 1/-4
b₁/b₂ = 2/-8 = 1/-4
c₁/c₂ = -5/20 = -¼
This shows:
a₁/a₂ = b₁/b₂ = c₁/c₂
Therefore, the pair of equations has infinitely many solutions.

The condition for parallel lines is:
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Hence, 3/2 = 2k/5
k = 15/4

The condition for dependent linear equations is:
a₁/a₂ = b₁/b₂ = c₁/c₂
For option a,
a₁/a₂ = b₁/b₂ ≠ c₁/c₂ =  ½