RD Sharma Test: Pair of Linear Equations in Two Variables - Year 10 MCQ

A system of two linear equations in two variables has infinitely many solutions, if their graphs coincide with each other

Here are the two solutions of each of the given equations. 2x+y = 6
2x+y=0


The area bounded by the given lines and x−axis has been shaded in the graph. 
∴

Given

Therefore, the system has infinite many solution.

The system of linear equations a₁x+b₁y+c₁ = 0 and a₂x+b₂y+c₂ = 0 has a unique solution if 

A system of two linear equations in two variables is consistent, if their graphs intersect only at a point, because it has a unique solution.

A system of two linear equations can have one solution, an infinite number of solutions, or no solution. ... If a system has at least one solution, it is said to be consistent . If a consistent system has exactly one solution, it is independent . If a consistent system has an infinite number of solutions, it is dependent .

Here are the two solutions of each of the given equations.x−y+1=03x+2y−12=0


The triangle formed by given two linear equations is shaded in the graph. The triangle formed by given lines is ΔAEF Therefore, the vertices of triangle AEF are A ( –1, 0), E (2, 3) and F (4, 0).

Given: a₁ = 5,a₂ = 3,b₁ = −15,b₂ = −9,c₁ = 8 and 

Therefore, the the pair of given linear equations has infinitely many solutions.

Given: x−y = 2…..(i)
And x+y = 4 ………(ii)
Adding eq. (i) and (ii) for elimination of y, we get
2x = 6 ⇒ x = 3
Putting the values of x in eq. (i), we get
3−y = 2 ⇒ y = 1
∴x = α = 3 and y = β = 1

Let the length of rectangular garden be x meters and breadth be y meters. According to question,
x = 4+y ⇒ x−y = 4 ……….(i)
Also half of the perimeter = 36 ⇒ x+y = 36 ……….(ii)
Adding eq. (i) and (ii), we get 2x = 40 ⇒ x = 20 meters
Subtracting eq. (i) from (ii), we get 2y = 32 ⇒ y = 16 meters 
∴ Area = Length x Breadth = 20 x 16 = 320 sq. meters

A system of two linear equations in two variables is dependentconsistent, if their graphs coincide with each other.

If equation of a pair of dependent linear equations, then 
For satisfying the condition of dependent linear equations, the values of a₂,b₂ and c₂ should be the multiples of the values of a₁,b₁ and c₁. 
∴ the values would be  and c₁ = 2×(−2)=−4
∴ the second equation can be 10x−14y = −4

Here a unique solution of each variable of a pair of linear equations is given, therefore, it has one solution of a system of linear quations.

Let the two numbers be x and y. 
According to question, x−y = 26 and x = 3y 
Putting the value of x in x−y = 26 we get 3y−y = 26 ⇒ y = 13 And x = 3 × 13 = 39

If equation of a pair of inconsistent linear equations, then 
 
For satisfying the condition of dependent linear equations, the values of a₂,b₂ and c₂ should be the multiples of the values of a₁,b₁ and c₁.
∴ the values would be
 ∴ the second equation can be 4x−6y = 8

Let the numbers be x and .y. 
According to question, x+y = 35 ………(i) and x−y = 13 ………..(ii)
Adding eq. (i) and (ii), we get  2x = 48 ⇒ x = 24
Putting the value of x in eq. (i), we get 12+y = 35 ⇒ y = 13
Therefore, the numbers are 24 and 13.

Here are the two solutions of each of the given equations. x+3y=62x−3y=12

Given: a₁ = a,a₂ = p,b₁ = b,b₂ = q,c₁ = c and c₂ = r
Since the pair of given linear equations has a unique