Chapter 4: Linear Equations in Two Variables

True.

Justification:

We have the equation, 3x + 4y = 12.

Substituting the values of x = 0 and y = 3 from the point (0, 3) in the equation,

We get,

3(0) + 4(3) = 12 = RHS.

Hence, the point (0, 3) lies on the graph of the linear equation 3x + 4y = 12.

False.

Justification:

We have the equation, x + 2y = 7.

Substituting the values of x = 0 and y = 7 from the point (0, 7) in the equation,

We get,

0 + 2(7) = 14 ≠RHS

Hence, the graph of the linear equation x + 2y = 7 passes through the point (0, 7).

True.

Justification:

We have the equation, x + y = 0.

x + y = 0

x = – y

from the graph, we get the points ( – 3, 3) and ( – 1, 1),

Considering the point ( – 3, 3)

x = – 3 and y = 3

Hence, substituting ( – 3, 3) in equation,

We get,

– 3 = 3 which satisfies the equation x = – y

Considering the point ( – 1, 1)

x = – 1 and y = 1

Hence, substituting ( – 1, 1) in equation,

We get,

– 1 = 1 which satisfies the equation x = – y

Therefore, the given solution: ( – 3, 3) and ( – 1, 1) satisfies the given equation x = – y.

Hence, the given graph represents the linear equation x + y = 0.