Class 9 Maths Chapter 4 Previous Year Questions - Linear Equations in Two Variables

(v) The graph of the equation y = mx + c passes through the origin.
Sol: 

(i) False. [Because ax + by + c = 0 is a linear equation in two variables if both ‘a‘ and ‘b’ are non-zero.]

(ii) False. [Because a linear equation in two variables has infinitely many solutions.]

(iii) False. [Because the points (2, 0) and (–3, 0) lie on the x-axis, (0, 5) lie on the y-axis whereas the point (4, 2) lies in the first quadrant.]

(iv) True.

(v) False. [Because the point (0, 0) i.e., x = 0 + y = 0 does not satisfy the equation]

Sol:

 True. [Since, on looking at the given coordinates, we observe that each y-coordinate is two units more than double the x-coordinate.]

Sol: The point (6, 0) does not lie on the graph.
∴ The point (6, 0) is not the solution of the equation.

Sol:

 The given equation is 2x + 3y = 12  ...(1)

Here Solution =

⇒ x = 2 and y = (8/3)
Substituting x = 2 and y =(8/3)  in (1), we get

⇒ 4 + 8 = 12⇒ 12 = 12∵ L.H.S. = R.H.S

∴ is a solution of 2x + 3y = 12.

Sol:

 We have 3x + y = 8
For x = 0, we have 3(0) + y = 8
⇒ 0 x y= 8 ⇒ y = 8

  .e. (0, 8) is a solution.For x = 1, we have 3(1) + y = 8⇒ 3 + y = 8⇒ y = 8 – 3 = 5 

    i.e. (1, 5) is another solution.

Sol:

We have kx + 3y = 7                          ...(1)
∴ Putting x = –1 and y = 2 in (1), we get
k(–1) + 3(2) = 7
⇒ –k + 6 = 7
⇒ –k = 7 – 6 = 1
⇒ k= –1
Thus, the required value of k = –1.

Sol:

 We have 2x + 3y = 7                       ...(1)
Since, x = 2 and y = 1 satisfy the equation (1).
∴ Substituting x = 2 and y = 1 in (1), we get
L.H.S. = 2(2) + 3(1) = 4 + 3 = 7
= R.H.S.
Since, L.H.S. = R.H.S.
∴ x = 2 and y = 1 satisfy the given equation.

 Sol:
∵ Total distance is x km.
Total fare = ₹y
∴ x = 1 + (x – 1) = First km + Subsequent distance
Since, fare the first km = ₹10
∴ Fare for the remaining distance = ₹6 x (x – 1) = ₹6x – ₹6
⇒ Total fare = ₹10 + ₹6x – ₹6
= ₹4 + ₹6x
∴ y = 4 + 6x

⇒ y – 6x = 4
⇒ 6x – y + 4 = 0
Which is the required equation.

Now, total fare for 15 km:
6 x 15 – y + 4 = 0   [Substituting x = 15]
⇒ 90 – y + 4 = 0
⇒ 94 – y = 0
⇒ y = 94
∴ Total fare = ₹94.

Sol:

We have: x + 2y = 6
⇒  When x = 0, then When x = 2, then When x = 4, then  

We get the following table of values of x and y.

Plotting the ordered pairs (0, 3), (2, 2) and (4, 1) and then joining them, we get the graph of x + 2y = 6 as shown below:

From the graph, we find that for y = – 3, the value of x = 12.