Previous Year Questions : Linear Equations in Two Variables

(i) False. [As written, ax + by + c is an expression, not an equation. The linear equation in two variables is usually written as ax + by + c = 0 with a and b not both zero.]

(ii) False.  [A single linear equation in two variables (like 2x + 3y = 5) represents a line in the plane, so it has infinitely many solutions (every point on that line is a solution). It would have a unique solution only if a second independent equation is given.]

(iii) False. [Points on the x-axis must have y = 0. Only (2, 0) and (-3, 0) lie on the x-axis; (4, 2) and (0, 5) do not.]

(iv) True. [The vertical line 4 units to the left of the y-axis has equation x = -4 (all points with x = -4).]

(v) False. [The graph of y = mx + c passes through the origin only when c = 0. For general c it does not pass through (0,0).]

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

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

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.

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.

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.

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.

∵ 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.

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.