Question 1. The following observed values of x and y are thought to satisfy a linear equation.
Draw the graph using the values of x and y as given in the above table.
At what points the graph of the linear equation cuts the x-axis?
Solution. Plotting the points (6, –2) and (–6, 6) and joining them we get the graph AB which is the required graph.
The graph AB cuts the x-axis at point C (3, 0) and the y-axis at point D (0, 2).
Question 2. The taxi fare in a town is ₹10 for the first kilometre and ₹ 6 per km for the subsequent distance. Taking the distance as ‘x’ km and total fare as ₹y, write a linear equation for this information, what will be the total fare for 15 km?
Solution:
∵ 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.
Question 3. Draw the graph of the equation x – y = 4. From the graph, find the coordinates of the point when the graph line meets the x-axis.
Solution: We have x – y = 4 or y = x – 4
When x = 0, then y = 0 – 4 = –4
When x = 1, then y = 1 – 4 = –3
When x = –1, then y = – 1 – 4 = – 5
We get the following table of values of x and y.
x | 0 | 1 | -1 |
y | -4 | -3 | -5 |
(x, y) | (0, –4) | (1, –3) | (–1, –5) |
∴ We have the ordered pairs of solution for x – y = 4 as (0, –4), (1, –3) and (–1, –5).
Now, plotting the points (0, –4), (1, –3) and (–1, –5) and then joining them, we get the following graph of x – y = 4.
From the graph, we find that the graph line meets the x-axis at (4, 0).
Question 4. Draw the graph x + 2y = 6 and from the graph, find the value of x when y = – 3.
Solution: 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.
x | 0 | 2 | 4 |
y | 3 | 2 | 1 |
(x, y) | (0, 3) | (2, 2) | (4, 1) |
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.
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