Advertisements
Advertisements
प्रश्न
Draw a graph of the equation 2x - 3y = 15. From the graph find the value of:
(i) x, when y = 3
(ii) y, when x = 0
Advertisements
उत्तर
We have
2x - 3y = 15
⇒ -3y = 15 - 2x
⇒ 3y = 2x - 15
⇒ y = `(2x - 15)/(3)`
When x = -2
⇒ y = `-(19)/(3)`
= -6.34
When x = 0
⇒ y = `-(15)/(3)` = -5
When x = 2
⇒ y = `-(11)/(3)` = -3.66
| x | -2 | -1 | 0 | 1 | 2 |
| y | -6.34 | -5.66 | -5 | -4.34 | -3.66 |
Thus ordered pairs of 2x - 3y = 15 are {(-2, -6.34), (-1, -5.66), (0, -5),(1, - 4.34), (2, -3.66)}. Hence graph is a below.
(i) x, when y = 3
From graph we find that x = 12, when y = 3
(ii) y, when x = 0
Fro graph we find that y = -5, when x = 0.
APPEARS IN
संबंधित प्रश्न
Draw the graph of the equation given below.
3x - y = 0
Draw the graph for the linear equation given below:
y = x
Draw the graph for the linear equation given below:
3x + 2y = 0
Draw the graph for the linear equation given below:
y = - x + 4
For the linear equation, given above, draw the graph and then use the graph drawn (in the following case) to find the area of a triangle enclosed by the graph and the co-ordinates axes:
3x − (5 − y) = 7
For the linear equation, given above, draw the graph and then use the graph drawn (in the following case) to find the area of a triangle enclosed by the graph and the co-ordinates axes:
7 - 3 (1 - y) = -5 + 2x
Draw the graph of equation `x/(4) + y/(5) = 1` Use the graph drawn to find:
(i) x1, the value of x, when y = 10
(ii) y1, the value of y, when x = 8.
Find if the following points are collinear or not by using a graph:
(i) (-2, -1), (0, 3) and (1, 5)
(ii) (1, 3), (-2, -4) and (3, 5)
(iii) (2, -1), (2, 5) and (2, 7)
(iv) (4, -1), (-5, -1) and (3, -1)
Draw a graph of the equation 3x - y = 7. From the graph find the value of:
(i) y, when x = 1
(ii) x, when y = 8
Draw a graph of the equation 2x + 3y + 5 = 0, from the graph find the value of:
(i) x, when y = -3
(ii) y, when x = 8
