Draw the graph for each equation given below; hence find the co-ordinates of the points where the graph drawn meets the co-ordinate axes :

(i) $\dfrac{1}{3}x + \dfrac{1}{5}y = 1$

(ii) $\dfrac{2x + 15}{3} = y - 1$

(i) $\dfrac{1}{3}x + \dfrac{1}{5}y = 1$

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -3, then $\dfrac{1}{3} \times (-3) + \dfrac{1}{5}y = 1$ ⇒ y = 10

Let x = 0, then $\dfrac{1}{3} \times 0 + \dfrac{1}{5}y = 1$ ⇒ y = 5

Let x = 3, then $\dfrac{1}{3} \times 3 + \dfrac{1}{5}y = 1$ ⇒ y = 0

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Hence, the co-ordinates of the points where the graph drawn meets the co-ordinate axes are (0, 5) and (3, 0).

(ii) $\dfrac{2x + 15}{3} = y - 1$

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -9, then $\dfrac{2 \times (-9) + 15}{3} = y - 1$ ⇒ y = 0

Let x = -3, then $\dfrac{2 \times (-3) + 15}{3} = y - 1$ ⇒ y = 4

Let x = 0, then $\dfrac{2 \times 0 + 15}{3} = y - 1$ ⇒ y = 6

Let x = 3, then $\dfrac{2 \times 3 + 15}{3} = y - 1$ ⇒ y = 8

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Hence, the co-ordinates of the points where the graph drawn meets the co-ordinate axes are (0, 6) and (-9, 0).