# Draw the graph of

3x + 5y – 15 = 0

5y = 15 – 3x

When, x = 0 then,

y = 3

When, x = 5 then,

y = 0

Plotting (0, 3) and (5, 0) we get the following graph,

The blue line indicates the required graph of 3x + 5y – 15 = 0

Now, to show that (1, 2) is not the solution of

3x + 5y – 15 = 0

We put x = 1 and y = 2 in

Since, LHS ≠ RHS therefore, x = 1, y = 2 is not a solution 3x + 5y – 15 = 0.

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