Question 18

If $$\mid x + 1 \mid + (y + 2)^2 = 0$$ and $$ax - 3ay = 1$$, Then the value of a is

Solution

If $$\mid x + 1 \mid + (y + 2)^2 = 0$$

That means x = -1 and y = -2

Then, only LHS can be equal to zero.

And, $$ax - 3ay = 1$$

$$-a +6a = 1$$

$$5a=1$$

Thus, $$a=\dfrac{1}{5}$$

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