Solve the error in the simultaneous equation. The question is: 
$$\begin{align}
\tag{1} 2x - 3y &= 3\\
\tag{2} 4x^2 - 9y^2 &= 3
\end{align}$$
From equation (1): 
$$\tag{3} 2x = 3 - 3y.$$
Substitute equation (3) in (2):
$$\begin{align}
4x^2 - 9y^2 &= 3\\
(2x)^2 - 9y^2 &= 3\\
(3 - 3y)^2 - 9y^2 &= 3\\
 [(3)^2 - 2\cdot 3\cdot 3y + (-3y)^2] - 9y^2 - 3 &= 0\\
 9 +18y + 9y^2 - 9y^2 - 3 &= 0\\
 -18y + 6 &= 0\\
-18y &= -6\\
y &= \frac 1 3
\end{align}$$
But the answer for y in my book is  $ - \frac 1 3\ $ 
Which one is right?

There is other way:
$\ 2x - 3y = 3 \tag{1}$
$(2x - 3y)(2x + 3y) = 3 \tag{2} $
Using $(1)$ we have
$$ 2x - 3y = 3 $$
$$ 2x + 3y = 1 $$
 $$(1)+(2)$$
$$4x=4$$
