Question: 232.
$$(x+y)^2 = (x-y)^2 - y + 4x(y-1)$$
$$2x + y = 3$$
1. **State the problem:**
Solve the system of equations:
$$ (x+y)^2 = (x-y)^2 - y + 4x(y-1) $$
$$ 2x + y = 3 $$
2. **Expand and simplify the first equation:**
Recall the expansion formulas:
$$ (x+y)^2 = x^2 + 2xy + y^2 $$
$$ (x-y)^2 = x^2 - 2xy + y^2 $$
Substitute these into the first equation:
$$ x^2 + 2xy + y^2 = x^2 - 2xy + y^2 - y + 4x(y-1) $$
3. **Simplify the right side:**
Distribute $4x$:
$$ 4x(y-1) = 4xy - 4x $$
So the equation becomes:
$$ x^2 + 2xy + y^2 = x^2 - 2xy + y^2 - y + 4xy - 4x $$
4. **Bring all terms to the left side:**
$$ x^2 + 2xy + y^2 - x^2 + 2xy - y^2 + y - 4xy + 4x = 0 $$
5. **Simplify by canceling terms:**
- $x^2 - x^2 = 0$
- $y^2 - y^2 = 0$
- Combine $2xy + 2xy - 4xy = 0$
So the equation reduces to:
$$ y + 4x = 0 $$
6. **Use the second equation:**
$$ 2x + y = 3 $$
7. **Solve the system:**
From step 5, we have:
$$ y = -4x $$
Substitute into the second equation:
$$ 2x + (-4x) = 3 $$
$$ 2x - 4x = 3 $$
$$ -2x = 3 $$
$$ x = -\frac{3}{2} $$
8. **Find $y$:**
$$ y = -4x = -4 \times \left(-\frac{3}{2}\right) = 6 $$
**Final answer:**
$$ x = -\frac{3}{2}, \quad y = 6 $$