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The image contains a system of two equations to be solved. The equations are as follows:

1. \( x y^2 + 2x - 4y = -1 \)
2. \( x^2 y^3 + 2 x y^2 - 4x + 3y = 2 \)

To solve this system, let's use algebraic methods. Here are the steps:

### Step 1: Solve for \(x\) or \(y\) in one equation
First, solve for one variable in terms of the other in one of the equations. Let's start with the first equation:
\[ x y^2 + 2x - 4y = -1 \]
Solve for \(x\) in terms of \(y\):
\[ x(y^2 + 2) = 4y - 1 \]
\[ x = \frac{4y - 1}{y^2 + 2} \]

### Step 2: Substitute into the second equation
Next, substitute this expression for \(x\) into the second equation:
\[ x^2 y^3 + 2 x y^2 - 4x + 3y = 2 \]
\[ \left(\frac{4y - 1}{y^2 + 2}\right)^2 y^3 + 2 \left(\frac{4y - 1}{y^2 + 2}\right) y^2

Giải hệ phương trình: x*y^2 + 2x - 4y = -1 và x^2*y^3 + 2xy^2 - 4x + 3y = 2

Learn how to solve the system of nonlinear equations: x*y^2 + 2x - 4y = -1 and x^2*y^3 + 2xy^2 - 4x + 3y = 2. This problem involves algebraic manipulation and substitution methods. Suitable for high school students.

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