xy+x=0.  x2+y2=4. solve these equations anywhere

We are given the system of equations:

\[
1. \ xy + x = 0 
\]
\[
2. \ x^2 + y^2 = 4
\]

### Step 1: Solve the first equation for \(y\)
From the first equation:
\[
xy + x = 0 \implies x(y + 1) = 0
\]
This gives us two possible cases:
1. \(x = 0\)
2. \(y + 1 = 0\) (i.e. \(y = -1\))

### Case 1: \(x = 0\)
Substitute \(x = 0\) into the second equation:
\[
x^2 + y^2 = 4 \implies 0^2 + y^2 = 4 \implies y^2 = 4
\]
Thus, \(y = \pm 2\).

Therefore, for \(x = 0\), the solutions are:
\[
(0, 2) \quad \text{or} \quad (0, -2)
\]

### Case 2: \(y = -1\)
Substitute \(y = -1\) into the second equation:
\[
x^2 + (-1)^2 = 4 \implies x^2 + 1 = 4 \implies x^2 = 3
\]
Thus, \(x = \pm \sqrt{3}\).

Therefore, for \(y = -1\), the solutions are:
\[
(\sqrt{3}, -1) \quad \text{or} \quad (-\sqrt{3}, -1)
\]

### Final Solution
The solutions to the system of equations are:
\[
(0, 2), \ (0, -2), \ (\sqrt{3}, -1), \ (-\sqrt{3}, -1)
\]

Would you like further details on any step or clarifications?

### Related Questions:
1. What is the general method for solving systems of nonlinear equations?
2. How does factoring help in solving algebraic equations?
3. What are the geometric interpretations of these solutions?
4. How can you apply substitution or elimination methods to nonlinear systems?
5. How would the solutions change if the second equation were \(x^2 + y^2 = r^2\)?

#### Tip:
For solving nonlinear systems, always check for multiple solutions by considering all possible factorizations.

xy + x = 0. x^2 + y^2 = 4. solve these equations anywhere

Learn how to solve the system of nonlinear equations xy + x = 0 and x^2 + y^2 = 4 using algebraic methods like factoring and substitution. This detailed step-by-step solution covers factoring techniques and geometric interpretation. Suitable for students in Grades 10-12.

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