Does the equation specify a function with independent variable x? If so, find the domain
of the function. If not, find a value of x to which there corresponds more than one value of y.

a) 2x + 5y = 10
b) y(x + y) = 4
c) x(x + y) = 4

Let's analyze each equation to determine if it specifies a function with an independent variable \(x\), and if so, find the domain of the function. If it doesn't specify a function, we'll look for values of \(x\) that correspond to more than one value of \(y\).

### a) \(2x + 5y = 10\)

To determine if this equation represents a function of \(x\), we solve for \(y\) in terms of \(x\):

\[
5y = 10 - 2x
\]
\[
y = \frac{10 - 2x}{5}
\]

This is a linear equation, so it specifies \(y\) as a function of \(x\). The domain of this function is all real numbers, since there are no restrictions on \(x\).

**Domain**: \( (-\infty, \infty) \)

### b) \(y(x + y) = 4\)

First, let's try to isolate \(y\) to see if it represents \(y\) as a function of \(x\):

\[
y(x + y) = 4
\]

Expanding the left-hand side:

\[
yx + y^2 = 4
\]

This is a quadratic equation in \(y\) (i.e., \(y^2 + xy - 4 = 0\)). Solving this quadratic for \(y\) using the quadratic formula:

\[
y = \frac{-x \pm \sqrt{x^2 + 16}}{2}
\]

Since there is a "±" sign in the solution, this means that for certain values of \(x\), there are two possible values of \(y\). Thus, this equation does **not** define \(y\) as a function of \(x\) because for some values of \(x\), there are multiple values of \(y\).

For instance, if \(x = 0\):

\[
y = \frac{-0 \pm \sqrt{0^2 + 16}}{2} = \frac{\pm 4}{2} = \pm 2
\]

So for \(x = 0\), there are two values of \(y\) (\(y = 2\) and \(y = -2\)).

### c) \(x(x + y) = 4\)

Let's try to solve for \(y\) in terms of \(x\):

\[
x(x + y) = 4
\]
\[
x^2 + xy = 4
\]
\[
xy = 4 - x^2
\]
\[
y = \frac{4 - x^2}{x}
\]

This expression for \(y\) is defined as long as \(x \neq 0\) (since division by zero is undefined). So, this equation does define \(y\) as a function of \(x\), but with the restriction that \(x \neq 0\).

**Domain**: \( (-\infty, 0) \cup (0, \infty) \)

---

### Summary:
- a) \(2x + 5y = 10\) defines \(y\) as a function of \(x\). Domain: \( (-\infty, \infty) \)
- b) \(y(x + y) = 4\) does not define \(y\) as a function of \(x\). Example: \(x = 0\) gives two values of \(y\).
- c) \(x(x + y) = 4\) defines \(y\) as a function of \(x\) with the domain \(x \neq 0\), or \( (-\infty, 0) \cup (0, \infty) \).

Do you want more details, or do you have any questions?

### Related Questions:
1. How do you determine if an equation represents a function?
2. What is the quadratic formula and how is it used to solve quadratic equations?
3. How do you find the domain of a function involving a fraction?
4. Why can't the domain include values where division by zero occurs?
5. What are the key differences between linear and quadratic equations?

**Tip**: When determining if an equation represents a function, always check if each value of the independent variable corresponds to exactly one value of the dependent variable.

Does the equation specify a function with independent variable x? If so, find the domain of the function. If not, find a value of x to which there corresponds more than one value of y.
a) 2x + 5y = 10
b) y(x + y) = 4
c) x(x + y) = 4

Explore how to determine if an equation defines y as a function of x, with examples involving linear and quadratic equations. Includes step-by-step solutions to find the domain and identify cases where multiple values of y correspond to the same x. Suitable for students in Grades 9-12.

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