Math Problem Statement

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

Solution

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

a) 2x+5y=102x + 5y = 10

To determine if this equation represents a function of xx, we solve for yy in terms of xx:

5y=102x5y = 10 - 2x y=102x5y = \frac{10 - 2x}{5}

This is a linear equation, so it specifies yy as a function of xx. The domain of this function is all real numbers, since there are no restrictions on xx.

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

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

First, let's try to isolate yy to see if it represents yy as a function of xx:

y(x+y)=4y(x + y) = 4

Expanding the left-hand side:

yx+y2=4yx + y^2 = 4

This is a quadratic equation in yy (i.e., y2+xy4=0y^2 + xy - 4 = 0). Solving this quadratic for yy using the quadratic formula:

y=x±x2+162y = \frac{-x \pm \sqrt{x^2 + 16}}{2}

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

For instance, if x=0x = 0:

y=0±02+162=±42=±2y = \frac{-0 \pm \sqrt{0^2 + 16}}{2} = \frac{\pm 4}{2} = \pm 2

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

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

Let's try to solve for yy in terms of xx:

x(x+y)=4x(x + y) = 4 x2+xy=4x^2 + xy = 4 xy=4x2xy = 4 - x^2 y=4x2xy = \frac{4 - x^2}{x}

This expression for yy is defined as long as x0x \neq 0 (since division by zero is undefined). So, this equation does define yy as a function of xx, but with the restriction that x0x \neq 0.

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


Summary:

  • a) 2x+5y=102x + 5y = 10 defines yy as a function of xx. Domain: (,)(-\infty, \infty)
  • b) y(x+y)=4y(x + y) = 4 does not define yy as a function of xx. Example: x=0x = 0 gives two values of yy.
  • c) x(x+y)=4x(x + y) = 4 defines yy as a function of xx with the domain x0x \neq 0, or (,0)(0,)(-\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.

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Math Problem Analysis

Mathematical Concepts

Linear Equations
Quadratic Equations
Functions
Domain of a Function

Formulas

Linear equation formula: ax + by = c
Quadratic formula: y = [-b ± sqrt(b² - 4ac)] / 2a
Solving for the domain: restrictions such as division by zero

Theorems

Quadratic Formula
Function Definition Theorem

Suitable Grade Level

Grades 9-12