Pre-Algebra Examples

$R (x) = \frac{17}{x^{2} - 7 x}$

Step 1

Set the denominator in $\frac{17}{x^{2} - 7 x}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 7 x = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor $x$ out of $x^{2} - 7 x$ .

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Step 2.1.1

Factor $x$ out of $x^{2}$ .

$x \cdot x - 7 x = 0$

Step 2.1.2

Factor $x$ out of $- 7 x$ .

$x \cdot x + x \cdot - 7 = 0$

Step 2.1.3

Factor $x$ out of $x \cdot x + x \cdot - 7$ .

$x (x - 7) = 0$

Step 2.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 7 = 0$

Step 2.3

Set $x$ equal to $0$ .

$x = 0$

Step 2.4

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 2.4.2

Add $7$ to both sides of the equation.

$x = 7$

Step 2.5

The final solution is all the values that make $x (x - 7) = 0$ true.

$x = 0, 7$

Step 3

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, 0) \cup (0, 7) \cup (7, \infty)$

Set-Builder Notation:

${x | x \neq 0, 7}$

Step 4

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, - \frac{68}{49}] \cup (0, \infty)$

Set-Builder Notation:

${y | y \leq - \frac{68}{49}, y > 0}$

Step 5

Determine the domain and range.

Domain: $(- \infty, 0) \cup (0, 7) \cup (7, \infty), {x | x \neq 0, 7}$

Range: $(- \infty, - \frac{68}{49}] \cup (0, \infty), {y | y \leq - \frac{68}{49}, y > 0}$

Step 6