Precalculus Examples

$\frac{x^{2}}{144} - \frac{y^{2}}{25} = 1$

Step 1

Subtract $\frac{x^{2}}{144}$ from both sides of the equation.

$- \frac{y^{2}}{25} = 1 - \frac{x^{2}}{144}$

Step 2

Multiply both sides of the equation by $- 25$ .

$- 25 (- \frac{y^{2}}{25}) = - 25 (1 - \frac{x^{2}}{144})$

Step 3

Simplify both sides of the equation.

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Step 3.1

Simplify the left side.

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Step 3.1.1

Simplify $- 25 (- \frac{y^{2}}{25})$ .

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Step 3.1.1.1

Cancel the common factor of $25$ .

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Step 3.1.1.1.1

Move the leading negative in $- \frac{y^{2}}{25}$ into the numerator.

$- 25 \frac{- y^{2}}{25} = - 25 (1 - \frac{x^{2}}{144})$

Step 3.1.1.1.2

Factor $25$ out of $- 25$ .

$25 (- 1) \frac{- y^{2}}{25} = - 25 (1 - \frac{x^{2}}{144})$

Step 3.1.1.1.3

Cancel the common factor.

$25 \cdot - 1 \frac{- y^{2}}{25} = - 25 (1 - \frac{x^{2}}{144})$

Step 3.1.1.1.4

Rewrite the expression.

$- - y^{2} = - 25 (1 - \frac{x^{2}}{144})$

Step 3.1.1.2

Multiply.

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Step 3.1.1.2.1

Multiply $- 1$ by $- 1$ .

$1 y^{2} = - 25 (1 - \frac{x^{2}}{144})$

Step 3.1.1.2.2

Multiply $y^{2}$ by $1$ .

$y^{2} = - 25 (1 - \frac{x^{2}}{144})$

Step 3.2

Simplify the right side.

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Step 3.2.1

Simplify $- 25 (1 - \frac{x^{2}}{144})$ .

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Step 3.2.1.1

Apply the distributive property.

$y^{2} = - 25 \cdot 1 - 25 (- \frac{x^{2}}{144})$

Step 3.2.1.2

Multiply $- 25$ by $1$ .

$y^{2} = - 25 - 25 (- \frac{x^{2}}{144})$

Step 3.2.1.3

Multiply $- 25 (- \frac{x^{2}}{144})$ .

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Step 3.2.1.3.1

Multiply $- 1$ by $- 25$ .

$y^{2} = - 25 + 25 \frac{x^{2}}{144}$

Step 3.2.1.3.2

Combine $25$ and $\frac{x^{2}}{144}$ .

$y^{2} = - 25 + \frac{25 x^{2}}{144}$

Step 4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{- 25 + \frac{25 x^{2}}{144}}$

Step 5

Simplify $\pm \sqrt{- 25 + \frac{25 x^{2}}{144}}$ .

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Step 5.1

Rewrite $25 x^{2}$ as ${(5 x)}^{2}$ .

$y = \pm \sqrt{- 25 + \frac{{(5 x)}^{2}}{144}}$

Step 5.2

Rewrite $144$ as $12^{2}$ .

$y = \pm \sqrt{- 25 + \frac{{(5 x)}^{2}}{12^{2}}}$

Step 5.3

Rewrite $\frac{{(5 x)}^{2}}{12^{2}}$ as ${(\frac{5 x}{12})}^{2}$ .

$y = \pm \sqrt{- 25 + {(\frac{5 x}{12})}^{2}}$

Step 5.4

Simplify the expression.

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Step 5.4.1

Rewrite $25$ as $5^{2}$ .

$y = \pm \sqrt{- 5^{2} + {(\frac{5 x}{12})}^{2}}$

Step 5.4.2

Reorder $- 5^{2}$ and ${(\frac{5 x}{12})}^{2}$ .

$y = \pm \sqrt{{(\frac{5 x}{12})}^{2} - 5^{2}}$

Step 5.5

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = \frac{5 x}{12}$ and $b = 5$ .

$y = \pm \sqrt{(\frac{5 x}{12} + 5) (\frac{5 x}{12} - 5)}$

Step 5.6

Simplify.

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Step 5.6.1

Factor $5$ out of $\frac{5 x}{12} + 5$ .

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Step 5.6.1.1

Factor $5$ out of $\frac{5 x}{12}$ .

$y = \pm \sqrt{(5 \frac{x}{12} + 5) (\frac{5 x}{12} - 5)}$

Step 5.6.1.2

Factor $5$ out of $5$ .

$y = \pm \sqrt{(5 \frac{x}{12} + 5 (1)) (\frac{5 x}{12} - 5)}$

Step 5.6.1.3

Factor $5$ out of $5 \frac{x}{12} + 5 (1)$ .

$y = \pm \sqrt{5 (\frac{x}{12} + 1) (\frac{5 x}{12} - 5)}$

Step 5.6.2

Factor $5$ out of $\frac{5 x}{12} - 5$ .

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Step 5.6.2.1

Factor $5$ out of $\frac{5 x}{12}$ .

$y = \pm \sqrt{5 (\frac{x}{12} + 1) (5 \frac{x}{12} - 5)}$

Step 5.6.2.2

Factor $5$ out of $- 5$ .

$y = \pm \sqrt{5 (\frac{x}{12} + 1) (5 \frac{x}{12} + 5 (- 1))}$

Step 5.6.2.3

Factor $5$ out of $5 \frac{x}{12} + 5 (- 1)$ .

$y = \pm \sqrt{5 (\frac{x}{12} + 1) (5 (\frac{x}{12} - 1))}$

$y = \pm \sqrt{5 (\frac{x}{12} + 1) \cdot 5 (\frac{x}{12} - 1)}$

Step 5.6.3

Multiply $5$ by $5$ .

$y = \pm \sqrt{25 (\frac{x}{12} + 1) (\frac{x}{12} - 1)}$

Step 5.7

Write $1$ as a fraction with a common denominator.

$y = \pm \sqrt{25 (\frac{x}{12} + \frac{12}{12}) (\frac{x}{12} - 1)}$

Step 5.8

Combine the numerators over the common denominator.

$y = \pm \sqrt{25 \frac{x + 12}{12} (\frac{x}{12} - 1)}$

Step 5.9

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{12}{12}$ .

$y = \pm \sqrt{25 \frac{x + 12}{12} (\frac{x}{12} - 1 \cdot \frac{12}{12})}$

Step 5.10

Combine $- 1$ and $\frac{12}{12}$ .

$y = \pm \sqrt{25 \frac{x + 12}{12} (\frac{x}{12} + \frac{- 1 \cdot 12}{12})}$

Step 5.11

Combine the numerators over the common denominator.

$y = \pm \sqrt{25 \frac{x + 12}{12} \frac{x - 1 \cdot 12}{12}}$

Step 5.12

Multiply $- 1$ by $12$ .

$y = \pm \sqrt{25 \frac{x + 12}{12} \frac{x - 12}{12}}$

Step 5.13

Combine exponents.

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Step 5.13.1

Combine $25$ and $\frac{x + 12}{12}$ .

$y = \pm \sqrt{\frac{25 (x + 12)}{12} \cdot \frac{x - 12}{12}}$

Step 5.13.2

Multiply $\frac{25 (x + 12)}{12}$ by $\frac{x - 12}{12}$ .

$y = \pm \sqrt{\frac{25 (x + 12) (x - 12)}{12 \cdot 12}}$

Step 5.13.3

Multiply $12$ by $12$ .

$y = \pm \sqrt{\frac{25 (x + 12) (x - 12)}{144}}$

Step 5.14

Rewrite $\frac{25 (x + 12) (x - 12)}{144}$ as ${(\frac{5}{12})}^{2} ((x + 12) (x - 12))$ .

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Step 5.14.1

Factor the perfect power $5^{2}$ out of $25 (x + 12) (x - 12)$ .

$y = \pm \sqrt{\frac{5^{2} ((x + 12) (x - 12))}{144}}$

Step 5.14.2

Factor the perfect power $12^{2}$ out of $144$ .

$y = \pm \sqrt{\frac{5^{2} ((x + 12) (x - 12))}{12^{2} \cdot 1}}$

Step 5.14.3

Rearrange the fraction $\frac{5^{2} ((x + 12) (x - 12))}{12^{2} \cdot 1}$ .

$y = \pm \sqrt{{(\frac{5}{12})}^{2} ((x + 12) (x - 12))}$

Step 5.15

Pull terms out from under the radical.

$y = \pm \frac{5}{12} \sqrt{(x + 12) (x - 12)}$

Step 5.16

Combine $\frac{5}{12}$ and $\sqrt{(x + 12) (x - 12)}$ .

$y = \pm \frac{5 \sqrt{(x + 12) (x - 12)}}{12}$

Step 6

The complete solution is the result of both the positive and negative portions of the solution.

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Step 6.1

First, use the positive value of the $\pm$ to find the first solution.

$y = \frac{5 \sqrt{(x + 12) (x - 12)}}{12}$

Step 6.2

Next, use the negative value of the $\pm$ to find the second solution.

$y = - \frac{5 \sqrt{(x + 12) (x - 12)}}{12}$

Step 6.3

The complete solution is the result of both the positive and negative portions of the solution.

$y = \frac{5 \sqrt{(x + 12) (x - 12)}}{12}$

$y = - \frac{5 \sqrt{(x + 12) (x - 12)}}{12}$

$y = \frac{5 \sqrt{(x + 12) (x - 12)}}{12}$

$y = - \frac{5 \sqrt{(x + 12) (x - 12)}}{12}$

Step 7

Set the radicand in $\sqrt{(x + 12) (x - 12)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 12) (x - 12) \geq 0$

Step 8

Solve for $x$ .

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Step 8.1

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

$x + 12 = 0$

$x - 12 = 0$

Step 8.2

Set $x + 12$ equal to $0$ and solve for $x$ .

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Step 8.2.1

Set $x + 12$ equal to $0$ .

$x + 12 = 0$

Step 8.2.2

Subtract $12$ from both sides of the equation.

$x = - 12$

Step 8.3

Set $x - 12$ equal to $0$ and solve for $x$ .

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Step 8.3.1

Set $x - 12$ equal to $0$ .

$x - 12 = 0$

Step 8.3.2

Add $12$ to both sides of the equation.

$x = 12$

Step 8.4

The final solution is all the values that make $(x + 12) (x - 12) \geq 0$ true.

$x = - 12, 12$

Step 8.5

Use each root to create test intervals.

$x < - 12$

$- 12 < x < 12$

$x > 12$

Step 8.6

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 8.6.1

Test a value on the interval $x < - 12$ to see if it makes the inequality true.

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Step 8.6.1.1

Choose a value on the interval $x < - 12$ and see if this value makes the original inequality true.

$x = - 14$

Step 8.6.1.2

Replace $x$ with $- 14$ in the original inequality.

$((- 14) + 12) ((- 14) - 12) \geq 0$

Step 8.6.1.3

The left side $52$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 8.6.2

Test a value on the interval $- 12 < x < 12$ to see if it makes the inequality true.

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Step 8.6.2.1

Choose a value on the interval $- 12 < x < 12$ and see if this value makes the original inequality true.

$x = 0$

Step 8.6.2.2

Replace $x$ with $0$ in the original inequality.

$((0) + 12) ((0) - 12) \geq 0$

Step 8.6.2.3

The left side $- 144$ is less than the right side $0$ , which means that the given statement is false.

False

Step 8.6.3

Test a value on the interval $x > 12$ to see if it makes the inequality true.

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Step 8.6.3.1

Choose a value on the interval $x > 12$ and see if this value makes the original inequality true.

$x = 14$

Step 8.6.3.2

Replace $x$ with $14$ in the original inequality.

$((14) + 12) ((14) - 12) \geq 0$

Step 8.6.3.3

The left side $52$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 8.6.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 12$ True

$- 12 < x < 12$ False

$x > 12$ True

$x < - 12$ True

$- 12 < x < 12$ False

$x > 12$ True

Step 8.7

The solution consists of all of the true intervals.

$x \leq - 12$ or $x \geq 12$

Step 9

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

Interval Notation:

$(- \infty, - 12] \cup [12, \infty)$

Set-Builder Notation:

${x | x \leq - 12, x \geq 12}$

Step 10

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

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${y | y \in ℝ}$

Step 11

Determine the domain and range.

Domain: $(- \infty, - 12] \cup [12, \infty), {x | x \leq - 12, x \geq 12}$

Range: $(- \infty, \infty), {y | y \in ℝ}$

Step 12