Finite Math Examples

$x^{2} - y^{2} = 1$

Step 1

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

$- y^{2} = 1 - x^{2}$

Step 2

Divide each term in $- y^{2} = 1 - x^{2}$ by $- 1$ and simplify.

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Step 2.1

Divide each term in $- y^{2} = 1 - x^{2}$ by $- 1$ .

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

Step 2.2

Simplify the left side.

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Step 2.2.1

Dividing two negative values results in a positive value.

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

Step 2.2.2

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

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

Step 2.3

Simplify the right side.

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Step 2.3.1

Simplify each term.

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Step 2.3.1.1

Divide $1$ by $- 1$ .

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

Step 2.3.1.2

Dividing two negative values results in a positive value.

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

Step 2.3.1.3

Divide $x^{2}$ by $1$ .

$y^{2} = - 1 + x^{2}$

Step 3

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

$y = \pm \sqrt{- 1 + x^{2}}$

Step 4

Simplify $\pm \sqrt{- 1 + x^{2}}$ .

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Step 4.1

Simplify the expression.

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Step 4.1.1

Rewrite $1$ as $1^{2}$ .

$y = \pm \sqrt{- 1^{2} + x^{2}}$

Step 4.1.2

Reorder $- 1^{2}$ and $x^{2}$ .

$y = \pm \sqrt{x^{2} - 1^{2}}$

Step 4.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

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

Step 5

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

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Step 5.1

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

$y = \sqrt{(x + 1) (x - 1)}$

Step 5.2

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

$y = - \sqrt{(x + 1) (x - 1)}$

Step 5.3

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

$y = \sqrt{(x + 1) (x - 1)}$

$y = - \sqrt{(x + 1) (x - 1)}$

$y = \sqrt{(x + 1) (x - 1)}$

$y = - \sqrt{(x + 1) (x - 1)}$

Step 6

Set the radicand in $\sqrt{(x + 1) (x - 1)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 1) (x - 1) \geq 0$

Step 7

Solve for $x$ .

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Step 7.1

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

$x + 1 = 0$

$x - 1 = 0$

Step 7.2

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 7.2.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 7.2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 7.3

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 7.3.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 7.3.2

Add $1$ to both sides of the equation.

$x = 1$

Step 7.4

The final solution is all the values that make $(x + 1) (x - 1) \geq 0$ true.

$x = - 1, 1$

Step 7.5

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 1$

$x > 1$

Step 7.6

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 7.6.1

Test a value on the interval $x < - 1$ to see if it makes the inequality true.

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Step 7.6.1.1

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

Step 7.6.1.2

Replace $x$ with $- 4$ in the original inequality.

$((- 4) + 1) ((- 4) - 1) \geq 0$

Step 7.6.1.3

The left side $15$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 7.6.2

Test a value on the interval $- 1 < x < 1$ to see if it makes the inequality true.

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Step 7.6.2.1

Choose a value on the interval $- 1 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 7.6.2.2

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

$((0) + 1) ((0) - 1) \geq 0$

Step 7.6.2.3

The left side $- 1$ is less than the right side $0$ , which means that the given statement is false.

False

Step 7.6.3

Test a value on the interval $x > 1$ to see if it makes the inequality true.

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Step 7.6.3.1

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Step 7.6.3.2

Replace $x$ with $4$ in the original inequality.

$((4) + 1) ((4) - 1) \geq 0$

Step 7.6.3.3

The left side $15$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 7.6.4

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

$x < - 1$ True

$- 1 < x < 1$ False

$x > 1$ True

$x < - 1$ True

$- 1 < x < 1$ False

$x > 1$ True

Step 7.7

The solution consists of all of the true intervals.

$x \leq - 1$ or $x \geq 1$

Step 8

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

Interval Notation:

$(- \infty, - 1] \cup [1, \infty)$

Set-Builder Notation:

${x | x \leq - 1, x \geq 1}$

Step 9