Linear Algebra Examples

$9 x^{2} - 4 y^{2} = 36$

Step 1

Subtract $9 x^{2}$ from both sides of the equation.

$- 4 y^{2} = 36 - 9 x^{2}$

Step 2

Divide each term in $- 4 y^{2} = 36 - 9 x^{2}$ by $- 4$ and simplify.

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Step 2.1

Divide each term in $- 4 y^{2} = 36 - 9 x^{2}$ by $- 4$ .

$\frac{- 4 y^{2}}{- 4} = \frac{36}{- 4} + \frac{- 9 x^{2}}{- 4}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $- 4$ .

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Step 2.2.1.1

Cancel the common factor.

$\frac{- 4 y^{2}}{- 4} = \frac{36}{- 4} + \frac{- 9 x^{2}}{- 4}$

Step 2.2.1.2

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

$y^{2} = \frac{36}{- 4} + \frac{- 9 x^{2}}{- 4}$

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 $36$ by $- 4$ .

$y^{2} = - 9 + \frac{- 9 x^{2}}{- 4}$

Step 2.3.1.2

Dividing two negative values results in a positive value.

$y^{2} = - 9 + \frac{9 x^{2}}{4}$

Step 3

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

$y = \pm \sqrt{- 9 + \frac{9 x^{2}}{4}}$

Step 4

Simplify $\pm \sqrt{- 9 + \frac{9 x^{2}}{4}}$ .

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Step 4.1

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

$y = \pm \sqrt{- 9 + \frac{{(3 x)}^{2}}{4}}$

Step 4.2

Rewrite $4$ as $2^{2}$ .

$y = \pm \sqrt{- 9 + \frac{{(3 x)}^{2}}{2^{2}}}$

Step 4.3

Rewrite $\frac{{(3 x)}^{2}}{2^{2}}$ as ${(\frac{3 x}{2})}^{2}$ .

$y = \pm \sqrt{- 9 + {(\frac{3 x}{2})}^{2}}$

Step 4.4

Simplify the expression.

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Step 4.4.1

Rewrite $9$ as $3^{2}$ .

$y = \pm \sqrt{- 3^{2} + {(\frac{3 x}{2})}^{2}}$

Step 4.4.2

Reorder $- 3^{2}$ and ${(\frac{3 x}{2})}^{2}$ .

$y = \pm \sqrt{{(\frac{3 x}{2})}^{2} - 3^{2}}$

Step 4.5

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = \frac{3 x}{2}$ and $b = 3$ .

$y = \pm \sqrt{(\frac{3 x}{2} + 3) (\frac{3 x}{2} - 3)}$

Step 4.6

Simplify.

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Step 4.6.1

Factor $3$ out of $\frac{3 x}{2} + 3$ .

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Step 4.6.1.1

Factor $3$ out of $\frac{3 x}{2}$ .

$y = \pm \sqrt{(3 \frac{x}{2} + 3) (\frac{3 x}{2} - 3)}$

Step 4.6.1.2

Factor $3$ out of $3$ .

$y = \pm \sqrt{(3 \frac{x}{2} + 3 (1)) (\frac{3 x}{2} - 3)}$

Step 4.6.1.3

Factor $3$ out of $3 \frac{x}{2} + 3 (1)$ .

$y = \pm \sqrt{3 (\frac{x}{2} + 1) (\frac{3 x}{2} - 3)}$

Step 4.6.2

Factor $3$ out of $\frac{3 x}{2} - 3$ .

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Step 4.6.2.1

Factor $3$ out of $\frac{3 x}{2}$ .

$y = \pm \sqrt{3 (\frac{x}{2} + 1) (3 \frac{x}{2} - 3)}$

Step 4.6.2.2

Factor $3$ out of $- 3$ .

$y = \pm \sqrt{3 (\frac{x}{2} + 1) (3 \frac{x}{2} + 3 (- 1))}$

Step 4.6.2.3

Factor $3$ out of $3 \frac{x}{2} + 3 (- 1)$ .

$y = \pm \sqrt{3 (\frac{x}{2} + 1) (3 (\frac{x}{2} - 1))}$

$y = \pm \sqrt{3 (\frac{x}{2} + 1) \cdot 3 (\frac{x}{2} - 1)}$

Step 4.6.3

Multiply $3$ by $3$ .

$y = \pm \sqrt{9 (\frac{x}{2} + 1) (\frac{x}{2} - 1)}$

Step 4.7

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

$y = \pm \sqrt{9 (\frac{x}{2} + \frac{2}{2}) (\frac{x}{2} - 1)}$

Step 4.8

Combine the numerators over the common denominator.

$y = \pm \sqrt{9 \frac{x + 2}{2} (\frac{x}{2} - 1)}$

Step 4.9

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$y = \pm \sqrt{9 \frac{x + 2}{2} (\frac{x}{2} - 1 \cdot \frac{2}{2})}$

Step 4.10

Combine $- 1$ and $\frac{2}{2}$ .

$y = \pm \sqrt{9 \frac{x + 2}{2} (\frac{x}{2} + \frac{- 1 \cdot 2}{2})}$

Step 4.11

Combine the numerators over the common denominator.

$y = \pm \sqrt{9 \frac{x + 2}{2} \frac{x - 1 \cdot 2}{2}}$

Step 4.12

Multiply $- 1$ by $2$ .

$y = \pm \sqrt{9 \frac{x + 2}{2} \frac{x - 2}{2}}$

Step 4.13

Combine exponents.

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Step 4.13.1

Combine $9$ and $\frac{x + 2}{2}$ .

$y = \pm \sqrt{\frac{9 (x + 2)}{2} \cdot \frac{x - 2}{2}}$

Step 4.13.2

Multiply $\frac{9 (x + 2)}{2}$ by $\frac{x - 2}{2}$ .

$y = \pm \sqrt{\frac{9 (x + 2) (x - 2)}{2 \cdot 2}}$

Step 4.13.3

Multiply $2$ by $2$ .

$y = \pm \sqrt{\frac{9 (x + 2) (x - 2)}{4}}$

Step 4.14

Rewrite $\frac{9 (x + 2) (x - 2)}{4}$ as ${(\frac{3}{2})}^{2} ((x + 2) (x - 2))$ .

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Step 4.14.1

Factor the perfect power $3^{2}$ out of $9 (x + 2) (x - 2)$ .

$y = \pm \sqrt{\frac{3^{2} ((x + 2) (x - 2))}{4}}$

Step 4.14.2

Factor the perfect power $2^{2}$ out of $4$ .

$y = \pm \sqrt{\frac{3^{2} ((x + 2) (x - 2))}{2^{2} \cdot 1}}$

Step 4.14.3

Rearrange the fraction $\frac{3^{2} ((x + 2) (x - 2))}{2^{2} \cdot 1}$ .

$y = \pm \sqrt{{(\frac{3}{2})}^{2} ((x + 2) (x - 2))}$

Step 4.15

Pull terms out from under the radical.

$y = \pm \frac{3}{2} \sqrt{(x + 2) (x - 2)}$

Step 4.16

Combine $\frac{3}{2}$ and $\sqrt{(x + 2) (x - 2)}$ .

$y = \pm \frac{3 \sqrt{(x + 2) (x - 2)}}{2}$

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.

$y = \frac{3 \sqrt{(x + 2) (x - 2)}}{2}$

Step 5.2

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

$y = - \frac{3 \sqrt{(x + 2) (x - 2)}}{2}$

Step 5.3

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

$y = \frac{3 \sqrt{(x + 2) (x - 2)}}{2}$

$y = - \frac{3 \sqrt{(x + 2) (x - 2)}}{2}$

$y = \frac{3 \sqrt{(x + 2) (x - 2)}}{2}$

$y = - \frac{3 \sqrt{(x + 2) (x - 2)}}{2}$

Step 6

Set the radicand in $\sqrt{(x + 2) (x - 2)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 2) (x - 2) \geq 0$

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 $0$ .

$x + 2 = 0$

$x - 2 = 0$

Step 7.2

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 7.2.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 7.2.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 7.3

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 7.3.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 7.3.2

Add $2$ to both sides of the equation.

$x = 2$

Step 7.4

The final solution is all the values that make $(x + 2) (x - 2) \geq 0$ true.

$x = - 2, 2$

Step 7.5

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 2$

$x > 2$

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 $x < - 2$ to see if it makes the inequality true.

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Step 7.6.1.1

Choose a value on the interval $x < - 2$ 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.

$((- 4) + 2) ((- 4) - 2) \geq 0$

Step 7.6.1.3

The left side $12$ 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 $- 2 < x < 2$ to see if it makes the inequality true.

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Step 7.6.2.1

Choose a value on the interval $- 2 < x < 2$ 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.

$((0) + 2) ((0) - 2) \geq 0$

Step 7.6.2.3

The left side $- 4$ 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 $x > 2$ to see if it makes the inequality true.

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Step 7.6.3.1

Choose a value on the interval $x > 2$ 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.

$((4) + 2) ((4) - 2) \geq 0$

Step 7.6.3.3

The left side $12$ 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.

$x < - 2$ True

$- 2 < x < 2$ False

$x > 2$ True

$x < - 2$ True

$- 2 < x < 2$ False

$x > 2$ True

Step 7.7

The solution consists of all of the true intervals.

$x \leq - 2$ or $x \geq 2$

Step 8

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

Interval Notation:

$(- \infty, - 2] \cup [2, \infty)$

Set-Builder Notation:

${x | x \leq - 2, x \geq 2}$

Step 9