Linear Algebra Examples

$\frac{1}{49} x^{2} + \frac{1}{9} y^{2} = 1$

Step 1

Simplify each term.

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Step 1.1

Combine $\frac{1}{49}$ and $x^{2}$ .

$\frac{x^{2}}{49} + \frac{1}{9} y^{2} = 1$

Step 1.2

Combine $\frac{1}{9}$ and $y^{2}$ .

$\frac{x^{2}}{49} + \frac{y^{2}}{9} = 1$

Step 2

Subtract $\frac{x^{2}}{49}$ from both sides of the equation.

$\frac{y^{2}}{9} = 1 - \frac{x^{2}}{49}$

Step 3

Multiply both sides of the equation by $9$ .

$9 \frac{y^{2}}{9} = 9 (1 - \frac{x^{2}}{49})$

Step 4

Simplify both sides of the equation.

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Step 4.1

Simplify the left side.

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Step 4.1.1

Cancel the common factor of $9$ .

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Step 4.1.1.1

Cancel the common factor.

$9 \frac{y^{2}}{9} = 9 (1 - \frac{x^{2}}{49})$

Step 4.1.1.2

Rewrite the expression.

$y^{2} = 9 (1 - \frac{x^{2}}{49})$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $9 (1 - \frac{x^{2}}{49})$ .

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Step 4.2.1.1

Apply the distributive property.

$y^{2} = 9 \cdot 1 + 9 (- \frac{x^{2}}{49})$

Step 4.2.1.2

Multiply $9$ by $1$ .

$y^{2} = 9 + 9 (- \frac{x^{2}}{49})$

Step 4.2.1.3

Multiply $9 (- \frac{x^{2}}{49})$ .

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Step 4.2.1.3.1

Multiply $- 1$ by $9$ .

$y^{2} = 9 - 9 \frac{x^{2}}{49}$

Step 4.2.1.3.2

Combine $- 9$ and $\frac{x^{2}}{49}$ .

$y^{2} = 9 + \frac{- 9 x^{2}}{49}$

Step 4.2.1.4

Move the negative in front of the fraction.

$y^{2} = 9 - \frac{9 x^{2}}{49}$

Step 5

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

$y = \pm \sqrt{9 - \frac{9 x^{2}}{49}}$

Step 6

Simplify $\pm \sqrt{9 - \frac{9 x^{2}}{49}}$ .

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Step 6.1

Write the expression using exponents.

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Step 6.1.1

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

$y = \pm \sqrt{3^{2} - \frac{9 x^{2}}{49}}$

Step 6.1.2

Rewrite $\frac{9 x^{2}}{49}$ as ${(\frac{3 x}{7})}^{2}$ .

$y = \pm \sqrt{3^{2} - {(\frac{3 x}{7})}^{2}}$

Step 6.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3$ and $b = \frac{3 x}{7}$ .

$y = \pm \sqrt{(3 + \frac{3 x}{7}) (3 - \frac{3 x}{7})}$

Step 6.3

To write $3$ as a fraction with a common denominator, multiply by $\frac{7}{7}$ .

$y = \pm \sqrt{(3 \cdot \frac{7}{7} + \frac{3 x}{7}) (3 - \frac{3 x}{7})}$

Step 6.4

Combine $3$ and $\frac{7}{7}$ .

$y = \pm \sqrt{(\frac{3 \cdot 7}{7} + \frac{3 x}{7}) (3 - \frac{3 x}{7})}$

Step 6.5

Combine the numerators over the common denominator.

$y = \pm \sqrt{\frac{3 \cdot 7 + 3 x}{7} (3 - \frac{3 x}{7})}$

Step 6.6

Factor $3$ out of $3 \cdot 7 + 3 x$ .

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Step 6.6.1

Factor $3$ out of $3 \cdot 7$ .

$y = \pm \sqrt{\frac{3 (7) + 3 x}{7} (3 - \frac{3 x}{7})}$

Step 6.6.2

Factor $3$ out of $3 x$ .

$y = \pm \sqrt{\frac{3 (7) + 3 (x)}{7} (3 - \frac{3 x}{7})}$

Step 6.6.3

Factor $3$ out of $3 (7) + 3 (x)$ .

$y = \pm \sqrt{\frac{3 (7 + x)}{7} (3 - \frac{3 x}{7})}$

Step 6.7

To write $3$ as a fraction with a common denominator, multiply by $\frac{7}{7}$ .

$y = \pm \sqrt{\frac{3 (7 + x)}{7} (3 \cdot \frac{7}{7} - \frac{3 x}{7})}$

Step 6.8

Combine $3$ and $\frac{7}{7}$ .

$y = \pm \sqrt{\frac{3 (7 + x)}{7} (\frac{3 \cdot 7}{7} - \frac{3 x}{7})}$

Step 6.9

Combine the numerators over the common denominator.

$y = \pm \sqrt{\frac{3 (7 + x)}{7} \cdot \frac{3 \cdot 7 - 3 x}{7}}$

Step 6.10

Factor $3$ out of $3 \cdot 7 - 3 x$ .

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Step 6.10.1

Factor $3$ out of $3 \cdot 7$ .

$y = \pm \sqrt{\frac{3 (7 + x)}{7} \cdot \frac{3 (7) - 3 x}{7}}$

Step 6.10.2

Factor $3$ out of $- 3 x$ .

$y = \pm \sqrt{\frac{3 (7 + x)}{7} \cdot \frac{3 (7) + 3 (- x)}{7}}$

Step 6.10.3

Factor $3$ out of $3 (7) + 3 (- x)$ .

$y = \pm \sqrt{\frac{3 (7 + x)}{7} \cdot \frac{3 (7 - x)}{7}}$

Step 6.11

Multiply $\frac{3 (7 + x)}{7}$ by $\frac{3 (7 - x)}{7}$ .

$y = \pm \sqrt{\frac{3 (7 + x) (3 (7 - x))}{7 \cdot 7}}$

Step 6.12

Multiply.

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Step 6.12.1

Multiply $3$ by $3$ .

$y = \pm \sqrt{\frac{9 (7 + x) (7 - x)}{7 \cdot 7}}$

Step 6.12.2

Multiply $7$ by $7$ .

$y = \pm \sqrt{\frac{9 (7 + x) (7 - x)}{49}}$

Step 6.13

Rewrite $\frac{9 (7 + x) (7 - x)}{49}$ as ${(\frac{3}{7})}^{2} ((7 + x) (7 - x))$ .

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Step 6.13.1

Factor the perfect power $3^{2}$ out of $9 (7 + x) (7 - x)$ .

$y = \pm \sqrt{\frac{3^{2} ((7 + x) (7 - x))}{49}}$

Step 6.13.2

Factor the perfect power $7^{2}$ out of $49$ .

$y = \pm \sqrt{\frac{3^{2} ((7 + x) (7 - x))}{7^{2} \cdot 1}}$

Step 6.13.3

Rearrange the fraction $\frac{3^{2} ((7 + x) (7 - x))}{7^{2} \cdot 1}$ .

$y = \pm \sqrt{{(\frac{3}{7})}^{2} ((7 + x) (7 - x))}$

Step 6.14

Pull terms out from under the radical.

$y = \pm \frac{3}{7} \sqrt{(7 + x) (7 - x)}$

Step 6.15

Combine $\frac{3}{7}$ and $\sqrt{(7 + x) (7 - x)}$ .

$y = \pm \frac{3 \sqrt{(7 + x) (7 - x)}}{7}$

Step 7

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

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Step 7.1

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

$y = \frac{3 \sqrt{(7 + x) (7 - x)}}{7}$

Step 7.2

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

$y = - \frac{3 \sqrt{(7 + x) (7 - x)}}{7}$

Step 7.3

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

$y = \frac{3 \sqrt{(7 + x) (7 - x)}}{7}$

$y = - \frac{3 \sqrt{(7 + x) (7 - x)}}{7}$

$y = \frac{3 \sqrt{(7 + x) (7 - x)}}{7}$

$y = - \frac{3 \sqrt{(7 + x) (7 - x)}}{7}$

Step 8

Set the radicand in $\sqrt{(7 + x) (7 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(7 + x) (7 - x) \geq 0$

Step 9

Solve for $x$ .

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Step 9.1

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

$7 + x = 0$

$7 - x = 0$

Step 9.2

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

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Step 9.2.1

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

$7 + x = 0$

Step 9.2.2

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 9.3

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

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Step 9.3.1

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

$7 - x = 0$

Step 9.3.2

Solve $7 - x = 0$ for $x$ .

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Step 9.3.2.1

Subtract $7$ from both sides of the equation.

$- x = - 7$

Step 9.3.2.2

Divide each term in $- x = - 7$ by $- 1$ and simplify.

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Step 9.3.2.2.1

Divide each term in $- x = - 7$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 7}{- 1}$

Step 9.3.2.2.2

Simplify the left side.

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Step 9.3.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 7}{- 1}$

Step 9.3.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 7}{- 1}$

Step 9.3.2.2.3

Simplify the right side.

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Step 9.3.2.2.3.1

Divide $- 7$ by $- 1$ .

$x = 7$

Step 9.4

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

$x = - 7, 7$

Step 9.5

Use each root to create test intervals.

$x < - 7$

$- 7 < x < 7$

$x > 7$

Step 9.6

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 9.6.1

Test a value on the interval $x < - 7$ to see if it makes the inequality true.

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Step 9.6.1.1

Choose a value on the interval $x < - 7$ and see if this value makes the original inequality true.

$x = - 10$

Step 9.6.1.2

Replace $x$ with $- 10$ in the original inequality.

$(7 - 10) (7 - (- 10)) \geq 0$

Step 9.6.1.3

The left side $- 51$ is less than the right side $0$ , which means that the given statement is false.

False

Step 9.6.2

Test a value on the interval $- 7 < x < 7$ to see if it makes the inequality true.

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Step 9.6.2.1

Choose a value on the interval $- 7 < x < 7$ and see if this value makes the original inequality true.

$x = 0$

Step 9.6.2.2

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

$(7 + 0) (7 - (0)) \geq 0$

Step 9.6.2.3

The left side $49$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 9.6.3

Test a value on the interval $x > 7$ to see if it makes the inequality true.

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Step 9.6.3.1

Choose a value on the interval $x > 7$ and see if this value makes the original inequality true.

$x = 10$

Step 9.6.3.2

Replace $x$ with $10$ in the original inequality.

$(7 + 10) (7 - (10)) \geq 0$

Step 9.6.3.3

The left side $- 51$ is less than the right side $0$ , which means that the given statement is false.

False

Step 9.6.4

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

$x < - 7$ False

$- 7 < x < 7$ True

$x > 7$ False

$x < - 7$ False

$- 7 < x < 7$ True

$x > 7$ False

Step 9.7

The solution consists of all of the true intervals.

$- 7 \leq x \leq 7$

Step 10

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

Interval Notation:

$[- 7, 7]$

Set-Builder Notation:

${x | - 7 \leq x \leq 7}$

Step 11