Finite Math Examples

$x^{2} + 25 y^{2} < 25$ , $x^{2} + y^{2} < 9$

Step 1

Solve $x^{2} + 25 y^{2} < 25$ for $y$ .

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Step 1.1

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

$25 y^{2} < 25 - x^{2}$

Step 1.2

Divide each term in $25 y^{2} < 25 - x^{2}$ by $25$ and simplify.

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Step 1.2.1

Divide each term in $25 y^{2} < 25 - x^{2}$ by $25$ .

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

Step 1.2.2

Simplify the left side.

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Step 1.2.2.1

Cancel the common factor of $25$ .

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Step 1.2.2.1.1

Cancel the common factor.

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

Step 1.2.2.1.2

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

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

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

Simplify each term.

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Step 1.2.3.1.1

Divide $25$ by $25$ .

$y^{2} < 1 + \frac{- x^{2}}{25}$

Step 1.2.3.1.2

Move the negative in front of the fraction.

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

Step 1.3

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

$\sqrt{y^{2}} < \sqrt{1 - \frac{x^{2}}{25}}$

Step 1.4

Simplify the equation.

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Step 1.4.1

Simplify the left side.

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Step 1.4.1.1

Pull terms out from under the radical.

$| y | < \sqrt{1 - \frac{x^{2}}{25}}$

Step 1.4.2

Simplify the right side.

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Step 1.4.2.1

Simplify $\sqrt{1 - \frac{x^{2}}{25}}$ .

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Step 1.4.2.1.1

Write the expression using exponents.

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Step 1.4.2.1.1.1

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

$| y | < \sqrt{1^{2} - \frac{x^{2}}{25}}$

Step 1.4.2.1.1.2

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

$| y | < \sqrt{1^{2} - {(\frac{x}{5})}^{2}}$

Step 1.4.2.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = \frac{x}{5}$ .

$| y | < \sqrt{(1 + \frac{x}{5}) (1 - \frac{x}{5})}$

Step 1.4.2.1.3

Simplify terms.

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Step 1.4.2.1.3.1

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

$| y | < \sqrt{(\frac{5}{5} + \frac{x}{5}) (1 - \frac{x}{5})}$

Step 1.4.2.1.3.2

Combine the numerators over the common denominator.

$| y | < \sqrt{\frac{5 + x}{5} (1 - \frac{x}{5})}$

Step 1.4.2.1.3.3

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

$| y | < \sqrt{\frac{5 + x}{5} (\frac{5}{5} - \frac{x}{5})}$

Step 1.4.2.1.3.4

Combine the numerators over the common denominator.

$| y | < \sqrt{\frac{5 + x}{5} \cdot \frac{5 - x}{5}}$

Step 1.4.2.1.3.5

Multiply $\frac{5 + x}{5}$ by $\frac{5 - x}{5}$ .

$| y | < \sqrt{\frac{(5 + x) (5 - x)}{5 \cdot 5}}$

Step 1.4.2.1.3.6

Multiply $5$ by $5$ .

$| y | < \sqrt{\frac{(5 + x) (5 - x)}{25}}$

Step 1.4.2.1.4

Rewrite $\frac{(5 + x) (5 - x)}{25}$ as ${(\frac{1}{5})}^{2} ((5 + x) (5 - x))$ .

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Step 1.4.2.1.4.1

Factor the perfect power $1^{2}$ out of $(5 + x) (5 - x)$ .

$| y | < \sqrt{\frac{1^{2} ((5 + x) (5 - x))}{25}}$

Step 1.4.2.1.4.2

Factor the perfect power $5^{2}$ out of $25$ .

$| y | < \sqrt{\frac{1^{2} ((5 + x) (5 - x))}{5^{2} \cdot 1}}$

Step 1.4.2.1.4.3

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

$| y | < \sqrt{{(\frac{1}{5})}^{2} ((5 + x) (5 - x))}$

Step 1.4.2.1.5

Pull terms out from under the radical.

$| y | < | \frac{1}{5} | \sqrt{(5 + x) (5 - x)}$

Step 1.4.2.1.6

$\frac{1}{5}$ is approximately $0.2$ which is positive so remove the absolute value

$| y | < \frac{1}{5} \sqrt{(5 + x) (5 - x)}$

Step 1.4.2.1.7

Combine $\frac{1}{5}$ and $\sqrt{(5 + x) (5 - x)}$ .

$| y | < \frac{\sqrt{(5 + x) (5 - x)}}{5}$

Step 1.5

Write $| y | < \frac{\sqrt{(5 + x) (5 - x)}}{5}$ as a piecewise.

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Step 1.5.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$y \geq 0$

Step 1.5.2

In the piece where $y$ is non-negative, remove the absolute value.

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

Step 1.5.3

Find the domain of $y < \frac{\sqrt{(5 + x) (5 - x)}}{5}$ and find the intersection with $y \geq 0$ .

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Step 1.5.3.1

Find the domain of $y < \frac{\sqrt{(5 + x) (5 - x)}}{5}$ .

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Step 1.5.3.1.1

Set the radicand in $\sqrt{(5 + x) (5 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(5 + x) (5 - x) \geq 0$

Step 1.5.3.1.2

Solve for $y$ .

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Step 1.5.3.1.2.1

Simplify $(5 + x) (5 - x)$ .

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Step 1.5.3.1.2.1.1

Expand $(5 + x) (5 - x)$ using the FOIL Method.

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Step 1.5.3.1.2.1.1.1

Apply the distributive property.

$5 (5 - x) + x (5 - x) \geq 0$

Step 1.5.3.1.2.1.1.2

Apply the distributive property.

$5 \cdot 5 + 5 (- x) + x (5 - x) \geq 0$

Step 1.5.3.1.2.1.1.3

Apply the distributive property.

$5 \cdot 5 + 5 (- x) + x \cdot 5 + x (- x) \geq 0$

Step 1.5.3.1.2.1.2

Simplify and combine like terms.

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Step 1.5.3.1.2.1.2.1

Simplify each term.

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Step 1.5.3.1.2.1.2.1.1

Multiply $5$ by $5$ .

$25 + 5 (- x) + x \cdot 5 + x (- x) \geq 0$

Step 1.5.3.1.2.1.2.1.2

Multiply $- 1$ by $5$ .

$25 - 5 x + x \cdot 5 + x (- x) \geq 0$

Step 1.5.3.1.2.1.2.1.3

Move $5$ to the left of $x$ .

$25 - 5 x + 5 \cdot x + x (- x) \geq 0$

Step 1.5.3.1.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$25 - 5 x + 5 x - x \cdot x \geq 0$

Step 1.5.3.1.2.1.2.1.5

Multiply $x$ by $x$ by adding the exponents.

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Step 1.5.3.1.2.1.2.1.5.1

Move $x$ .

$25 - 5 x + 5 x - (x \cdot x) \geq 0$

Step 1.5.3.1.2.1.2.1.5.2

Multiply $x$ by $x$ .

$25 - 5 x + 5 x - x^{2} \geq 0$

Step 1.5.3.1.2.1.2.2

Add $- 5 x$ and $5 x$ .

$25 + 0 - x^{2} \geq 0$

Step 1.5.3.1.2.1.2.3

Add $25$ and $0$ .

$25 - x^{2} \geq 0$

Step 1.5.3.1.2.2

Subtract $25$ from both sides of the inequality.

$- x^{2} \geq - 25$

Step 1.5.3.1.2.3

Divide each term in $- x^{2} \geq - 25$ by $- 1$ and simplify.

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Step 1.5.3.1.2.3.1

Divide each term in $- x^{2} \geq - 25$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 25}{- 1}$

Step 1.5.3.1.2.3.2

Simplify the left side.

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Step 1.5.3.1.2.3.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 25}{- 1}$

Step 1.5.3.1.2.3.2.2

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

$x^{2} \leq \frac{- 25}{- 1}$

Step 1.5.3.1.2.3.3

Simplify the right side.

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Step 1.5.3.1.2.3.3.1

Divide $- 25$ by $- 1$ .

$x^{2} \leq 25$

Step 1.5.3.1.2.4

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

$\sqrt{x^{2}} \leq \sqrt{25}$

Step 1.5.3.1.2.5

Simplify the equation.

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Step 1.5.3.1.2.5.1

Simplify the left side.

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Step 1.5.3.1.2.5.1.1

Pull terms out from under the radical.

$| x | \leq \sqrt{25}$

Step 1.5.3.1.2.5.2

Simplify the right side.

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Step 1.5.3.1.2.5.2.1

Simplify $\sqrt{25}$ .

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Step 1.5.3.1.2.5.2.1.1

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

$| x | \leq \sqrt{5^{2}}$

Step 1.5.3.1.2.5.2.1.2

Pull terms out from under the radical.

$| x | \leq | 5 |$

Step 1.5.3.1.2.5.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $5$ is $5$ .

$| x | \leq 5$

Step 1.5.3.1.2.6

Write $| x | \leq 5$ as a piecewise.

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Step 1.5.3.1.2.6.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 1.5.3.1.2.6.2

In the piece where $x$ is non-negative, remove the absolute value.

$x \leq 5$

Step 1.5.3.1.2.6.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 1.5.3.1.2.6.4

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \leq 5$

Step 1.5.3.1.2.6.5

Write as a piecewise.

${\begin{cases} x \leq 5 & x \geq 0 \\ - x \leq 5 & x < 0 \end{cases}$

Step 1.5.3.1.2.7

Find the intersection of $x \leq 5$ and $x \geq 0$ .

$0 \leq x \leq 5$

Step 1.5.3.1.2.8

Solve $- x \leq 5$ when $x < 0$ .

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Step 1.5.3.1.2.8.1

Divide each term in $- x \leq 5$ by $- 1$ and simplify.

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Step 1.5.3.1.2.8.1.1

Divide each term in $- x \leq 5$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{5}{- 1}$

Step 1.5.3.1.2.8.1.2

Simplify the left side.

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Step 1.5.3.1.2.8.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{5}{- 1}$

Step 1.5.3.1.2.8.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{5}{- 1}$

Step 1.5.3.1.2.8.1.3

Simplify the right side.

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Step 1.5.3.1.2.8.1.3.1

Divide $5$ by $- 1$ .

$x \geq - 5$

Step 1.5.3.1.2.8.2

Find the intersection of $x \geq - 5$ and $x < 0$ .

$- 5 \leq x < 0$

Step 1.5.3.1.2.9

Find the union of the solutions.

$- 5 \leq x \leq 5$

Step 1.5.3.1.3

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

$[- 5, 5]$

Step 1.5.3.2

Find the intersection of $y \geq 0$ and $[- 5, 5]$ .

$0 \leq y \leq 5$

Step 1.5.4

To find the interval for the second piece, find where the inside of the absolute value is negative.

$y < 0$

Step 1.5.5

In the piece where $y$ is negative, remove the absolute value and multiply by $- 1$ .

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

Step 1.5.6

Find the domain of $- y < \frac{\sqrt{(5 + x) (5 - x)}}{5}$ and find the intersection with $y < 0$ .

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Step 1.5.6.1

Find the domain of $- y < \frac{\sqrt{(5 + x) (5 - x)}}{5}$ .

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Step 1.5.6.1.1

Set the radicand in $\sqrt{(5 + x) (5 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(5 + x) (5 - x) \geq 0$

Step 1.5.6.1.2

Solve for $y$ .

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Step 1.5.6.1.2.1

Simplify $(5 + x) (5 - x)$ .

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Step 1.5.6.1.2.1.1

Expand $(5 + x) (5 - x)$ using the FOIL Method.

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Step 1.5.6.1.2.1.1.1

Apply the distributive property.

$5 (5 - x) + x (5 - x) \geq 0$

Step 1.5.6.1.2.1.1.2

Apply the distributive property.

$5 \cdot 5 + 5 (- x) + x (5 - x) \geq 0$

Step 1.5.6.1.2.1.1.3

Apply the distributive property.

$5 \cdot 5 + 5 (- x) + x \cdot 5 + x (- x) \geq 0$

Step 1.5.6.1.2.1.2

Simplify and combine like terms.

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Step 1.5.6.1.2.1.2.1

Simplify each term.

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Step 1.5.6.1.2.1.2.1.1

Multiply $5$ by $5$ .

$25 + 5 (- x) + x \cdot 5 + x (- x) \geq 0$

Step 1.5.6.1.2.1.2.1.2

Multiply $- 1$ by $5$ .

$25 - 5 x + x \cdot 5 + x (- x) \geq 0$

Step 1.5.6.1.2.1.2.1.3

Move $5$ to the left of $x$ .

$25 - 5 x + 5 \cdot x + x (- x) \geq 0$

Step 1.5.6.1.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$25 - 5 x + 5 x - x \cdot x \geq 0$

Step 1.5.6.1.2.1.2.1.5

Multiply $x$ by $x$ by adding the exponents.

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Step 1.5.6.1.2.1.2.1.5.1

Move $x$ .

$25 - 5 x + 5 x - (x \cdot x) \geq 0$

Step 1.5.6.1.2.1.2.1.5.2

Multiply $x$ by $x$ .

$25 - 5 x + 5 x - x^{2} \geq 0$

Step 1.5.6.1.2.1.2.2

Add $- 5 x$ and $5 x$ .

$25 + 0 - x^{2} \geq 0$

Step 1.5.6.1.2.1.2.3

Add $25$ and $0$ .

$25 - x^{2} \geq 0$

Step 1.5.6.1.2.2

Subtract $25$ from both sides of the inequality.

$- x^{2} \geq - 25$

Step 1.5.6.1.2.3

Divide each term in $- x^{2} \geq - 25$ by $- 1$ and simplify.

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Step 1.5.6.1.2.3.1

Divide each term in $- x^{2} \geq - 25$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 25}{- 1}$

Step 1.5.6.1.2.3.2

Simplify the left side.

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Step 1.5.6.1.2.3.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 25}{- 1}$

Step 1.5.6.1.2.3.2.2

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

$x^{2} \leq \frac{- 25}{- 1}$

Step 1.5.6.1.2.3.3

Simplify the right side.

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Step 1.5.6.1.2.3.3.1

Divide $- 25$ by $- 1$ .

$x^{2} \leq 25$

Step 1.5.6.1.2.4

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

$\sqrt{x^{2}} \leq \sqrt{25}$

Step 1.5.6.1.2.5

Simplify the equation.

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Step 1.5.6.1.2.5.1

Simplify the left side.

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Step 1.5.6.1.2.5.1.1

Pull terms out from under the radical.

$| x | \leq \sqrt{25}$

Step 1.5.6.1.2.5.2

Simplify the right side.

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Step 1.5.6.1.2.5.2.1

Simplify $\sqrt{25}$ .

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Step 1.5.6.1.2.5.2.1.1

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

$| x | \leq \sqrt{5^{2}}$

Step 1.5.6.1.2.5.2.1.2

Pull terms out from under the radical.

$| x | \leq | 5 |$

Step 1.5.6.1.2.5.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $5$ is $5$ .

$| x | \leq 5$

Step 1.5.6.1.2.6

Write $| x | \leq 5$ as a piecewise.

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Step 1.5.6.1.2.6.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 1.5.6.1.2.6.2

In the piece where $x$ is non-negative, remove the absolute value.

$x \leq 5$

Step 1.5.6.1.2.6.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 1.5.6.1.2.6.4

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \leq 5$

Step 1.5.6.1.2.6.5

Write as a piecewise.

${\begin{cases} x \leq 5 & x \geq 0 \\ - x \leq 5 & x < 0 \end{cases}$

Step 1.5.6.1.2.7

Find the intersection of $x \leq 5$ and $x \geq 0$ .

$0 \leq x \leq 5$

Step 1.5.6.1.2.8

Solve $- x \leq 5$ when $x < 0$ .

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Step 1.5.6.1.2.8.1

Divide each term in $- x \leq 5$ by $- 1$ and simplify.

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Step 1.5.6.1.2.8.1.1

Divide each term in $- x \leq 5$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{5}{- 1}$

Step 1.5.6.1.2.8.1.2

Simplify the left side.

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Step 1.5.6.1.2.8.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{5}{- 1}$

Step 1.5.6.1.2.8.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{5}{- 1}$

Step 1.5.6.1.2.8.1.3

Simplify the right side.

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Step 1.5.6.1.2.8.1.3.1

Divide $5$ by $- 1$ .

$x \geq - 5$

Step 1.5.6.1.2.8.2

Find the intersection of $x \geq - 5$ and $x < 0$ .

$- 5 \leq x < 0$

Step 1.5.6.1.2.9

Find the union of the solutions.

$- 5 \leq x \leq 5$

Step 1.5.6.1.3

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

$[- 5, 5]$

Step 1.5.6.2

Find the intersection of $y < 0$ and $[- 5, 5]$ .

$- 5 \leq y < 0$

Step 1.5.7

Write as a piecewise.

${\begin{cases} y < \frac{\sqrt{(5 + x) (5 - x)}}{5} & 0 \leq y \leq 5 \\ - y < \frac{\sqrt{(5 + x) (5 - x)}}{5} & - 5 \leq y < 0 \end{cases}$

Step 1.6

Find the intersection of $y < \frac{\sqrt{(5 + x) (5 - x)}}{5}$ and $0 \leq y \leq 5$ .

$0 \leq y \leq N o (Min i m u m)$

Step 1.7

Solve $- y < \frac{\sqrt{(5 + x) (5 - x)}}{5}$ when $- 5 \leq y < 0$ .

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Step 1.7.1

Divide each term in $- y < \frac{\sqrt{(5 + x) (5 - x)}}{5}$ by $- 1$ and simplify.

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Step 1.7.1.1

Divide each term in $- y < \frac{\sqrt{(5 + x) (5 - x)}}{5}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} > \frac{\frac{\sqrt{(5 + x) (5 - x)}}{5}}{- 1}$

Step 1.7.1.2

Simplify the left side.

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Step 1.7.1.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} > \frac{\frac{\sqrt{(5 + x) (5 - x)}}{5}}{- 1}$

Step 1.7.1.2.2

Divide $y$ by $1$ .

$y > \frac{\frac{\sqrt{(5 + x) (5 - x)}}{5}}{- 1}$

Step 1.7.1.3

Simplify the right side.

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Step 1.7.1.3.1

Move the negative one from the denominator of $\frac{\frac{\sqrt{(5 + x) (5 - x)}}{5}}{- 1}$ .

$y > - 1 \cdot \frac{\sqrt{(5 + x) (5 - x)}}{5}$

Step 1.7.1.3.2

Rewrite $- 1 \cdot \frac{\sqrt{(5 + x) (5 - x)}}{5}$ as $- \frac{\sqrt{(5 + x) (5 - x)}}{5}$ .

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

Step 1.7.2

Find the intersection of $y > - \frac{\sqrt{(5 + x) (5 - x)}}{5}$ and $- 5 \leq y < 0$ .

No solution

Step 1.8

Find the union of the solutions.

$0 \leq y \leq i N o Min m^{2} u$

Step 2

Solve $x^{2} + y^{2} < 9$ for $y$ .

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Step 2.1

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

$y^{2} < 9 - x^{2}$

Step 2.2

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

$\sqrt{y^{2}} < \sqrt{9 - x^{2}}$

Step 2.3

Simplify the equation.

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Step 2.3.1

Simplify the left side.

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Step 2.3.1.1

Pull terms out from under the radical.

$| y | < \sqrt{9 - x^{2}}$

Step 2.3.2

Simplify the right side.

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Step 2.3.2.1

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

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Step 2.3.2.1.1

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

$| y | < \sqrt{3^{2} - x^{2}}$

Step 2.3.2.1.2

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 $b = x$ .

$| y | < \sqrt{(3 + x) (3 - x)}$

Step 2.4

Write $| y | < \sqrt{(3 + x) (3 - x)}$ as a piecewise.

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Step 2.4.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$y \geq 0$

Step 2.4.2

In the piece where $y$ is non-negative, remove the absolute value.

$y < \sqrt{(3 + x) (3 - x)}$

Step 2.4.3

Find the domain of $y < \sqrt{(3 + x) (3 - x)}$ and find the intersection with $y \geq 0$ .

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Step 2.4.3.1

Find the domain of $y < \sqrt{(3 + x) (3 - x)}$ .

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Step 2.4.3.1.1

Set the radicand in $\sqrt{(3 + x) (3 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(3 + x) (3 - x) \geq 0$

Step 2.4.3.1.2

Solve for $y$ .

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Step 2.4.3.1.2.1

Simplify $(3 + x) (3 - x)$ .

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Step 2.4.3.1.2.1.1

Expand $(3 + x) (3 - x)$ using the FOIL Method.

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Step 2.4.3.1.2.1.1.1

Apply the distributive property.

$3 (3 - x) + x (3 - x) \geq 0$

Step 2.4.3.1.2.1.1.2

Apply the distributive property.

$3 \cdot 3 + 3 (- x) + x (3 - x) \geq 0$

Step 2.4.3.1.2.1.1.3

Apply the distributive property.

$3 \cdot 3 + 3 (- x) + x \cdot 3 + x (- x) \geq 0$

Step 2.4.3.1.2.1.2

Simplify and combine like terms.

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Step 2.4.3.1.2.1.2.1

Simplify each term.

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Step 2.4.3.1.2.1.2.1.1

Multiply $3$ by $3$ .

$9 + 3 (- x) + x \cdot 3 + x (- x) \geq 0$

Step 2.4.3.1.2.1.2.1.2

Multiply $- 1$ by $3$ .

$9 - 3 x + x \cdot 3 + x (- x) \geq 0$

Step 2.4.3.1.2.1.2.1.3

Move $3$ to the left of $x$ .

$9 - 3 x + 3 \cdot x + x (- x) \geq 0$

Step 2.4.3.1.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$9 - 3 x + 3 x - x \cdot x \geq 0$

Step 2.4.3.1.2.1.2.1.5

Multiply $x$ by $x$ by adding the exponents.

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Step 2.4.3.1.2.1.2.1.5.1

Move $x$ .

$9 - 3 x + 3 x - (x \cdot x) \geq 0$

Step 2.4.3.1.2.1.2.1.5.2

Multiply $x$ by $x$ .

$9 - 3 x + 3 x - x^{2} \geq 0$

Step 2.4.3.1.2.1.2.2

Add $- 3 x$ and $3 x$ .

$9 + 0 - x^{2} \geq 0$

Step 2.4.3.1.2.1.2.3

Add $9$ and $0$ .

$9 - x^{2} \geq 0$

Step 2.4.3.1.2.2

Subtract $9$ from both sides of the inequality.

$- x^{2} \geq - 9$

Step 2.4.3.1.2.3

Divide each term in $- x^{2} \geq - 9$ by $- 1$ and simplify.

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Step 2.4.3.1.2.3.1

Divide each term in $- x^{2} \geq - 9$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 9}{- 1}$

Step 2.4.3.1.2.3.2

Simplify the left side.

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Step 2.4.3.1.2.3.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 9}{- 1}$

Step 2.4.3.1.2.3.2.2

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

$x^{2} \leq \frac{- 9}{- 1}$

Step 2.4.3.1.2.3.3

Simplify the right side.

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Step 2.4.3.1.2.3.3.1

Divide $- 9$ by $- 1$ .

$x^{2} \leq 9$

Step 2.4.3.1.2.4

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

$\sqrt{x^{2}} \leq \sqrt{9}$

Step 2.4.3.1.2.5

Simplify the equation.

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Step 2.4.3.1.2.5.1

Simplify the left side.

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Step 2.4.3.1.2.5.1.1

Pull terms out from under the radical.

$| x | \leq \sqrt{9}$

Step 2.4.3.1.2.5.2

Simplify the right side.

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Step 2.4.3.1.2.5.2.1

Simplify $\sqrt{9}$ .

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Step 2.4.3.1.2.5.2.1.1

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

$| x | \leq \sqrt{3^{2}}$

Step 2.4.3.1.2.5.2.1.2

Pull terms out from under the radical.

$| x | \leq | 3 |$

Step 2.4.3.1.2.5.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$| x | \leq 3$

Step 2.4.3.1.2.6

Write $| x | \leq 3$ as a piecewise.

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Step 2.4.3.1.2.6.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 2.4.3.1.2.6.2

In the piece where $x$ is non-negative, remove the absolute value.

$x \leq 3$

Step 2.4.3.1.2.6.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 2.4.3.1.2.6.4

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \leq 3$

Step 2.4.3.1.2.6.5

Write as a piecewise.

${\begin{cases} x \leq 3 & x \geq 0 \\ - x \leq 3 & x < 0 \end{cases}$

Step 2.4.3.1.2.7

Find the intersection of $x \leq 3$ and $x \geq 0$ .

$0 \leq x \leq 3$

Step 2.4.3.1.2.8

Solve $- x \leq 3$ when $x < 0$ .

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Step 2.4.3.1.2.8.1

Divide each term in $- x \leq 3$ by $- 1$ and simplify.

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Step 2.4.3.1.2.8.1.1

Divide each term in $- x \leq 3$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{3}{- 1}$

Step 2.4.3.1.2.8.1.2

Simplify the left side.

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Step 2.4.3.1.2.8.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{3}{- 1}$

Step 2.4.3.1.2.8.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{3}{- 1}$

Step 2.4.3.1.2.8.1.3

Simplify the right side.

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Step 2.4.3.1.2.8.1.3.1

Divide $3$ by $- 1$ .

$x \geq - 3$

Step 2.4.3.1.2.8.2

Find the intersection of $x \geq - 3$ and $x < 0$ .

$- 3 \leq x < 0$

Step 2.4.3.1.2.9

Find the union of the solutions.

$- 3 \leq x \leq 3$

Step 2.4.3.1.3

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

$[- 3, 3]$

Step 2.4.3.2

Find the intersection of $y \geq 0$ and $[- 3, 3]$ .

$0 \leq y \leq 3$

Step 2.4.4

To find the interval for the second piece, find where the inside of the absolute value is negative.

$y < 0$

Step 2.4.5

In the piece where $y$ is negative, remove the absolute value and multiply by $- 1$ .

$- y < \sqrt{(3 + x) (3 - x)}$

Step 2.4.6

Find the domain of $- y < \sqrt{(3 + x) (3 - x)}$ and find the intersection with $y < 0$ .

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Step 2.4.6.1

Find the domain of $- y < \sqrt{(3 + x) (3 - x)}$ .

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Step 2.4.6.1.1

Set the radicand in $\sqrt{(3 + x) (3 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(3 + x) (3 - x) \geq 0$

Step 2.4.6.1.2

Solve for $y$ .

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Step 2.4.6.1.2.1

Simplify $(3 + x) (3 - x)$ .

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Step 2.4.6.1.2.1.1

Expand $(3 + x) (3 - x)$ using the FOIL Method.

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Step 2.4.6.1.2.1.1.1

Apply the distributive property.

$3 (3 - x) + x (3 - x) \geq 0$

Step 2.4.6.1.2.1.1.2

Apply the distributive property.

$3 \cdot 3 + 3 (- x) + x (3 - x) \geq 0$

Step 2.4.6.1.2.1.1.3

Apply the distributive property.

$3 \cdot 3 + 3 (- x) + x \cdot 3 + x (- x) \geq 0$

Step 2.4.6.1.2.1.2

Simplify and combine like terms.

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Step 2.4.6.1.2.1.2.1

Simplify each term.

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Step 2.4.6.1.2.1.2.1.1

Multiply $3$ by $3$ .

$9 + 3 (- x) + x \cdot 3 + x (- x) \geq 0$

Step 2.4.6.1.2.1.2.1.2

Multiply $- 1$ by $3$ .

$9 - 3 x + x \cdot 3 + x (- x) \geq 0$

Step 2.4.6.1.2.1.2.1.3

Move $3$ to the left of $x$ .

$9 - 3 x + 3 \cdot x + x (- x) \geq 0$

Step 2.4.6.1.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$9 - 3 x + 3 x - x \cdot x \geq 0$

Step 2.4.6.1.2.1.2.1.5

Multiply $x$ by $x$ by adding the exponents.

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Step 2.4.6.1.2.1.2.1.5.1

Move $x$ .

$9 - 3 x + 3 x - (x \cdot x) \geq 0$

Step 2.4.6.1.2.1.2.1.5.2

Multiply $x$ by $x$ .

$9 - 3 x + 3 x - x^{2} \geq 0$

Step 2.4.6.1.2.1.2.2

Add $- 3 x$ and $3 x$ .

$9 + 0 - x^{2} \geq 0$

Step 2.4.6.1.2.1.2.3

Add $9$ and $0$ .

$9 - x^{2} \geq 0$

Step 2.4.6.1.2.2

Subtract $9$ from both sides of the inequality.

$- x^{2} \geq - 9$

Step 2.4.6.1.2.3

Divide each term in $- x^{2} \geq - 9$ by $- 1$ and simplify.

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Step 2.4.6.1.2.3.1

Divide each term in $- x^{2} \geq - 9$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 9}{- 1}$

Step 2.4.6.1.2.3.2

Simplify the left side.

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Step 2.4.6.1.2.3.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 9}{- 1}$

Step 2.4.6.1.2.3.2.2

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

$x^{2} \leq \frac{- 9}{- 1}$

Step 2.4.6.1.2.3.3

Simplify the right side.

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Step 2.4.6.1.2.3.3.1

Divide $- 9$ by $- 1$ .

$x^{2} \leq 9$

Step 2.4.6.1.2.4

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

$\sqrt{x^{2}} \leq \sqrt{9}$

Step 2.4.6.1.2.5

Simplify the equation.

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Step 2.4.6.1.2.5.1

Simplify the left side.

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Step 2.4.6.1.2.5.1.1

Pull terms out from under the radical.

$| x | \leq \sqrt{9}$

Step 2.4.6.1.2.5.2

Simplify the right side.

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Step 2.4.6.1.2.5.2.1

Simplify $\sqrt{9}$ .

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Step 2.4.6.1.2.5.2.1.1

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

$| x | \leq \sqrt{3^{2}}$

Step 2.4.6.1.2.5.2.1.2

Pull terms out from under the radical.

$| x | \leq | 3 |$

Step 2.4.6.1.2.5.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$| x | \leq 3$

Step 2.4.6.1.2.6

Write $| x | \leq 3$ as a piecewise.

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Step 2.4.6.1.2.6.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 2.4.6.1.2.6.2

In the piece where $x$ is non-negative, remove the absolute value.

$x \leq 3$

Step 2.4.6.1.2.6.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 2.4.6.1.2.6.4

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \leq 3$

Step 2.4.6.1.2.6.5

Write as a piecewise.

${\begin{cases} x \leq 3 & x \geq 0 \\ - x \leq 3 & x < 0 \end{cases}$

Step 2.4.6.1.2.7

Find the intersection of $x \leq 3$ and $x \geq 0$ .

$0 \leq x \leq 3$

Step 2.4.6.1.2.8

Solve $- x \leq 3$ when $x < 0$ .

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Step 2.4.6.1.2.8.1

Divide each term in $- x \leq 3$ by $- 1$ and simplify.

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Step 2.4.6.1.2.8.1.1

Divide each term in $- x \leq 3$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{3}{- 1}$

Step 2.4.6.1.2.8.1.2

Simplify the left side.

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Step 2.4.6.1.2.8.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{3}{- 1}$

Step 2.4.6.1.2.8.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{3}{- 1}$

Step 2.4.6.1.2.8.1.3

Simplify the right side.

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Step 2.4.6.1.2.8.1.3.1

Divide $3$ by $- 1$ .

$x \geq - 3$

Step 2.4.6.1.2.8.2

Find the intersection of $x \geq - 3$ and $x < 0$ .

$- 3 \leq x < 0$

Step 2.4.6.1.2.9

Find the union of the solutions.

$- 3 \leq x \leq 3$

Step 2.4.6.1.3

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

$[- 3, 3]$

Step 2.4.6.2

Find the intersection of $y < 0$ and $[- 3, 3]$ .

$- 3 \leq y < 0$

Step 2.4.7

Write as a piecewise.

${\begin{cases} y < \sqrt{(3 + x) (3 - x)} & 0 \leq y \leq 3 \\ - y < \sqrt{(3 + x) (3 - x)} & - 3 \leq y < 0 \end{cases}$

Step 2.5

Find the intersection of $y < \sqrt{(3 + x) (3 - x)}$ and $0 \leq y \leq 3$ .

$0 \leq y \leq N o (Min i m u m)$

Step 2.6

Solve $- y < \sqrt{(3 + x) (3 - x)}$ when $- 3 \leq y < 0$ .

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Step 2.6.1

Divide each term in $- y < \sqrt{(3 + x) (3 - x)}$ by $- 1$ and simplify.

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Step 2.6.1.1

Divide each term in $- y < \sqrt{(3 + x) (3 - x)}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} > \frac{\sqrt{(3 + x) (3 - x)}}{- 1}$

Step 2.6.1.2

Simplify the left side.

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Step 2.6.1.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} > \frac{\sqrt{(3 + x) (3 - x)}}{- 1}$

Step 2.6.1.2.2

Divide $y$ by $1$ .

$y > \frac{\sqrt{(3 + x) (3 - x)}}{- 1}$

Step 2.6.1.3

Simplify the right side.

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Step 2.6.1.3.1

Move the negative one from the denominator of $\frac{\sqrt{(3 + x) (3 - x)}}{- 1}$ .

$y > - 1 \cdot \sqrt{(3 + x) (3 - x)}$

Step 2.6.1.3.2

Rewrite $- 1 \cdot \sqrt{(3 + x) (3 - x)}$ as $- \sqrt{(3 + x) (3 - x)}$ .

$y > - \sqrt{(3 + x) (3 - x)}$

Step 2.6.2

Find the intersection of $y > - \sqrt{(3 + x) (3 - x)}$ and $- 3 \leq y < 0$ .

No solution

Step 2.7

Find the union of the solutions.

$0 \leq y \leq i N o Min m^{2} u$

Step 3

Find the intersection of $0 \leq y \leq i N o Min m^{2} u$ and $0 \leq y \leq i N o Min m^{2} u$ .

$0 \leq y \leq N o (Min i m u m)$

Step 4