Algebra Examples

$- y^{2} + x \geq - 8$

Step 1

Subtract $x$ from both sides of the inequality.

$- y^{2} \geq - 8 - x$

Step 2

Divide each term in $- y^{2} \geq - 8 - x$ by $- 1$ and simplify.

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Step 2.1

Divide each term in $- y^{2} \geq - 8 - x$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y^{2}}{- 1} \leq \frac{- 8}{- 1} + \frac{- x}{- 1}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Dividing two negative values results in a positive value.

$\frac{y^{2}}{1} \leq \frac{- 8}{- 1} + \frac{- x}{- 1}$

Step 2.2.2

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

$y^{2} \leq \frac{- 8}{- 1} + \frac{- x}{- 1}$

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 $- 8$ by $- 1$ .

$y^{2} \leq 8 + \frac{- x}{- 1}$

Step 2.3.1.2

Dividing two negative values results in a positive value.

$y^{2} \leq 8 + \frac{x}{1}$

Step 2.3.1.3

Divide $x$ by $1$ .

$y^{2} \leq 8 + x$

Step 3

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

$\sqrt{y^{2}} \leq \sqrt{8 + x}$

Step 4

Simplify the left side.

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Step 4.1

Pull terms out from under the radical.

$| y | \leq \sqrt{8 + x}$

Step 5

Write $| y | \leq \sqrt{8 + x}$ as a piecewise.

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Step 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 5.2

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

$y \leq \sqrt{8 + x}$

Step 5.3

Find the domain of $y \leq \sqrt{8 + x}$ and find the intersection with $y \geq 0$ .

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Step 5.3.1

Find the domain of $y \leq \sqrt{8 + x}$ .

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Step 5.3.1.1

Set the radicand in $\sqrt{8 + x}$ greater than or equal to $0$ to find where the expression is defined.

$8 + x \geq 0$

Step 5.3.1.2

Subtract $8$ from both sides of the inequality.

$x \geq - 8$

Step 5.3.1.3

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

$[- 8, \infty)$

Step 5.3.2

Find the intersection of $y \geq 0$ and $[- 8, \infty)$ .

$y \geq 0$

Step 5.4

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

$y < 0$

Step 5.5

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

$- y \leq \sqrt{8 + x}$

Step 5.6

Find the domain of $- y \leq \sqrt{8 + x}$ and find the intersection with $y < 0$ .

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Step 5.6.1

Find the domain of $- y \leq \sqrt{8 + x}$ .

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Step 5.6.1.1

Set the radicand in $\sqrt{8 + x}$ greater than or equal to $0$ to find where the expression is defined.

$8 + x \geq 0$

Step 5.6.1.2

Subtract $8$ from both sides of the inequality.

$x \geq - 8$

Step 5.6.1.3

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

$[- 8, \infty)$

Step 5.6.2

Find the intersection of $y < 0$ and $[- 8, \infty)$ .

$- 8 \leq y < 0$

Step 5.7

Write as a piecewise.

${\begin{cases} y \leq \sqrt{8 + x} & y \geq 0 \\ - y \leq \sqrt{8 + x} & - 8 \leq y < 0 \end{cases}$

Step 6

Find the intersection of $y \leq \sqrt{8 + x}$ and $y \geq 0$ .

$0 \leq y \leq \sqrt{8 + x}$

Step 7

Solve $- y \leq \sqrt{8 + x}$ when $- 8 \leq y < 0$ .

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Step 7.1

Divide each term in $- y \leq \sqrt{8 + x}$ by $- 1$ and simplify.

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Step 7.1.1

Divide each term in $- y \leq \sqrt{8 + x}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} \geq \frac{\sqrt{8 + x}}{- 1}$

Step 7.1.2

Simplify the left side.

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Step 7.1.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} \geq \frac{\sqrt{8 + x}}{- 1}$

Step 7.1.2.2

Divide $y$ by $1$ .

$y \geq \frac{\sqrt{8 + x}}{- 1}$

Step 7.1.3

Simplify the right side.

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Step 7.1.3.1

Move the negative one from the denominator of $\frac{\sqrt{8 + x}}{- 1}$ .

$y \geq - 1 \cdot \sqrt{8 + x}$

Step 7.1.3.2

Rewrite $- 1 \cdot \sqrt{8 + x}$ as $- \sqrt{8 + x}$ .

$y \geq - \sqrt{8 + x}$

Step 7.2

Find the intersection of $y \geq - \sqrt{8 + x}$ and $- 8 \leq y < 0$ .

No solution

Step 8

Find the union of the solutions.

$0 \leq y \leq \sqrt{8 + x}$

Step 9