Algebra Examples

$x^{2} + y^{2} \leq 49$ $y \geq | x^{2} + 1 |$

Step 1

Solve for $y$ .

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Step 1.1

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

$y^{2} \leq 49 - x^{2}$

Step 1.2

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

$\sqrt{y^{2}} \leq \sqrt{49 - x^{2}}$

Step 1.3

Simplify the equation.

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Step 1.3.1

Simplify the left side.

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Step 1.3.1.1

Pull terms out from under the radical.

$| y | \leq \sqrt{49 - x^{2}}$

Step 1.3.2

Simplify the right side.

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Step 1.3.2.1

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

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Step 1.3.2.1.1

Rewrite $49$ as $7^{2}$ .

$| y | \leq \sqrt{7^{2} - x^{2}}$

Step 1.3.2.1.2

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

$| y | \leq \sqrt{(7 + x) (7 - x)}$

Step 1.4

Write $| y | \leq \sqrt{(7 + x) (7 - x)}$ as a piecewise.

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Step 1.4.1

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

$y \geq 0$

Step 1.4.2

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

$y \leq \sqrt{(7 + x) (7 - x)}$

Step 1.4.3

Find the domain of $y \leq \sqrt{(7 + x) (7 - x)}$ and find the intersection with $y \geq 0$ .

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Step 1.4.3.1

Find the domain of $y \leq \sqrt{(7 + x) (7 - x)}$ .

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Step 1.4.3.1.1

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 1.4.3.1.2

Solve for $y$ .

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Step 1.4.3.1.2.1

Simplify $(7 + x) (7 - x)$ .

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Step 1.4.3.1.2.1.1

Expand $(7 + x) (7 - x)$ using the FOIL Method.

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Step 1.4.3.1.2.1.1.1

Apply the distributive property.

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

Step 1.4.3.1.2.1.1.2

Apply the distributive property.

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

Step 1.4.3.1.2.1.1.3

Apply the distributive property.

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

Step 1.4.3.1.2.1.2

Simplify and combine like terms.

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Step 1.4.3.1.2.1.2.1

Simplify each term.

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Step 1.4.3.1.2.1.2.1.1

Multiply $7$ by $7$ .

$49 + 7 (- x) + x \cdot 7 + x (- x) \geq 0$

Step 1.4.3.1.2.1.2.1.2

Multiply $- 1$ by $7$ .

$49 - 7 x + x \cdot 7 + x (- x) \geq 0$

Step 1.4.3.1.2.1.2.1.3

Move $7$ to the left of $x$ .

$49 - 7 x + 7 \cdot x + x (- x) \geq 0$

Step 1.4.3.1.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$49 - 7 x + 7 x - x \cdot x \geq 0$

Step 1.4.3.1.2.1.2.1.5

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

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Step 1.4.3.1.2.1.2.1.5.1

Move $x$ .

$49 - 7 x + 7 x - (x \cdot x) \geq 0$

Step 1.4.3.1.2.1.2.1.5.2

Multiply $x$ by $x$ .

$49 - 7 x + 7 x - x^{2} \geq 0$

Step 1.4.3.1.2.1.2.2

Add $- 7 x$ and $7 x$ .

$49 + 0 - x^{2} \geq 0$

Step 1.4.3.1.2.1.2.3

Add $49$ and $0$ .

$49 - x^{2} \geq 0$

Step 1.4.3.1.2.2

Subtract $49$ from both sides of the inequality.

$- x^{2} \geq - 49$

Step 1.4.3.1.2.3

Divide each term in $- x^{2} \geq - 49$ by $- 1$ and simplify.

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Step 1.4.3.1.2.3.1

Divide each term in $- x^{2} \geq - 49$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 49}{- 1}$

Step 1.4.3.1.2.3.2

Simplify the left side.

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Step 1.4.3.1.2.3.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 49}{- 1}$

Step 1.4.3.1.2.3.2.2

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

$x^{2} \leq \frac{- 49}{- 1}$

Step 1.4.3.1.2.3.3

Simplify the right side.

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Step 1.4.3.1.2.3.3.1

Divide $- 49$ by $- 1$ .

$x^{2} \leq 49$

Step 1.4.3.1.2.4

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

$\sqrt{x^{2}} \leq \sqrt{49}$

Step 1.4.3.1.2.5

Simplify the equation.

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Step 1.4.3.1.2.5.1

Simplify the left side.

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Step 1.4.3.1.2.5.1.1

Pull terms out from under the radical.

$| x | \leq \sqrt{49}$

Step 1.4.3.1.2.5.2

Simplify the right side.

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Step 1.4.3.1.2.5.2.1

Simplify $\sqrt{49}$ .

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Step 1.4.3.1.2.5.2.1.1

Rewrite $49$ as $7^{2}$ .

$| x | \leq \sqrt{7^{2}}$

Step 1.4.3.1.2.5.2.1.2

Pull terms out from under the radical.

$| x | \leq | 7 |$

Step 1.4.3.1.2.5.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $7$ is $7$ .

$| x | \leq 7$

Step 1.4.3.1.2.6

Write $| x | \leq 7$ as a piecewise.

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Step 1.4.3.1.2.6.1

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

$x \geq 0$

Step 1.4.3.1.2.6.2

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

$x \leq 7$

Step 1.4.3.1.2.6.3

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

$x < 0$

Step 1.4.3.1.2.6.4

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

$- x \leq 7$

Step 1.4.3.1.2.6.5

Write as a piecewise.

${\begin{cases} x \leq 7 & x \geq 0 \\ - x \leq 7 & x < 0 \end{cases}$

Step 1.4.3.1.2.7

Find the intersection of $x \leq 7$ and $x \geq 0$ .

$0 \leq x \leq 7$

Step 1.4.3.1.2.8

Solve $- x \leq 7$ when $x < 0$ .

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Step 1.4.3.1.2.8.1

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

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Step 1.4.3.1.2.8.1.1

Divide each term in $- x \leq 7$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

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

Step 1.4.3.1.2.8.1.2

Simplify the left side.

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Step 1.4.3.1.2.8.1.2.1

Dividing two negative values results in a positive value.

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

Step 1.4.3.1.2.8.1.2.2

Divide $x$ by $1$ .

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

Step 1.4.3.1.2.8.1.3

Simplify the right side.

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Step 1.4.3.1.2.8.1.3.1

Divide $7$ by $- 1$ .

$x \geq - 7$

Step 1.4.3.1.2.8.2

Find the intersection of $x \geq - 7$ and $x < 0$ .

$- 7 \leq x < 0$

Step 1.4.3.1.2.9

Find the union of the solutions.

$- 7 \leq x \leq 7$

Step 1.4.3.1.3

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

$[- 7, 7]$

Step 1.4.3.2

Find the intersection of $y \geq 0$ and $[- 7, 7]$ .

$0 \leq y \leq 7$

Step 1.4.4

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

$y < 0$

Step 1.4.5

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

$- y \leq \sqrt{(7 + x) (7 - x)}$

Step 1.4.6

Find the domain of $- y \leq \sqrt{(7 + x) (7 - x)}$ and find the intersection with $y < 0$ .

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Step 1.4.6.1

Find the domain of $- y \leq \sqrt{(7 + x) (7 - x)}$ .

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Step 1.4.6.1.1

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 1.4.6.1.2

Solve for $y$ .

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Step 1.4.6.1.2.1

Simplify $(7 + x) (7 - x)$ .

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Step 1.4.6.1.2.1.1

Expand $(7 + x) (7 - x)$ using the FOIL Method.

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Step 1.4.6.1.2.1.1.1

Apply the distributive property.

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

Step 1.4.6.1.2.1.1.2

Apply the distributive property.

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

Step 1.4.6.1.2.1.1.3

Apply the distributive property.

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

Step 1.4.6.1.2.1.2

Simplify and combine like terms.

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Step 1.4.6.1.2.1.2.1

Simplify each term.

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Step 1.4.6.1.2.1.2.1.1

Multiply $7$ by $7$ .

$49 + 7 (- x) + x \cdot 7 + x (- x) \geq 0$

Step 1.4.6.1.2.1.2.1.2

Multiply $- 1$ by $7$ .

$49 - 7 x + x \cdot 7 + x (- x) \geq 0$

Step 1.4.6.1.2.1.2.1.3

Move $7$ to the left of $x$ .

$49 - 7 x + 7 \cdot x + x (- x) \geq 0$

Step 1.4.6.1.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$49 - 7 x + 7 x - x \cdot x \geq 0$

Step 1.4.6.1.2.1.2.1.5

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

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Step 1.4.6.1.2.1.2.1.5.1

Move $x$ .

$49 - 7 x + 7 x - (x \cdot x) \geq 0$

Step 1.4.6.1.2.1.2.1.5.2

Multiply $x$ by $x$ .

$49 - 7 x + 7 x - x^{2} \geq 0$

Step 1.4.6.1.2.1.2.2

Add $- 7 x$ and $7 x$ .

$49 + 0 - x^{2} \geq 0$

Step 1.4.6.1.2.1.2.3

Add $49$ and $0$ .

$49 - x^{2} \geq 0$

Step 1.4.6.1.2.2

Subtract $49$ from both sides of the inequality.

$- x^{2} \geq - 49$

Step 1.4.6.1.2.3

Divide each term in $- x^{2} \geq - 49$ by $- 1$ and simplify.

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Step 1.4.6.1.2.3.1

Divide each term in $- x^{2} \geq - 49$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 49}{- 1}$

Step 1.4.6.1.2.3.2

Simplify the left side.

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Step 1.4.6.1.2.3.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 49}{- 1}$

Step 1.4.6.1.2.3.2.2

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

$x^{2} \leq \frac{- 49}{- 1}$

Step 1.4.6.1.2.3.3

Simplify the right side.

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Step 1.4.6.1.2.3.3.1

Divide $- 49$ by $- 1$ .

$x^{2} \leq 49$

Step 1.4.6.1.2.4

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

$\sqrt{x^{2}} \leq \sqrt{49}$

Step 1.4.6.1.2.5

Simplify the equation.

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Step 1.4.6.1.2.5.1

Simplify the left side.

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Step 1.4.6.1.2.5.1.1

Pull terms out from under the radical.

$| x | \leq \sqrt{49}$

Step 1.4.6.1.2.5.2

Simplify the right side.

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Step 1.4.6.1.2.5.2.1

Simplify $\sqrt{49}$ .

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Step 1.4.6.1.2.5.2.1.1

Rewrite $49$ as $7^{2}$ .

$| x | \leq \sqrt{7^{2}}$

Step 1.4.6.1.2.5.2.1.2

Pull terms out from under the radical.

$| x | \leq | 7 |$

Step 1.4.6.1.2.5.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $7$ is $7$ .

$| x | \leq 7$

Step 1.4.6.1.2.6

Write $| x | \leq 7$ as a piecewise.

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Step 1.4.6.1.2.6.1

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

$x \geq 0$

Step 1.4.6.1.2.6.2

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

$x \leq 7$

Step 1.4.6.1.2.6.3

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

$x < 0$

Step 1.4.6.1.2.6.4

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

$- x \leq 7$

Step 1.4.6.1.2.6.5

Write as a piecewise.

${\begin{cases} x \leq 7 & x \geq 0 \\ - x \leq 7 & x < 0 \end{cases}$

Step 1.4.6.1.2.7

Find the intersection of $x \leq 7$ and $x \geq 0$ .

$0 \leq x \leq 7$

Step 1.4.6.1.2.8

Solve $- x \leq 7$ when $x < 0$ .

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Step 1.4.6.1.2.8.1

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

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Step 1.4.6.1.2.8.1.1

Divide each term in $- x \leq 7$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

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

Step 1.4.6.1.2.8.1.2

Simplify the left side.

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Step 1.4.6.1.2.8.1.2.1

Dividing two negative values results in a positive value.

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

Step 1.4.6.1.2.8.1.2.2

Divide $x$ by $1$ .

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

Step 1.4.6.1.2.8.1.3

Simplify the right side.

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Step 1.4.6.1.2.8.1.3.1

Divide $7$ by $- 1$ .

$x \geq - 7$

Step 1.4.6.1.2.8.2

Find the intersection of $x \geq - 7$ and $x < 0$ .

$- 7 \leq x < 0$

Step 1.4.6.1.2.9

Find the union of the solutions.

$- 7 \leq x \leq 7$

Step 1.4.6.1.3

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

$[- 7, 7]$

Step 1.4.6.2

Find the intersection of $y < 0$ and $[- 7, 7]$ .

$- 7 \leq y < 0$

Step 1.4.7

Write as a piecewise.

${\begin{cases} y \leq \sqrt{(7 + x) (7 - x)} & 0 \leq y \leq 7 \\ - y \leq \sqrt{(7 + x) (7 - x)} & - 7 \leq y < 0 \end{cases}$

Step 1.5

Find the intersection of $y \leq \sqrt{(7 + x) (7 - x)}$ and $0 \leq y \leq 7$ .

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

Step 1.6

Solve $- y \leq \sqrt{(7 + x) (7 - x)}$ when $- 7 \leq y < 0$ .

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Step 1.6.1

Divide each term in $- y \leq \sqrt{(7 + x) (7 - x)}$ by $- 1$ and simplify.

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Step 1.6.1.1

Divide each term in $- y \leq \sqrt{(7 + x) (7 - x)}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} \geq \frac{\sqrt{(7 + x) (7 - x)}}{- 1}$

Step 1.6.1.2

Simplify the left side.

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Step 1.6.1.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} \geq \frac{\sqrt{(7 + x) (7 - x)}}{- 1}$

Step 1.6.1.2.2

Divide $y$ by $1$ .

$y \geq \frac{\sqrt{(7 + x) (7 - x)}}{- 1}$

Step 1.6.1.3

Simplify the right side.

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Step 1.6.1.3.1

Move the negative one from the denominator of $\frac{\sqrt{(7 + x) (7 - x)}}{- 1}$ .

$y \geq - 1 \cdot \sqrt{(7 + x) (7 - x)}$

Step 1.6.1.3.2

Rewrite $- 1 \cdot \sqrt{(7 + x) (7 - x)}$ as $- \sqrt{(7 + x) (7 - x)}$ .

$y \geq - \sqrt{(7 + x) (7 - x)}$

Step 1.6.2

Find the intersection of $y \geq - \sqrt{(7 + x) (7 - x)}$ and $- 7 \leq y < 0$ .

No solution

Step 1.7

Find the union of the solutions.

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

Step 2

Graph $y \geq | x^{2} + 1 |$ .

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Step 2.1

The equation is not linear, so a constant slope does not exist.

Not Linear

Step 2.2

Graph a solid line, then shade the area above the boundary line since $y$ is greater than $| x^{2} + 1 |$ .

$y \geq | x^{2} + 1 |$

Step 3

Plot each graph on the same coordinate system.

$x^{2} + y^{2} \leq 49$

$y \geq | x^{2} + 1 |$

Step 4