Algebra Examples

$6 x^{2} + 9 {(y - 2)}^{2} \leq 36$ $x^{2} + {(y + 3)}^{2} \leq 25$

Step 1

Subtract $6 x^{2}$ from both sides of the inequality.

$9 {(y - 2)}^{2} \leq 36 - 6 x^{2}$

Step 2

Solve for $y$ .

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Step 2.1

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

${(y + 3)}^{2} \leq 25 - x^{2}$

Step 2.2

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

$\sqrt{{(y + 3)}^{2}} \leq \sqrt{25 - x^{2}}$

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.

$| y + 3 | \leq \sqrt{25 - x^{2}}$

Step 2.3.2

Simplify the right side.

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Step 2.3.2.1

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

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Step 2.3.2.1.1

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

$| y + 3 | \leq \sqrt{5^{2} - x^{2}}$

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 $a = 5$ and $b = x$ .

$| y + 3 | \leq \sqrt{(5 + x) (5 - x)}$

Step 2.4

Write $| y + 3 | \leq \sqrt{(5 + x) (5 - x)}$ 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.

$y + 3 \geq 0$

Step 2.4.2

Subtract $3$ from both sides of the inequality.

$y \geq - 3$

Step 2.4.3

In the piece where $y + 3$ is non-negative, remove the absolute value.

$y + 3 \leq \sqrt{(5 + x) (5 - x)}$

Step 2.4.4

Find the domain of $y + 3 \leq \sqrt{(5 + x) (5 - x)}$ and find the intersection with $y \geq - 3$ .

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Step 2.4.4.1

Find the domain of $y + 3 \leq \sqrt{(5 + x) (5 - x)}$ .

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Step 2.4.4.1.1

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 2.4.4.1.2

Solve for $y$ .

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Step 2.4.4.1.2.1

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

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Step 2.4.4.1.2.1.1

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

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Step 2.4.4.1.2.1.1.1

Apply the distributive property.

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

Step 2.4.4.1.2.1.1.2

Apply the distributive property.

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

Step 2.4.4.1.2.1.1.3

Apply the distributive property.

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

Step 2.4.4.1.2.1.2

Simplify and combine like terms.

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Step 2.4.4.1.2.1.2.1

Simplify each term.

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Step 2.4.4.1.2.1.2.1.1

Multiply $5$ by $5$ .

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

Step 2.4.4.1.2.1.2.1.2

Multiply $- 1$ by $5$ .

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

Step 2.4.4.1.2.1.2.1.3

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

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

Step 2.4.4.1.2.1.2.1.4

Rewrite using the commutative property of multiplication.

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

Step 2.4.4.1.2.1.2.1.5

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

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Step 2.4.4.1.2.1.2.1.5.1

Move $x$ .

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

Step 2.4.4.1.2.1.2.1.5.2

Multiply $x$ by $x$ .

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

Step 2.4.4.1.2.1.2.2

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

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

Step 2.4.4.1.2.1.2.3

Add $25$ and $0$ .

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

Step 2.4.4.1.2.2

Subtract $25$ from both sides of the inequality.

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

Step 2.4.4.1.2.3

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

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Step 2.4.4.1.2.3.1

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 2.4.4.1.2.3.2

Simplify the left side.

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Step 2.4.4.1.2.3.2.1

Dividing two negative values results in a positive value.

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

Step 2.4.4.1.2.3.2.2

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

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

Step 2.4.4.1.2.3.3

Simplify the right side.

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Step 2.4.4.1.2.3.3.1

Divide $- 25$ by $- 1$ .

$x^{2} \leq 25$

Step 2.4.4.1.2.4

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 2.4.4.1.2.5

Simplify the equation.

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Step 2.4.4.1.2.5.1

Simplify the left side.

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Step 2.4.4.1.2.5.1.1

Pull terms out from under the radical.

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

Step 2.4.4.1.2.5.2

Simplify the right side.

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Step 2.4.4.1.2.5.2.1

Simplify $\sqrt{25}$ .

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Step 2.4.4.1.2.5.2.1.1

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

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

Step 2.4.4.1.2.5.2.1.2

Pull terms out from under the radical.

$| x | \leq | 5 |$

Step 2.4.4.1.2.5.2.1.3

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

$| x | \leq 5$

Step 2.4.4.1.2.6

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

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Step 2.4.4.1.2.6.1

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

$x \geq 0$

Step 2.4.4.1.2.6.2

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

$x \leq 5$

Step 2.4.4.1.2.6.3

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

$x < 0$

Step 2.4.4.1.2.6.4

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

$- x \leq 5$

Step 2.4.4.1.2.6.5

Write as a piecewise.

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

Step 2.4.4.1.2.7

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

$0 \leq x \leq 5$

Step 2.4.4.1.2.8

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

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Step 2.4.4.1.2.8.1

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

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Step 2.4.4.1.2.8.1.1

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 2.4.4.1.2.8.1.2

Simplify the left side.

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Step 2.4.4.1.2.8.1.2.1

Dividing two negative values results in a positive value.

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

Step 2.4.4.1.2.8.1.2.2

Divide $x$ by $1$ .

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

Step 2.4.4.1.2.8.1.3

Simplify the right side.

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Step 2.4.4.1.2.8.1.3.1

Divide $5$ by $- 1$ .

$x \geq - 5$

Step 2.4.4.1.2.8.2

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

$- 5 \leq x < 0$

Step 2.4.4.1.2.9

Find the union of the solutions.

$- 5 \leq x \leq 5$

Step 2.4.4.1.3

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

$[- 5, 5]$

Step 2.4.4.2

Find the intersection of $y \geq - 3$ and $[- 5, 5]$ .

$- 3 \leq y \leq 5$

Step 2.4.5

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

$y + 3 < 0$

Step 2.4.6

Subtract $3$ from both sides of the inequality.

$y < - 3$

Step 2.4.7

In the piece where $y + 3$ is negative, remove the absolute value and multiply by $- 1$ .

$- (y + 3) \leq \sqrt{(5 + x) (5 - x)}$

Step 2.4.8

Find the domain of $- (y + 3) \leq \sqrt{(5 + x) (5 - x)}$ and find the intersection with $y < - 3$ .

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Step 2.4.8.1

Find the domain of $- (y + 3) \leq \sqrt{(5 + x) (5 - x)}$ .

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Step 2.4.8.1.1

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 2.4.8.1.2

Solve for $y$ .

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Step 2.4.8.1.2.1

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

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Step 2.4.8.1.2.1.1

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

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Step 2.4.8.1.2.1.1.1

Apply the distributive property.

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

Step 2.4.8.1.2.1.1.2

Apply the distributive property.

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

Step 2.4.8.1.2.1.1.3

Apply the distributive property.

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

Step 2.4.8.1.2.1.2

Simplify and combine like terms.

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Step 2.4.8.1.2.1.2.1

Simplify each term.

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Step 2.4.8.1.2.1.2.1.1

Multiply $5$ by $5$ .

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

Step 2.4.8.1.2.1.2.1.2

Multiply $- 1$ by $5$ .

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

Step 2.4.8.1.2.1.2.1.3

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

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

Step 2.4.8.1.2.1.2.1.4

Rewrite using the commutative property of multiplication.

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

Step 2.4.8.1.2.1.2.1.5

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

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Step 2.4.8.1.2.1.2.1.5.1

Move $x$ .

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

Step 2.4.8.1.2.1.2.1.5.2

Multiply $x$ by $x$ .

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

Step 2.4.8.1.2.1.2.2

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

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

Step 2.4.8.1.2.1.2.3

Add $25$ and $0$ .

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

Step 2.4.8.1.2.2

Subtract $25$ from both sides of the inequality.

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

Step 2.4.8.1.2.3

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

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Step 2.4.8.1.2.3.1

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 2.4.8.1.2.3.2

Simplify the left side.

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Step 2.4.8.1.2.3.2.1

Dividing two negative values results in a positive value.

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

Step 2.4.8.1.2.3.2.2

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

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

Step 2.4.8.1.2.3.3

Simplify the right side.

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Step 2.4.8.1.2.3.3.1

Divide $- 25$ by $- 1$ .

$x^{2} \leq 25$

Step 2.4.8.1.2.4

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 2.4.8.1.2.5

Simplify the equation.

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Step 2.4.8.1.2.5.1

Simplify the left side.

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Step 2.4.8.1.2.5.1.1

Pull terms out from under the radical.

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

Step 2.4.8.1.2.5.2

Simplify the right side.

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Step 2.4.8.1.2.5.2.1

Simplify $\sqrt{25}$ .

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Step 2.4.8.1.2.5.2.1.1

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

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

Step 2.4.8.1.2.5.2.1.2

Pull terms out from under the radical.

$| x | \leq | 5 |$

Step 2.4.8.1.2.5.2.1.3

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

$| x | \leq 5$

Step 2.4.8.1.2.6

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

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Step 2.4.8.1.2.6.1

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

$x \geq 0$

Step 2.4.8.1.2.6.2

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

$x \leq 5$

Step 2.4.8.1.2.6.3

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

$x < 0$

Step 2.4.8.1.2.6.4

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

$- x \leq 5$

Step 2.4.8.1.2.6.5

Write as a piecewise.

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

Step 2.4.8.1.2.7

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

$0 \leq x \leq 5$

Step 2.4.8.1.2.8

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

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Step 2.4.8.1.2.8.1

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

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Step 2.4.8.1.2.8.1.1

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 2.4.8.1.2.8.1.2

Simplify the left side.

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Step 2.4.8.1.2.8.1.2.1

Dividing two negative values results in a positive value.

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

Step 2.4.8.1.2.8.1.2.2

Divide $x$ by $1$ .

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

Step 2.4.8.1.2.8.1.3

Simplify the right side.

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Step 2.4.8.1.2.8.1.3.1

Divide $5$ by $- 1$ .

$x \geq - 5$

Step 2.4.8.1.2.8.2

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

$- 5 \leq x < 0$

Step 2.4.8.1.2.9

Find the union of the solutions.

$- 5 \leq x \leq 5$

Step 2.4.8.1.3

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

$[- 5, 5]$

Step 2.4.8.2

Find the intersection of $y < - 3$ and $[- 5, 5]$ .

$- 5 \leq y < - 3$

Step 2.4.9

Write as a piecewise.

${\begin{cases} y + 3 \leq \sqrt{(5 + x) (5 - x)} & - 3 \leq y \leq 5 \\ - (y + 3) \leq \sqrt{(5 + x) (5 - x)} & - 5 \leq y < - 3 \end{cases}$

Step 2.4.10

Simplify $- (y + 3) \leq \sqrt{(5 + x) (5 - x)}$ .

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Step 2.4.10.1

Apply the distributive property.

${\begin{cases} y + 3 \leq \sqrt{(5 + x) (5 - x)} & - 3 \leq y \leq 5 \\ - y - 1 \cdot 3 \leq \sqrt{(5 + x) (5 - x)} & - 5 \leq y < - 3 \end{cases}$

Step 2.4.10.2

Multiply $- 1$ by $3$ .

${\begin{cases} y + 3 \leq \sqrt{(5 + x) (5 - x)} & - 3 \leq y \leq 5 \\ - y - 3 \leq \sqrt{(5 + x) (5 - x)} & - 5 \leq y < - 3 \end{cases}$

Step 2.5

Solve $y + 3 \leq \sqrt{(5 + x) (5 - x)}$ when $- 3 \leq y \leq 5$ .

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Step 2.5.1

Subtract $3$ from both sides of the inequality.

$y \leq \sqrt{(5 + x) (5 - x)} - 3$

Step 2.5.2

Find the intersection of $y \leq \sqrt{(5 + x) (5 - x)} - 3$ and $- 3 \leq y \leq 5$ .

$- 3 \leq y \leq N o (Min i m u m)$

Step 2.6

Solve $- y - 3 \leq \sqrt{(5 + x) (5 - x)}$ when $- 5 \leq y < - 3$ .

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Step 2.6.1

Solve $- y - 3 \leq \sqrt{(5 + x) (5 - x)}$ for $y$ .

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Step 2.6.1.1

Add $3$ to both sides of the inequality.

$- y \leq \sqrt{(5 + x) (5 - x)} + 3$

Step 2.6.1.2

Divide each term in $- y \leq \sqrt{(5 + x) (5 - x)} + 3$ by $- 1$ and simplify.

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Step 2.6.1.2.1

Divide each term in $- y \leq \sqrt{(5 + x) (5 - x)} + 3$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} \geq \frac{\sqrt{(5 + x) (5 - x)}}{- 1} + \frac{3}{- 1}$

Step 2.6.1.2.2

Simplify the left side.

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Step 2.6.1.2.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} \geq \frac{\sqrt{(5 + x) (5 - x)}}{- 1} + \frac{3}{- 1}$

Step 2.6.1.2.2.2

Divide $y$ by $1$ .

$y \geq \frac{\sqrt{(5 + x) (5 - x)}}{- 1} + \frac{3}{- 1}$

Step 2.6.1.2.3

Simplify the right side.

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Step 2.6.1.2.3.1

Simplify each term.

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Step 2.6.1.2.3.1.1

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

$y \geq - 1 \cdot \sqrt{(5 + x) (5 - x)} + \frac{3}{- 1}$

Step 2.6.1.2.3.1.2

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

$y \geq - \sqrt{(5 + x) (5 - x)} + \frac{3}{- 1}$

Step 2.6.1.2.3.1.3

Divide $3$ by $- 1$ .

$y \geq - \sqrt{(5 + x) (5 - x)} - 3$

Step 2.6.2

Find the intersection of $y \geq - \sqrt{(5 + x) (5 - x)} - 3$ and $- 5 \leq y < - 3$ .

No solution

Step 2.7

Find the union of the solutions.

$- 3 \leq y \leq i N o Min m^{2} u$

Step 3

Plot each graph on the same coordinate system.

$6 x^{2} + 9 {(y - 2)}^{2} \leq 36$

$x^{2} + {(y + 3)}^{2} \leq 25$

Step 4