Algebra Examples

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

Step 1

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

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

Step 2

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

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

Step 3

Simplify the equation.

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Step 3.1

Simplify the left side.

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Step 3.1.1

Pull terms out from under the radical.

$| y | \leq \sqrt{9 - {(x + 2)}^{2}}$

Step 3.2

Simplify the right side.

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Step 3.2.1

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

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Step 3.2.1.1

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

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

Step 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 $b = x + 2$ .

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

Step 3.2.1.3

Simplify.

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Step 3.2.1.3.1

Add $3$ and $2$ .

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

Step 3.2.1.3.2

Apply the distributive property.

$| y | \leq \sqrt{(x + 5) (3 - x - 1 \cdot 2)}$

Step 3.2.1.3.3

Multiply $- 1$ by $2$ .

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

Step 3.2.1.3.4

Subtract $2$ from $3$ .

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

Step 4

Write $| y | \leq \sqrt{(x + 5) (- x + 1)}$ as a piecewise.

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Step 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 4.2

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

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

Step 4.3

Find the domain of $y \leq \sqrt{(x + 5) (- x + 1)}$ and find the intersection with $y \geq 0$ .

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Step 4.3.1

Find the domain of $y \leq \sqrt{(x + 5) (- x + 1)}$ .

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Step 4.3.1.1

Set the radicand in $\sqrt{(x + 5) (- x + 1)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 5) (- x + 1) \geq 0$

Step 4.3.1.2

Solve for $y$ .

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Step 4.3.1.2.1

Simplify $(x + 5) (- x + 1)$ .

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Step 4.3.1.2.1.1

Expand $(x + 5) (- x + 1)$ using the FOIL Method.

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Step 4.3.1.2.1.1.1

Apply the distributive property.

$x (- x + 1) + 5 (- x + 1) \geq 0$

Step 4.3.1.2.1.1.2

Apply the distributive property.

$x (- x) + x \cdot 1 + 5 (- x + 1) \geq 0$

Step 4.3.1.2.1.1.3

Apply the distributive property.

$x (- x) + x \cdot 1 + 5 (- x) + 5 \cdot 1 \geq 0$

Step 4.3.1.2.1.2

Simplify and combine like terms.

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Step 4.3.1.2.1.2.1

Simplify each term.

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Step 4.3.1.2.1.2.1.1

Rewrite using the commutative property of multiplication.

$- x \cdot x + x \cdot 1 + 5 (- x) + 5 \cdot 1 \geq 0$

Step 4.3.1.2.1.2.1.2

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

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Step 4.3.1.2.1.2.1.2.1

Move $x$ .

$- (x \cdot x) + x \cdot 1 + 5 (- x) + 5 \cdot 1 \geq 0$

Step 4.3.1.2.1.2.1.2.2

Multiply $x$ by $x$ .

$- x^{2} + x \cdot 1 + 5 (- x) + 5 \cdot 1 \geq 0$

Step 4.3.1.2.1.2.1.3

Multiply $x$ by $1$ .

$- x^{2} + x + 5 (- x) + 5 \cdot 1 \geq 0$

Step 4.3.1.2.1.2.1.4

Multiply $- 1$ by $5$ .

$- x^{2} + x - 5 x + 5 \cdot 1 \geq 0$

Step 4.3.1.2.1.2.1.5

Multiply $5$ by $1$ .

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

Step 4.3.1.2.1.2.2

Subtract $5 x$ from $x$ .

$- x^{2} - 4 x + 5 \geq 0$

Step 4.3.1.2.2

Convert the inequality to an equation.

$- x^{2} - 4 x + 5 = 0$

Step 4.3.1.2.3

Factor the left side of the equation.

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Step 4.3.1.2.3.1

Factor $- 1$ out of $- x^{2} - 4 x + 5$ .

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Step 4.3.1.2.3.1.1

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) - 4 x + 5 = 0$

Step 4.3.1.2.3.1.2

Factor $- 1$ out of $- 4 x$ .

$- (x^{2}) - (4 x) + 5 = 0$

Step 4.3.1.2.3.1.3

Rewrite $5$ as $- 1 (- 5)$ .

$- (x^{2}) - (4 x) - 1 \cdot - 5 = 0$

Step 4.3.1.2.3.1.4

Factor $- 1$ out of $- (x^{2}) - (4 x)$ .

$- (x^{2} + 4 x) - 1 \cdot - 5 = 0$

Step 4.3.1.2.3.1.5

Factor $- 1$ out of $- (x^{2} + 4 x) - 1 (- 5)$ .

$- (x^{2} + 4 x - 5) = 0$

Step 4.3.1.2.3.2

Factor.

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Step 4.3.1.2.3.2.1

Factor $x^{2} + 4 x - 5$ using the AC method.

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Step 4.3.1.2.3.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 5$ and whose sum is $4$ .

$- 1, 5$

Step 4.3.1.2.3.2.1.2

Write the factored form using these integers.

$- ((x - 1) (x + 5)) = 0$

Step 4.3.1.2.3.2.2

Remove unnecessary parentheses.

$- (x - 1) (x + 5) = 0$

Step 4.3.1.2.4

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

$x - 1 = 0$

$x + 5 = 0$

Step 4.3.1.2.5

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 4.3.1.2.5.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 4.3.1.2.5.2

Add $1$ to both sides of the equation.

$x = 1$

Step 4.3.1.2.6

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 4.3.1.2.6.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 4.3.1.2.6.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 4.3.1.2.7

The final solution is all the values that make $- (x - 1) (x + 5) = 0$ true.

$x = 1, - 5$

Step 4.3.1.2.8

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 1$

$x > 1$

Step 4.3.1.2.9

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 4.3.1.2.9.1

Test a value on the interval $x < - 5$ to see if it makes the inequality true.

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Step 4.3.1.2.9.1.1

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Step 4.3.1.2.9.1.2

Replace $x$ with $- 8$ in the original inequality.

$((- 8) + 5) (- (- 8) + 1) \geq 0$

Step 4.3.1.2.9.1.3

The left side $- 27$ is less than the right side $0$ , which means that the given statement is false.

False

Step 4.3.1.2.9.2

Test a value on the interval $- 5 < x < 1$ to see if it makes the inequality true.

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Step 4.3.1.2.9.2.1

Choose a value on the interval $- 5 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 4.3.1.2.9.2.2

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

$((0) + 5) (- (0) + 1) \geq 0$

Step 4.3.1.2.9.2.3

The left side $5$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 4.3.1.2.9.3

Test a value on the interval $x > 1$ to see if it makes the inequality true.

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Step 4.3.1.2.9.3.1

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Step 4.3.1.2.9.3.2

Replace $x$ with $4$ in the original inequality.

$((4) + 5) (- (4) + 1) \geq 0$

Step 4.3.1.2.9.3.3

The left side $- 27$ is less than the right side $0$ , which means that the given statement is false.

False

Step 4.3.1.2.9.4

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

$x < - 5$ False

$- 5 < x < 1$ True

$x > 1$ False

$x < - 5$ False

$- 5 < x < 1$ True

$x > 1$ False

Step 4.3.1.2.10

The solution consists of all of the true intervals.

$- 5 \leq x \leq 1$

Step 4.3.1.3

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

$[- 5, 1]$

Step 4.3.2

Find the intersection of $y \geq 0$ and $[- 5, 1]$ .

$0 \leq y \leq 1$

Step 4.4

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

$y < 0$

Step 4.5

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

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

Step 4.6

Find the domain of $- y \leq \sqrt{(x + 5) (- x + 1)}$ and find the intersection with $y < 0$ .

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Step 4.6.1

Find the domain of $- y \leq \sqrt{(x + 5) (- x + 1)}$ .

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Step 4.6.1.1

Set the radicand in $\sqrt{(x + 5) (- x + 1)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 5) (- x + 1) \geq 0$

Step 4.6.1.2

Solve for $y$ .

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Step 4.6.1.2.1

Simplify $(x + 5) (- x + 1)$ .

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Step 4.6.1.2.1.1

Expand $(x + 5) (- x + 1)$ using the FOIL Method.

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Step 4.6.1.2.1.1.1

Apply the distributive property.

$x (- x + 1) + 5 (- x + 1) \geq 0$

Step 4.6.1.2.1.1.2

Apply the distributive property.

$x (- x) + x \cdot 1 + 5 (- x + 1) \geq 0$

Step 4.6.1.2.1.1.3

Apply the distributive property.

$x (- x) + x \cdot 1 + 5 (- x) + 5 \cdot 1 \geq 0$

Step 4.6.1.2.1.2

Simplify and combine like terms.

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Step 4.6.1.2.1.2.1

Simplify each term.

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Step 4.6.1.2.1.2.1.1

Rewrite using the commutative property of multiplication.

$- x \cdot x + x \cdot 1 + 5 (- x) + 5 \cdot 1 \geq 0$

Step 4.6.1.2.1.2.1.2

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

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Step 4.6.1.2.1.2.1.2.1

Move $x$ .

$- (x \cdot x) + x \cdot 1 + 5 (- x) + 5 \cdot 1 \geq 0$

Step 4.6.1.2.1.2.1.2.2

Multiply $x$ by $x$ .

$- x^{2} + x \cdot 1 + 5 (- x) + 5 \cdot 1 \geq 0$

Step 4.6.1.2.1.2.1.3

Multiply $x$ by $1$ .

$- x^{2} + x + 5 (- x) + 5 \cdot 1 \geq 0$

Step 4.6.1.2.1.2.1.4

Multiply $- 1$ by $5$ .

$- x^{2} + x - 5 x + 5 \cdot 1 \geq 0$

Step 4.6.1.2.1.2.1.5

Multiply $5$ by $1$ .

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

Step 4.6.1.2.1.2.2

Subtract $5 x$ from $x$ .

$- x^{2} - 4 x + 5 \geq 0$

Step 4.6.1.2.2

Convert the inequality to an equation.

$- x^{2} - 4 x + 5 = 0$

Step 4.6.1.2.3

Factor the left side of the equation.

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Step 4.6.1.2.3.1

Factor $- 1$ out of $- x^{2} - 4 x + 5$ .

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Step 4.6.1.2.3.1.1

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) - 4 x + 5 = 0$

Step 4.6.1.2.3.1.2

Factor $- 1$ out of $- 4 x$ .

$- (x^{2}) - (4 x) + 5 = 0$

Step 4.6.1.2.3.1.3

Rewrite $5$ as $- 1 (- 5)$ .

$- (x^{2}) - (4 x) - 1 \cdot - 5 = 0$

Step 4.6.1.2.3.1.4

Factor $- 1$ out of $- (x^{2}) - (4 x)$ .

$- (x^{2} + 4 x) - 1 \cdot - 5 = 0$

Step 4.6.1.2.3.1.5

Factor $- 1$ out of $- (x^{2} + 4 x) - 1 (- 5)$ .

$- (x^{2} + 4 x - 5) = 0$

Step 4.6.1.2.3.2

Factor.

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Step 4.6.1.2.3.2.1

Factor $x^{2} + 4 x - 5$ using the AC method.

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Step 4.6.1.2.3.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 5$ and whose sum is $4$ .

$- 1, 5$

Step 4.6.1.2.3.2.1.2

Write the factored form using these integers.

$- ((x - 1) (x + 5)) = 0$

Step 4.6.1.2.3.2.2

Remove unnecessary parentheses.

$- (x - 1) (x + 5) = 0$

Step 4.6.1.2.4

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

$x - 1 = 0$

$x + 5 = 0$

Step 4.6.1.2.5

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 4.6.1.2.5.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 4.6.1.2.5.2

Add $1$ to both sides of the equation.

$x = 1$

Step 4.6.1.2.6

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 4.6.1.2.6.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 4.6.1.2.6.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 4.6.1.2.7

The final solution is all the values that make $- (x - 1) (x + 5) = 0$ true.

$x = 1, - 5$

Step 4.6.1.2.8

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 1$

$x > 1$

Step 4.6.1.2.9

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 4.6.1.2.9.1

Test a value on the interval $x < - 5$ to see if it makes the inequality true.

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Step 4.6.1.2.9.1.1

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Step 4.6.1.2.9.1.2

Replace $x$ with $- 8$ in the original inequality.

$((- 8) + 5) (- (- 8) + 1) \geq 0$

Step 4.6.1.2.9.1.3

The left side $- 27$ is less than the right side $0$ , which means that the given statement is false.

False

Step 4.6.1.2.9.2

Test a value on the interval $- 5 < x < 1$ to see if it makes the inequality true.

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Step 4.6.1.2.9.2.1

Choose a value on the interval $- 5 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 4.6.1.2.9.2.2

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

$((0) + 5) (- (0) + 1) \geq 0$

Step 4.6.1.2.9.2.3

The left side $5$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 4.6.1.2.9.3

Test a value on the interval $x > 1$ to see if it makes the inequality true.

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Step 4.6.1.2.9.3.1

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Step 4.6.1.2.9.3.2

Replace $x$ with $4$ in the original inequality.

$((4) + 5) (- (4) + 1) \geq 0$

Step 4.6.1.2.9.3.3

The left side $- 27$ is less than the right side $0$ , which means that the given statement is false.

False

Step 4.6.1.2.9.4

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

$x < - 5$ False

$- 5 < x < 1$ True

$x > 1$ False

$x < - 5$ False

$- 5 < x < 1$ True

$x > 1$ False

Step 4.6.1.2.10

The solution consists of all of the true intervals.

$- 5 \leq x \leq 1$

Step 4.6.1.3

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

$[- 5, 1]$

Step 4.6.2

Find the intersection of $y < 0$ and $[- 5, 1]$ .

$- 5 \leq y < 0$

Step 4.7

Write as a piecewise.

${\begin{cases} y \leq \sqrt{(x + 5) (- x + 1)} & 0 \leq y \leq 1 \\ - y \leq \sqrt{(x + 5) (- x + 1)} & - 5 \leq y < 0 \end{cases}$

Step 5

Find the intersection of $y \leq \sqrt{(x + 5) (- x + 1)}$ and $0 \leq y \leq 1$ .

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

Step 6

Solve $- y \leq \sqrt{(x + 5) (- x + 1)}$ when $- 5 \leq y < 0$ .

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Step 6.1

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

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Step 6.1.1

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

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

Step 6.1.2

Simplify the left side.

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Step 6.1.2.1

Dividing two negative values results in a positive value.

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

Step 6.1.2.2

Divide $y$ by $1$ .

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

Step 6.1.3

Simplify the right side.

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Step 6.1.3.1

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

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

Step 6.1.3.2

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

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

Step 6.2

Find the intersection of $y \geq - \sqrt{(x + 5) (- x + 1)}$ and $- 5 \leq y < 0$ .

No solution

Step 7

Find the union of the solutions.

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

Step 8