Algebra Examples

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

Step 1

Subtract ${(x - 3)}^{2}$ from both sides of the inequality.

$y^{2} \leq 9 - {(x - 3)}^{2}$

Step 2

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

$\sqrt{y^{2}} \leq \sqrt{9 - {(x - 3)}^{2}}$

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.

$| y | \leq \sqrt{9 - {(x - 3)}^{2}}$

Step 3.2

Simplify the right side.

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Step 3.2.1

Simplify $\sqrt{9 - {(x - 3)}^{2}}$ .

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Step 3.2.1.1

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

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

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 $b = x - 3$ .

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

Step 3.2.1.3

Simplify.

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Step 3.2.1.3.1

Subtract $3$ from $3$ .

$| y | \leq \sqrt{(x + 0) (3 - (x - 3))}$

Step 3.2.1.3.2

Add $x$ and $0$ .

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

Step 3.2.1.3.3

Apply the distributive property.

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

Step 3.2.1.3.4

Multiply $- 1$ by $- 3$ .

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

Step 3.2.1.3.5

Add $3$ and $3$ .

$| y | \leq \sqrt{x (- x + 6)}$

Step 4

Write $| y | \leq \sqrt{x (- x + 6)}$ 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.

$y \leq \sqrt{x (- x + 6)}$

Step 4.3

Find the domain of $y \leq \sqrt{x (- x + 6)}$ and find the intersection with $y \geq 0$ .

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Step 4.3.1

Find the domain of $y \leq \sqrt{x (- x + 6)}$ .

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Step 4.3.1.1

Set the radicand in $\sqrt{x (- x + 6)}$ greater than or equal to $0$ to find where the expression is defined.

$x (- x + 6) \geq 0$

Step 4.3.1.2

Solve for $y$ .

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Step 4.3.1.2.1

Simplify $x (- x + 6)$ .

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Step 4.3.1.2.1.1

Simplify by multiplying through.

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Step 4.3.1.2.1.1.1

Apply the distributive property.

$x (- x) + x \cdot 6 \geq 0$

Step 4.3.1.2.1.1.2

Reorder.

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Step 4.3.1.2.1.1.2.1

Rewrite using the commutative property of multiplication.

$- x \cdot x + x \cdot 6 \geq 0$

Step 4.3.1.2.1.1.2.2

Move $6$ to the left of $x$ .

$- x \cdot x + 6 \cdot x \geq 0$

Step 4.3.1.2.1.2

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

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Step 4.3.1.2.1.2.1

Move $x$ .

$- (x \cdot x) + 6 \cdot x \geq 0$

Step 4.3.1.2.1.2.2

Multiply $x$ by $x$ .

$- x^{2} + 6 \cdot x \geq 0$

$- x^{2} + 6 x \geq 0$

Step 4.3.1.2.2

Convert the inequality to an equation.

$- x^{2} + 6 x = 0$

Step 4.3.1.2.3

Factor $- x$ out of $- x^{2} + 6 x$ .

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Step 4.3.1.2.3.1

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

$- x \cdot x + 6 x = 0$

Step 4.3.1.2.3.2

Factor $- x$ out of $6 x$ .

$- x \cdot x - x \cdot - 6 = 0$

Step 4.3.1.2.3.3

Factor $- x$ out of $- x (x) - x (- 6)$ .

$- x (x - 6) = 0$

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 $0$ .

$x = 0$

$x - 6 = 0$

Step 4.3.1.2.5

Set $x$ equal to $0$ .

$x = 0$

Step 4.3.1.2.6

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 4.3.1.2.6.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 4.3.1.2.6.2

Add $6$ to both sides of the equation.

$x = 6$

Step 4.3.1.2.7

The final solution is all the values that make $- x (x - 6) = 0$ true.

$x = 0, 6$

Step 4.3.1.2.8

Use each root to create test intervals.

$x < 0$

$0 < x < 6$

$x > 6$

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 $x < 0$ 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 $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 4.3.1.2.9.1.2

Replace $x$ with $- 2$ in the original inequality.

$(- 2) (- (- 2) + 6) \geq 0$

Step 4.3.1.2.9.1.3

The left side $- 16$ 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 $0 < x < 6$ 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 $0 < x < 6$ and see if this value makes the original inequality true.

$x = 3$

Step 4.3.1.2.9.2.2

Replace $x$ with $3$ in the original inequality.

$(3) (- (3) + 6) \geq 0$

Step 4.3.1.2.9.2.3

The left side $9$ 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 $x > 6$ 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 $x > 6$ and see if this value makes the original inequality true.

$x = 8$

Step 4.3.1.2.9.3.2

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

$(8) (- (8) + 6) \geq 0$

Step 4.3.1.2.9.3.3

The left side $- 16$ 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.

$x < 0$ False

$0 < x < 6$ True

$x > 6$ False

$x < 0$ False

$0 < x < 6$ True

$x > 6$ False

Step 4.3.1.2.10

The solution consists of all of the true intervals.

$0 \leq x \leq 6$

Step 4.3.1.3

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

$[0, 6]$

Step 4.3.2

Find the intersection of $y \geq 0$ and $[0, 6]$ .

$0 \leq y \leq 6$

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 $- 1$ .

$- y \leq \sqrt{x (- x + 6)}$

Step 4.6

Find the domain of $- y \leq \sqrt{x (- x + 6)}$ and find the intersection with $y < 0$ .

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Step 4.6.1

Find the domain of $- y \leq \sqrt{x (- x + 6)}$ .

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Step 4.6.1.1

Set the radicand in $\sqrt{x (- x + 6)}$ greater than or equal to $0$ to find where the expression is defined.

$x (- x + 6) \geq 0$

Step 4.6.1.2

Solve for $y$ .

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Step 4.6.1.2.1

Simplify $x (- x + 6)$ .

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Step 4.6.1.2.1.1

Simplify by multiplying through.

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Step 4.6.1.2.1.1.1

Apply the distributive property.

$x (- x) + x \cdot 6 \geq 0$

Step 4.6.1.2.1.1.2

Reorder.

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Step 4.6.1.2.1.1.2.1

Rewrite using the commutative property of multiplication.

$- x \cdot x + x \cdot 6 \geq 0$

Step 4.6.1.2.1.1.2.2

Move $6$ to the left of $x$ .

$- x \cdot x + 6 \cdot x \geq 0$

Step 4.6.1.2.1.2

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

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Step 4.6.1.2.1.2.1

Move $x$ .

$- (x \cdot x) + 6 \cdot x \geq 0$

Step 4.6.1.2.1.2.2

Multiply $x$ by $x$ .

$- x^{2} + 6 \cdot x \geq 0$

$- x^{2} + 6 x \geq 0$

Step 4.6.1.2.2

Convert the inequality to an equation.

$- x^{2} + 6 x = 0$

Step 4.6.1.2.3

Factor $- x$ out of $- x^{2} + 6 x$ .

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Step 4.6.1.2.3.1

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

$- x \cdot x + 6 x = 0$

Step 4.6.1.2.3.2

Factor $- x$ out of $6 x$ .

$- x \cdot x - x \cdot - 6 = 0$

Step 4.6.1.2.3.3

Factor $- x$ out of $- x (x) - x (- 6)$ .

$- x (x - 6) = 0$

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 $0$ .

$x = 0$

$x - 6 = 0$

Step 4.6.1.2.5

Set $x$ equal to $0$ .

$x = 0$

Step 4.6.1.2.6

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 4.6.1.2.6.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 4.6.1.2.6.2

Add $6$ to both sides of the equation.

$x = 6$

Step 4.6.1.2.7

The final solution is all the values that make $- x (x - 6) = 0$ true.

$x = 0, 6$

Step 4.6.1.2.8

Use each root to create test intervals.

$x < 0$

$0 < x < 6$

$x > 6$

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 $x < 0$ 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 $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 4.6.1.2.9.1.2

Replace $x$ with $- 2$ in the original inequality.

$(- 2) (- (- 2) + 6) \geq 0$

Step 4.6.1.2.9.1.3

The left side $- 16$ 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 $0 < x < 6$ 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 $0 < x < 6$ and see if this value makes the original inequality true.

$x = 3$

Step 4.6.1.2.9.2.2

Replace $x$ with $3$ in the original inequality.

$(3) (- (3) + 6) \geq 0$

Step 4.6.1.2.9.2.3

The left side $9$ 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 $x > 6$ 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 $x > 6$ and see if this value makes the original inequality true.

$x = 8$

Step 4.6.1.2.9.3.2

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

$(8) (- (8) + 6) \geq 0$

Step 4.6.1.2.9.3.3

The left side $- 16$ 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.

$x < 0$ False

$0 < x < 6$ True

$x > 6$ False

$x < 0$ False

$0 < x < 6$ True

$x > 6$ False

Step 4.6.1.2.10

The solution consists of all of the true intervals.

$0 \leq x \leq 6$

Step 4.6.1.3

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

$[0, 6]$

Step 4.6.2

Find the intersection of $y < 0$ and $[0, 6]$ .

$No solution$

Step 4.7

Write as a piecewise.

${\begin{cases} y \leq \sqrt{x (- x + 6)} & 0 \leq y \leq 6 \end{cases}$

Step 5

Find the intersection of $y \leq \sqrt{x (- x + 6)}$ and $0 \leq y \leq 6$ .

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

Step 6

Find the union of the solutions.

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

Step 7