Algebra Examples

${(x - 1)}^{2} + {(y - 2)}^{2} \leq 4$

Step 1

Subtract ${(x - 1)}^{2}$ from both sides of the inequality.

${(y - 2)}^{2} \leq 4 - {(x - 1)}^{2}$

Step 2

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

$\sqrt{{(y - 2)}^{2}} \leq \sqrt{4 - {(x - 1)}^{2}}$

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.

$| y - 2 | \leq \sqrt{4 - {(x - 1)}^{2}}$

Step 3.2

Simplify the right side.

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Step 3.2.1

Simplify $\sqrt{4 - {(x - 1)}^{2}}$ .

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Step 3.2.1.1

Rewrite $4$ as $2^{2}$ .

$| y - 2 | \leq \sqrt{2^{2} - {(x - 1)}^{2}}$

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 $a = 2$ and $b = x - 1$ .

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

Step 3.2.1.3

Simplify.

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Step 3.2.1.3.1

Subtract $1$ from $2$ .

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

Step 3.2.1.3.2

Apply the distributive property.

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

Step 3.2.1.3.3

Multiply $- 1$ by $- 1$ .

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

Step 3.2.1.3.4

Add $2$ and $1$ .

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

Step 4

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

$y - 2 \geq 0$

Step 4.2

Add $2$ to both sides of the inequality.

$y \geq 2$

Step 4.3

In the piece where $y - 2$ is non-negative, remove the absolute value.

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

Step 4.4

Find the domain of $y - 2 \leq \sqrt{(x + 1) (- x + 3)}$ and find the intersection with $y \geq 2$ .

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Step 4.4.1

Find the domain of $y - 2 \leq \sqrt{(x + 1) (- x + 3)}$ .

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Step 4.4.1.1

Set the radicand in $\sqrt{(x + 1) (- x + 3)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 1) (- x + 3) \geq 0$

Step 4.4.1.2

Solve for $y$ .

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Step 4.4.1.2.1

Simplify $(x + 1) (- x + 3)$ .

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Step 4.4.1.2.1.1

Expand $(x + 1) (- x + 3)$ using the FOIL Method.

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Step 4.4.1.2.1.1.1

Apply the distributive property.

$x (- x + 3) + 1 (- x + 3) \geq 0$

Step 4.4.1.2.1.1.2

Apply the distributive property.

$x (- x) + x \cdot 3 + 1 (- x + 3) \geq 0$

Step 4.4.1.2.1.1.3

Apply the distributive property.

$x (- x) + x \cdot 3 + 1 (- x) + 1 \cdot 3 \geq 0$

Step 4.4.1.2.1.2

Simplify and combine like terms.

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Step 4.4.1.2.1.2.1

Simplify each term.

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Step 4.4.1.2.1.2.1.1

Rewrite using the commutative property of multiplication.

$- x \cdot x + x \cdot 3 + 1 (- x) + 1 \cdot 3 \geq 0$

Step 4.4.1.2.1.2.1.2

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

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Step 4.4.1.2.1.2.1.2.1

Move $x$ .

$- (x \cdot x) + x \cdot 3 + 1 (- x) + 1 \cdot 3 \geq 0$

Step 4.4.1.2.1.2.1.2.2

Multiply $x$ by $x$ .

$- x^{2} + x \cdot 3 + 1 (- x) + 1 \cdot 3 \geq 0$

Step 4.4.1.2.1.2.1.3

Move $3$ to the left of $x$ .

$- x^{2} + 3 \cdot x + 1 (- x) + 1 \cdot 3 \geq 0$

Step 4.4.1.2.1.2.1.4

Multiply $- x$ by $1$ .

$- x^{2} + 3 x - x + 1 \cdot 3 \geq 0$

Step 4.4.1.2.1.2.1.5

Multiply $3$ by $1$ .

$- x^{2} + 3 x - x + 3 \geq 0$

Step 4.4.1.2.1.2.2

Subtract $x$ from $3 x$ .

$- x^{2} + 2 x + 3 \geq 0$

Step 4.4.1.2.2

Convert the inequality to an equation.

$- x^{2} + 2 x + 3 = 0$

Step 4.4.1.2.3

Factor the left side of the equation.

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Step 4.4.1.2.3.1

Factor $- 1$ out of $- x^{2} + 2 x + 3$ .

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Step 4.4.1.2.3.1.1

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

$- (x^{2}) + 2 x + 3 = 0$

Step 4.4.1.2.3.1.2

Factor $- 1$ out of $2 x$ .

$- (x^{2}) - (- 2 x) + 3 = 0$

Step 4.4.1.2.3.1.3

Rewrite $3$ as $- 1 (- 3)$ .

$- (x^{2}) - (- 2 x) - 1 \cdot - 3 = 0$

Step 4.4.1.2.3.1.4

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

$- (x^{2} - 2 x) - 1 \cdot - 3 = 0$

Step 4.4.1.2.3.1.5

Factor $- 1$ out of $- (x^{2} - 2 x) - 1 (- 3)$ .

$- (x^{2} - 2 x - 3) = 0$

Step 4.4.1.2.3.2

Factor.

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Step 4.4.1.2.3.2.1

Factor $x^{2} - 2 x - 3$ using the AC method.

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Step 4.4.1.2.3.2.1.1

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 $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Step 4.4.1.2.3.2.1.2

Write the factored form using these integers.

$- ((x - 3) (x + 1)) = 0$

Step 4.4.1.2.3.2.2

Remove unnecessary parentheses.

$- (x - 3) (x + 1) = 0$

Step 4.4.1.2.4

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

$x - 3 = 0$

$x + 1 = 0$

Step 4.4.1.2.5

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 4.4.1.2.5.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 4.4.1.2.5.2

Add $3$ to both sides of the equation.

$x = 3$

Step 4.4.1.2.6

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

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Step 4.4.1.2.6.1

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

$x + 1 = 0$

Step 4.4.1.2.6.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 4.4.1.2.7

The final solution is all the values that make $- (x - 3) (x + 1) = 0$ true.

$x = 3, - 1$

Step 4.4.1.2.8

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 3$

$x > 3$

Step 4.4.1.2.9

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 4.4.1.2.9.1

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

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Step 4.4.1.2.9.1.1

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

$x = - 4$

Step 4.4.1.2.9.1.2

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

$((- 4) + 1) (- (- 4) + 3) \geq 0$

Step 4.4.1.2.9.1.3

The left side $- 21$ is less than the right side $0$ , which means that the given statement is false.

False

Step 4.4.1.2.9.2

Test a value on the interval $- 1 < x < 3$ to see if it makes the inequality true.

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Step 4.4.1.2.9.2.1

Choose a value on the interval $- 1 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Step 4.4.1.2.9.2.2

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

$((0) + 1) (- (0) + 3) \geq 0$

Step 4.4.1.2.9.2.3

The left side $3$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 4.4.1.2.9.3

Test a value on the interval $x > 3$ to see if it makes the inequality true.

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Step 4.4.1.2.9.3.1

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Step 4.4.1.2.9.3.2

Replace $x$ with $6$ in the original inequality.

$((6) + 1) (- (6) + 3) \geq 0$

Step 4.4.1.2.9.3.3

The left side $- 21$ is less than the right side $0$ , which means that the given statement is false.

False

Step 4.4.1.2.9.4

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

$x < - 1$ False

$- 1 < x < 3$ True

$x > 3$ False

$x < - 1$ False

$- 1 < x < 3$ True

$x > 3$ False

Step 4.4.1.2.10

The solution consists of all of the true intervals.

$- 1 \leq x \leq 3$

Step 4.4.1.3

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

$[- 1, 3]$

Step 4.4.2

Find the intersection of $y \geq 2$ and $[- 1, 3]$ .

$2 \leq y \leq 3$

Step 4.5

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

$y - 2 < 0$

Step 4.6

Add $2$ to both sides of the inequality.

$y < 2$

Step 4.7

In the piece where $y - 2$ is negative, remove the absolute value and multiply by $- 1$ .

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

Step 4.8

Find the domain of $- (y - 2) \leq \sqrt{(x + 1) (- x + 3)}$ and find the intersection with $y < 2$ .

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Step 4.8.1

Find the domain of $- (y - 2) \leq \sqrt{(x + 1) (- x + 3)}$ .

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Step 4.8.1.1

Set the radicand in $\sqrt{(x + 1) (- x + 3)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 1) (- x + 3) \geq 0$

Step 4.8.1.2

Solve for $y$ .

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Step 4.8.1.2.1

Simplify $(x + 1) (- x + 3)$ .

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Step 4.8.1.2.1.1

Expand $(x + 1) (- x + 3)$ using the FOIL Method.

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Step 4.8.1.2.1.1.1

Apply the distributive property.

$x (- x + 3) + 1 (- x + 3) \geq 0$

Step 4.8.1.2.1.1.2

Apply the distributive property.

$x (- x) + x \cdot 3 + 1 (- x + 3) \geq 0$

Step 4.8.1.2.1.1.3

Apply the distributive property.

$x (- x) + x \cdot 3 + 1 (- x) + 1 \cdot 3 \geq 0$

Step 4.8.1.2.1.2

Simplify and combine like terms.

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Step 4.8.1.2.1.2.1

Simplify each term.

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Step 4.8.1.2.1.2.1.1

Rewrite using the commutative property of multiplication.

$- x \cdot x + x \cdot 3 + 1 (- x) + 1 \cdot 3 \geq 0$

Step 4.8.1.2.1.2.1.2

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

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Step 4.8.1.2.1.2.1.2.1

Move $x$ .

$- (x \cdot x) + x \cdot 3 + 1 (- x) + 1 \cdot 3 \geq 0$

Step 4.8.1.2.1.2.1.2.2

Multiply $x$ by $x$ .

$- x^{2} + x \cdot 3 + 1 (- x) + 1 \cdot 3 \geq 0$

Step 4.8.1.2.1.2.1.3

Move $3$ to the left of $x$ .

$- x^{2} + 3 \cdot x + 1 (- x) + 1 \cdot 3 \geq 0$

Step 4.8.1.2.1.2.1.4

Multiply $- x$ by $1$ .

$- x^{2} + 3 x - x + 1 \cdot 3 \geq 0$

Step 4.8.1.2.1.2.1.5

Multiply $3$ by $1$ .

$- x^{2} + 3 x - x + 3 \geq 0$

Step 4.8.1.2.1.2.2

Subtract $x$ from $3 x$ .

$- x^{2} + 2 x + 3 \geq 0$

Step 4.8.1.2.2

Convert the inequality to an equation.

$- x^{2} + 2 x + 3 = 0$

Step 4.8.1.2.3

Factor the left side of the equation.

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Step 4.8.1.2.3.1

Factor $- 1$ out of $- x^{2} + 2 x + 3$ .

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Step 4.8.1.2.3.1.1

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

$- (x^{2}) + 2 x + 3 = 0$

Step 4.8.1.2.3.1.2

Factor $- 1$ out of $2 x$ .

$- (x^{2}) - (- 2 x) + 3 = 0$

Step 4.8.1.2.3.1.3

Rewrite $3$ as $- 1 (- 3)$ .

$- (x^{2}) - (- 2 x) - 1 \cdot - 3 = 0$

Step 4.8.1.2.3.1.4

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

$- (x^{2} - 2 x) - 1 \cdot - 3 = 0$

Step 4.8.1.2.3.1.5

Factor $- 1$ out of $- (x^{2} - 2 x) - 1 (- 3)$ .

$- (x^{2} - 2 x - 3) = 0$

Step 4.8.1.2.3.2

Factor.

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Step 4.8.1.2.3.2.1

Factor $x^{2} - 2 x - 3$ using the AC method.

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Step 4.8.1.2.3.2.1.1

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 $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Step 4.8.1.2.3.2.1.2

Write the factored form using these integers.

$- ((x - 3) (x + 1)) = 0$

Step 4.8.1.2.3.2.2

Remove unnecessary parentheses.

$- (x - 3) (x + 1) = 0$

Step 4.8.1.2.4

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

$x - 3 = 0$

$x + 1 = 0$

Step 4.8.1.2.5

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 4.8.1.2.5.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 4.8.1.2.5.2

Add $3$ to both sides of the equation.

$x = 3$

Step 4.8.1.2.6

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

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Step 4.8.1.2.6.1

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

$x + 1 = 0$

Step 4.8.1.2.6.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 4.8.1.2.7

The final solution is all the values that make $- (x - 3) (x + 1) = 0$ true.

$x = 3, - 1$

Step 4.8.1.2.8

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 3$

$x > 3$

Step 4.8.1.2.9

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 4.8.1.2.9.1

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

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Step 4.8.1.2.9.1.1

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

$x = - 4$

Step 4.8.1.2.9.1.2

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

$((- 4) + 1) (- (- 4) + 3) \geq 0$

Step 4.8.1.2.9.1.3

The left side $- 21$ is less than the right side $0$ , which means that the given statement is false.

False

Step 4.8.1.2.9.2

Test a value on the interval $- 1 < x < 3$ to see if it makes the inequality true.

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Step 4.8.1.2.9.2.1

Choose a value on the interval $- 1 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Step 4.8.1.2.9.2.2

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

$((0) + 1) (- (0) + 3) \geq 0$

Step 4.8.1.2.9.2.3

The left side $3$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 4.8.1.2.9.3

Test a value on the interval $x > 3$ to see if it makes the inequality true.

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Step 4.8.1.2.9.3.1

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Step 4.8.1.2.9.3.2

Replace $x$ with $6$ in the original inequality.

$((6) + 1) (- (6) + 3) \geq 0$

Step 4.8.1.2.9.3.3

The left side $- 21$ is less than the right side $0$ , which means that the given statement is false.

False

Step 4.8.1.2.9.4

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

$x < - 1$ False

$- 1 < x < 3$ True

$x > 3$ False

$x < - 1$ False

$- 1 < x < 3$ True

$x > 3$ False

Step 4.8.1.2.10

The solution consists of all of the true intervals.

$- 1 \leq x \leq 3$

Step 4.8.1.3

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

$[- 1, 3]$

Step 4.8.2

Find the intersection of $y < 2$ and $[- 1, 3]$ .

$- 1 \leq y < 2$

Step 4.9

Write as a piecewise.

${\begin{cases} y - 2 \leq \sqrt{(x + 1) (- x + 3)} & 2 \leq y \leq 3 \\ - (y - 2) \leq \sqrt{(x + 1) (- x + 3)} & - 1 \leq y < 2 \end{cases}$

Step 4.10

Simplify $- (y - 2) \leq \sqrt{(x + 1) (- x + 3)}$ .

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Step 4.10.1

Apply the distributive property.

${\begin{cases} y - 2 \leq \sqrt{(x + 1) (- x + 3)} & 2 \leq y \leq 3 \\ - y - - 2 \leq \sqrt{(x + 1) (- x + 3)} & - 1 \leq y < 2 \end{cases}$

Step 4.10.2

Multiply $- 1$ by $- 2$ .

${\begin{cases} y - 2 \leq \sqrt{(x + 1) (- x + 3)} & 2 \leq y \leq 3 \\ - y + 2 \leq \sqrt{(x + 1) (- x + 3)} & - 1 \leq y < 2 \end{cases}$

Step 5

Solve $y - 2 \leq \sqrt{(x + 1) (- x + 3)}$ when $2 \leq y \leq 3$ .

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Step 5.1

Add $2$ to both sides of the inequality.

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

Step 5.2

Find the intersection of $y \leq \sqrt{(x + 1) (- x + 3)} + 2$ and $2 \leq y \leq 3$ .

$2 \leq y \leq N o (Min i m u m)$

Step 6

Solve $- y + 2 \leq \sqrt{(x + 1) (- x + 3)}$ when $- 1 \leq y < 2$ .

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Step 6.1

Solve $- y + 2 \leq \sqrt{(x + 1) (- x + 3)}$ for $y$ .

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Step 6.1.1

Subtract $2$ from both sides of the inequality.

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

Step 6.1.2

Divide each term in $- y \leq \sqrt{(x + 1) (- x + 3)} - 2$ by $- 1$ and simplify.

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Step 6.1.2.1

Divide each term in $- y \leq \sqrt{(x + 1) (- x + 3)} - 2$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} \geq \frac{\sqrt{(x + 1) (- x + 3)}}{- 1} + \frac{- 2}{- 1}$

Step 6.1.2.2

Simplify the left side.

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Step 6.1.2.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} \geq \frac{\sqrt{(x + 1) (- x + 3)}}{- 1} + \frac{- 2}{- 1}$

Step 6.1.2.2.2

Divide $y$ by $1$ .

$y \geq \frac{\sqrt{(x + 1) (- x + 3)}}{- 1} + \frac{- 2}{- 1}$

Step 6.1.2.3

Simplify the right side.

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Step 6.1.2.3.1

Simplify each term.

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Step 6.1.2.3.1.1

Move the negative one from the denominator of $\frac{\sqrt{(x + 1) (- x + 3)}}{- 1}$ .

$y \geq - 1 \cdot \sqrt{(x + 1) (- x + 3)} + \frac{- 2}{- 1}$

Step 6.1.2.3.1.2

Rewrite $- 1 \cdot \sqrt{(x + 1) (- x + 3)}$ as $- \sqrt{(x + 1) (- x + 3)}$ .

$y \geq - \sqrt{(x + 1) (- x + 3)} + \frac{- 2}{- 1}$

Step 6.1.2.3.1.3

Divide $- 2$ by $- 1$ .

$y \geq - \sqrt{(x + 1) (- x + 3)} + 2$

Step 6.2

Find the intersection of $y \geq - \sqrt{(x + 1) (- x + 3)} + 2$ and $- 1 \leq y < 2$ .

No solution

Step 7

Find the union of the solutions.

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

Step 8