Algebra Examples

$\frac{{(x + 4)}^{2}}{9^{2}} + \frac{{(y + 1)}^{2}}{6^{2}} > 1$

Step 1

Subtract $\frac{{(x + 4)}^{2}}{9^{2}}$ from both sides of the inequality.

$\frac{{(y + 1)}^{2}}{6^{2}} > 1 - \frac{{(x + 4)}^{2}}{9^{2}}$

Step 2

Raise $6$ to the power of $2$ .

$\frac{{(y + 1)}^{2}}{36} > 1 - \frac{{(x + 4)}^{2}}{9^{2}}$

Step 3

Simplify $1 - \frac{{(x + 4)}^{2}}{9^{2}}$ .

Tap for more steps...

Step 3.1

Combine into one fraction.

Tap for more steps...

Step 3.1.1

Raise $9$ to the power of $2$ .

$\frac{{(y + 1)}^{2}}{36} > 1 - \frac{{(x + 4)}^{2}}{81}$

Step 3.1.2

Write $1$ as a fraction with a common denominator.

$\frac{{(y + 1)}^{2}}{36} > \frac{81}{81} - \frac{{(x + 4)}^{2}}{81}$

Step 3.1.3

Combine the numerators over the common denominator.

$\frac{{(y + 1)}^{2}}{36} > \frac{81 - {(x + 4)}^{2}}{81}$

Step 3.2

Simplify the numerator.

Tap for more steps...

Step 3.2.1

Rewrite $81$ as $9^{2}$ .

$\frac{{(y + 1)}^{2}}{36} > \frac{9^{2} - {(x + 4)}^{2}}{81}$

Step 3.2.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 9$ and $b = x + 4$ .

$\frac{{(y + 1)}^{2}}{36} > \frac{(9 + x + 4) (9 - (x + 4))}{81}$

Step 3.2.3

Simplify.

Tap for more steps...

Step 3.2.3.1

Add $9$ and $4$ .

$\frac{{(y + 1)}^{2}}{36} > \frac{(x + 13) (9 - (x + 4))}{81}$

Step 3.2.3.2

Apply the distributive property.

$\frac{{(y + 1)}^{2}}{36} > \frac{(x + 13) (9 - x - 1 \cdot 4)}{81}$

Step 3.2.3.3

Multiply $- 1$ by $4$ .

$\frac{{(y + 1)}^{2}}{36} > \frac{(x + 13) (9 - x - 4)}{81}$

Step 3.2.3.4

Subtract $4$ from $9$ .

$\frac{{(y + 1)}^{2}}{36} > \frac{(x + 13) (- x + 5)}{81}$

Step 3.3

Simplify with factoring out.

Tap for more steps...

Step 3.3.1

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

$\frac{{(y + 1)}^{2}}{36} > \frac{(x + 13) (- (x) + 5)}{81}$

Step 3.3.2

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

$\frac{{(y + 1)}^{2}}{36} > \frac{(x + 13) (- (x) - 1 (- 5))}{81}$

Step 3.3.3

Factor $- 1$ out of $- (x) - 1 (- 5)$ .

$\frac{{(y + 1)}^{2}}{36} > \frac{(x + 13) (- (x - 5))}{81}$

Step 3.3.4

Simplify the expression.

Tap for more steps...

Step 3.3.4.1

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

$\frac{{(y + 1)}^{2}}{36} > \frac{(x + 13) (- 1 (x - 5))}{81}$

Step 3.3.4.2

Move the negative in front of the fraction.

$\frac{{(y + 1)}^{2}}{36} > - \frac{(x + 13) (x - 5)}{81}$

Step 4

Multiply both sides by $36$ .

$\frac{{(y + 1)}^{2}}{36} \cdot 36 > - \frac{(x + 13) (x - 5)}{81} \cdot 36$

Step 5

Simplify.

Tap for more steps...

Step 5.1

Simplify the left side.

Tap for more steps...

Step 5.1.1

Cancel the common factor of $36$ .

Tap for more steps...

Step 5.1.1.1

Cancel the common factor.

$\frac{{(y + 1)}^{2}}{36} \cdot 36 > - \frac{(x + 13) (x - 5)}{81} \cdot 36$

Step 5.1.1.2

Rewrite the expression.

${(y + 1)}^{2} > - \frac{(x + 13) (x - 5)}{81} \cdot 36$

Step 5.2

Simplify the right side.

Tap for more steps...

Step 5.2.1

Simplify $- \frac{(x + 13) (x - 5)}{81} \cdot 36$ .

Tap for more steps...

Step 5.2.1.1

Cancel the common factor of $9$ .

Tap for more steps...

Step 5.2.1.1.1

Move the leading negative in $- \frac{(x + 13) (x - 5)}{81}$ into the numerator.

${(y + 1)}^{2} > \frac{- (x + 13) (x - 5)}{81} \cdot 36$

Step 5.2.1.1.2

Factor $9$ out of $81$ .

${(y + 1)}^{2} > \frac{- (x + 13) (x - 5)}{9 (9)} \cdot 36$

Step 5.2.1.1.3

Factor $9$ out of $36$ .

${(y + 1)}^{2} > \frac{- (x + 13) (x - 5)}{9 \cdot 9} \cdot (9 \cdot 4)$

Step 5.2.1.1.4

Cancel the common factor.

${(y + 1)}^{2} > \frac{- (x + 13) (x - 5)}{9 \cdot 9} \cdot (9 \cdot 4)$

Step 5.2.1.1.5

Rewrite the expression.

${(y + 1)}^{2} > \frac{- (x + 13) (x - 5)}{9} \cdot 4$

Step 5.2.1.2

Combine $\frac{- (x + 13) (x - 5)}{9}$ and $4$ .

${(y + 1)}^{2} > \frac{- (x + 13) (x - 5) \cdot 4}{9}$

Step 5.2.1.3

Simplify the expression.

Tap for more steps...

Step 5.2.1.3.1

Multiply $4$ by $- 1$ .

${(y + 1)}^{2} > \frac{- 4 (x + 13) (x - 5)}{9}$

Step 5.2.1.3.2

Move the negative in front of the fraction.

${(y + 1)}^{2} > - \frac{4 (x + 13) (x - 5)}{9}$

Step 6

Solve for $y$ .

Tap for more steps...

Step 6.1

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

$\sqrt{{(y + 1)}^{2}} > \sqrt{- \frac{4 (x + 13) (x - 5)}{9}}$

Step 6.2

Simplify the equation.

Tap for more steps...

Step 6.2.1

Simplify the left side.

Tap for more steps...

Step 6.2.1.1

Pull terms out from under the radical.

$| y + 1 | > \sqrt{- \frac{4 (x + 13) (x - 5)}{9}}$

Step 6.2.2

Simplify the right side.

Tap for more steps...

Step 6.2.2.1

Simplify $\sqrt{- \frac{4 (x + 13) (x - 5)}{9}}$ .

Tap for more steps...

Step 6.2.2.1.1

Rewrite $- \frac{4 (x + 13) (x - 5)}{9}$ as ${(\frac{2}{3})}^{2} \cdot (- (x + 13) (x - 5))$ .

Tap for more steps...

Step 6.2.2.1.1.1

Factor the perfect power $2^{2}$ out of $4 (x + 13) (x - 5)$ .

$| y + 1 | > \sqrt{- \frac{2^{2} ((x + 13) (x - 5))}{9}}$

Step 6.2.2.1.1.2

Factor the perfect power $3^{2}$ out of $9$ .

$| y + 1 | > \sqrt{- \frac{2^{2} ((x + 13) (x - 5))}{3^{2} \cdot 1}}$

Step 6.2.2.1.1.3

Rearrange the fraction $\frac{2^{2} ((x + 13) (x - 5))}{3^{2} \cdot 1}$ .

$| y + 1 | > \sqrt{- ({(\frac{2}{3})}^{2} ((x + 13) (x - 5)))}$

Step 6.2.2.1.1.4

Reorder $- 1$ and ${(\frac{2}{3})}^{2}$ .

$| y + 1 | > \sqrt{{(\frac{2}{3})}^{2} \cdot - 1 (x + 13) (x - 5)}$

Step 6.2.2.1.1.5

Add parentheses.

$| y + 1 | > \sqrt{{(\frac{2}{3})}^{2} \cdot - 1 ((x + 13) (x - 5))}$

Step 6.2.2.1.1.6

Add parentheses.

$| y + 1 | > \sqrt{{(\frac{2}{3})}^{2} \cdot (- (x + 13) (x - 5))}$

Step 6.2.2.1.2

Pull terms out from under the radical.

$| y + 1 | > | \frac{2}{3} | \sqrt{- (x + 13) (x - 5)}$

Step 6.2.2.1.3

$\frac{2}{3}$ is approximately $0.\overline{6}$ which is positive so remove the absolute value

$| y + 1 | > \frac{2}{3} \sqrt{- (x + 13) (x - 5)}$

Step 6.2.2.1.4

Combine $\frac{2}{3}$ and $\sqrt{- (x + 13) (x - 5)}$ .

$| y + 1 | > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3}$

Step 6.3

Write $| y + 1 | > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3}$ as a piecewise.

Tap for more steps...

Step 6.3.1

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

$y + 1 \geq 0$

Step 6.3.2

Subtract $1$ from both sides of the inequality.

$y \geq - 1$

Step 6.3.3

In the piece where $y + 1$ is non-negative, remove the absolute value.

$y + 1 > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3}$

Step 6.3.4

Find the domain of $y + 1 > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3}$ and find the intersection with $y \geq - 1$ .

Tap for more steps...

Step 6.3.4.1

Find the domain of $y + 1 > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3}$ .

Tap for more steps...

Step 6.3.4.1.1

Set the radicand in $\sqrt{- (x + 13) (x - 5)}$ greater than or equal to $0$ to find where the expression is defined.

$- (x + 13) (x - 5) \geq 0$

Step 6.3.4.1.2

Solve for $y$ .

Tap for more steps...

Step 6.3.4.1.2.1

Divide each term in $- (x + 13) (x - 5) \geq 0$ by $- 1$ and simplify.

Tap for more steps...

Step 6.3.4.1.2.1.1

Divide each term in $- (x + 13) (x - 5) \geq 0$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- (x + 13) (x - 5)}{- 1} \leq \frac{0}{- 1}$

Step 6.3.4.1.2.1.2

Simplify the left side.

Tap for more steps...

Step 6.3.4.1.2.1.2.1

Reduce the expression by cancelling the common factors.

Tap for more steps...

Step 6.3.4.1.2.1.2.1.1

Dividing two negative values results in a positive value.

$\frac{(x + 13) (x - 5)}{1} \leq \frac{0}{- 1}$

Step 6.3.4.1.2.1.2.1.2

Divide $(x + 13) (x - 5)$ by $1$ .

$(x + 13) (x - 5) \leq \frac{0}{- 1}$

Step 6.3.4.1.2.1.2.2

Expand $(x + 13) (x - 5)$ using the FOIL Method.

Tap for more steps...

Step 6.3.4.1.2.1.2.2.1

Apply the distributive property.

$x (x - 5) + 13 (x - 5) \leq \frac{0}{- 1}$

Step 6.3.4.1.2.1.2.2.2

Apply the distributive property.

$x \cdot x + x \cdot - 5 + 13 (x - 5) \leq \frac{0}{- 1}$

Step 6.3.4.1.2.1.2.2.3

Apply the distributive property.

$x \cdot x + x \cdot - 5 + 13 x + 13 \cdot - 5 \leq \frac{0}{- 1}$

Step 6.3.4.1.2.1.2.3

Simplify and combine like terms.

Tap for more steps...

Step 6.3.4.1.2.1.2.3.1

Simplify each term.

Tap for more steps...

Step 6.3.4.1.2.1.2.3.1.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 5 + 13 x + 13 \cdot - 5 \leq \frac{0}{- 1}$

Step 6.3.4.1.2.1.2.3.1.2

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

$x^{2} - 5 \cdot x + 13 x + 13 \cdot - 5 \leq \frac{0}{- 1}$

Step 6.3.4.1.2.1.2.3.1.3

Multiply $13$ by $- 5$ .

$x^{2} - 5 x + 13 x - 65 \leq \frac{0}{- 1}$

Step 6.3.4.1.2.1.2.3.2

Add $- 5 x$ and $13 x$ .

$x^{2} + 8 x - 65 \leq \frac{0}{- 1}$

Step 6.3.4.1.2.1.3

Simplify the right side.

Tap for more steps...

Step 6.3.4.1.2.1.3.1

Divide $0$ by $- 1$ .

$x^{2} + 8 x - 65 \leq 0$

Step 6.3.4.1.2.2

Convert the inequality to an equation.

$x^{2} + 8 x - 65 = 0$

Step 6.3.4.1.2.3

Factor $x^{2} + 8 x - 65$ using the AC method.

Tap for more steps...

Step 6.3.4.1.2.3.1

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 $- 65$ and whose sum is $8$ .

$- 5, 13$

Step 6.3.4.1.2.3.2

Write the factored form using these integers.

$(x - 5) (x + 13) = 0$

Step 6.3.4.1.2.4

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

$x - 5 = 0$

$x + 13 = 0$

Step 6.3.4.1.2.5

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

Tap for more steps...

Step 6.3.4.1.2.5.1

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

$x - 5 = 0$

Step 6.3.4.1.2.5.2

Add $5$ to both sides of the equation.

$x = 5$

Step 6.3.4.1.2.6

Set $x + 13$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 6.3.4.1.2.6.1

Set $x + 13$ equal to $0$ .

$x + 13 = 0$

Step 6.3.4.1.2.6.2

Subtract $13$ from both sides of the equation.

$x = - 13$

Step 6.3.4.1.2.7

The final solution is all the values that make $(x - 5) (x + 13) = 0$ true.

$x = 5, - 13$

Step 6.3.4.1.2.8

Use each root to create test intervals.

$x < - 13$

$- 13 < x < 5$

$x > 5$

Step 6.3.4.1.2.9

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Tap for more steps...

Step 6.3.4.1.2.9.1

Test a value on the interval $x < - 13$ to see if it makes the inequality true.

Tap for more steps...

Step 6.3.4.1.2.9.1.1

Choose a value on the interval $x < - 13$ and see if this value makes the original inequality true.

$x = - 16$

Step 6.3.4.1.2.9.1.2

Replace $x$ with $- 16$ in the original inequality.

$- ((- 16) + 13) ((- 16) - 5) \geq 0$

Step 6.3.4.1.2.9.1.3

The left side $- 63$ is less than the right side $0$ , which means that the given statement is false.

False

Step 6.3.4.1.2.9.2

Test a value on the interval $- 13 < x < 5$ to see if it makes the inequality true.

Tap for more steps...

Step 6.3.4.1.2.9.2.1

Choose a value on the interval $- 13 < x < 5$ and see if this value makes the original inequality true.

$x = 0$

Step 6.3.4.1.2.9.2.2

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

$- ((0) + 13) ((0) - 5) \geq 0$

Step 6.3.4.1.2.9.2.3

The left side $65$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 6.3.4.1.2.9.3

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

Tap for more steps...

Step 6.3.4.1.2.9.3.1

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

$x = 8$

Step 6.3.4.1.2.9.3.2

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

$- ((8) + 13) ((8) - 5) \geq 0$

Step 6.3.4.1.2.9.3.3

The left side $- 63$ is less than the right side $0$ , which means that the given statement is false.

False

Step 6.3.4.1.2.9.4

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

$x < - 13$ False

$- 13 < x < 5$ True

$x > 5$ False

$x < - 13$ False

$- 13 < x < 5$ True

$x > 5$ False

Step 6.3.4.1.2.10

The solution consists of all of the true intervals.

$- 13 \leq x \leq 5$

Step 6.3.4.1.3

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

$[- 13, 5]$

Step 6.3.4.2

Find the intersection of $y \geq - 1$ and $[- 13, 5]$ .

$- 1 \leq y \leq 5$

Step 6.3.5

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

$y + 1 < 0$

Step 6.3.6

Subtract $1$ from both sides of the inequality.

$y < - 1$

Step 6.3.7

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

$- (y + 1) > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3}$

Step 6.3.8

Find the domain of $- (y + 1) > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3}$ and find the intersection with $y < - 1$ .

Tap for more steps...

Step 6.3.8.1

Find the domain of $- (y + 1) > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3}$ .

Tap for more steps...

Step 6.3.8.1.1

Set the radicand in $\sqrt{- (x + 13) (x - 5)}$ greater than or equal to $0$ to find where the expression is defined.

$- (x + 13) (x - 5) \geq 0$

Step 6.3.8.1.2

Solve for $y$ .

Tap for more steps...

Step 6.3.8.1.2.1

Divide each term in $- (x + 13) (x - 5) \geq 0$ by $- 1$ and simplify.

Tap for more steps...

Step 6.3.8.1.2.1.1

Divide each term in $- (x + 13) (x - 5) \geq 0$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- (x + 13) (x - 5)}{- 1} \leq \frac{0}{- 1}$

Step 6.3.8.1.2.1.2

Simplify the left side.

Tap for more steps...

Step 6.3.8.1.2.1.2.1

Reduce the expression by cancelling the common factors.

Tap for more steps...

Step 6.3.8.1.2.1.2.1.1

Dividing two negative values results in a positive value.

$\frac{(x + 13) (x - 5)}{1} \leq \frac{0}{- 1}$

Step 6.3.8.1.2.1.2.1.2

Divide $(x + 13) (x - 5)$ by $1$ .

$(x + 13) (x - 5) \leq \frac{0}{- 1}$

Step 6.3.8.1.2.1.2.2

Expand $(x + 13) (x - 5)$ using the FOIL Method.

Tap for more steps...

Step 6.3.8.1.2.1.2.2.1

Apply the distributive property.

$x (x - 5) + 13 (x - 5) \leq \frac{0}{- 1}$

Step 6.3.8.1.2.1.2.2.2

Apply the distributive property.

$x \cdot x + x \cdot - 5 + 13 (x - 5) \leq \frac{0}{- 1}$

Step 6.3.8.1.2.1.2.2.3

Apply the distributive property.

$x \cdot x + x \cdot - 5 + 13 x + 13 \cdot - 5 \leq \frac{0}{- 1}$

Step 6.3.8.1.2.1.2.3

Simplify and combine like terms.

Tap for more steps...

Step 6.3.8.1.2.1.2.3.1

Simplify each term.

Tap for more steps...

Step 6.3.8.1.2.1.2.3.1.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 5 + 13 x + 13 \cdot - 5 \leq \frac{0}{- 1}$

Step 6.3.8.1.2.1.2.3.1.2

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

$x^{2} - 5 \cdot x + 13 x + 13 \cdot - 5 \leq \frac{0}{- 1}$

Step 6.3.8.1.2.1.2.3.1.3

Multiply $13$ by $- 5$ .

$x^{2} - 5 x + 13 x - 65 \leq \frac{0}{- 1}$

Step 6.3.8.1.2.1.2.3.2

Add $- 5 x$ and $13 x$ .

$x^{2} + 8 x - 65 \leq \frac{0}{- 1}$

Step 6.3.8.1.2.1.3

Simplify the right side.

Tap for more steps...

Step 6.3.8.1.2.1.3.1

Divide $0$ by $- 1$ .

$x^{2} + 8 x - 65 \leq 0$

Step 6.3.8.1.2.2

Convert the inequality to an equation.

$x^{2} + 8 x - 65 = 0$

Step 6.3.8.1.2.3

Factor $x^{2} + 8 x - 65$ using the AC method.

Tap for more steps...

Step 6.3.8.1.2.3.1

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 $- 65$ and whose sum is $8$ .

$- 5, 13$

Step 6.3.8.1.2.3.2

Write the factored form using these integers.

$(x - 5) (x + 13) = 0$

Step 6.3.8.1.2.4

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

$x - 5 = 0$

$x + 13 = 0$

Step 6.3.8.1.2.5

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

Tap for more steps...

Step 6.3.8.1.2.5.1

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

$x - 5 = 0$

Step 6.3.8.1.2.5.2

Add $5$ to both sides of the equation.

$x = 5$

Step 6.3.8.1.2.6

Set $x + 13$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 6.3.8.1.2.6.1

Set $x + 13$ equal to $0$ .

$x + 13 = 0$

Step 6.3.8.1.2.6.2

Subtract $13$ from both sides of the equation.

$x = - 13$

Step 6.3.8.1.2.7

The final solution is all the values that make $(x - 5) (x + 13) = 0$ true.

$x = 5, - 13$

Step 6.3.8.1.2.8

Use each root to create test intervals.

$x < - 13$

$- 13 < x < 5$

$x > 5$

Step 6.3.8.1.2.9

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Tap for more steps...

Step 6.3.8.1.2.9.1

Test a value on the interval $x < - 13$ to see if it makes the inequality true.

Tap for more steps...

Step 6.3.8.1.2.9.1.1

Choose a value on the interval $x < - 13$ and see if this value makes the original inequality true.

$x = - 16$

Step 6.3.8.1.2.9.1.2

Replace $x$ with $- 16$ in the original inequality.

$- ((- 16) + 13) ((- 16) - 5) \geq 0$

Step 6.3.8.1.2.9.1.3

The left side $- 63$ is less than the right side $0$ , which means that the given statement is false.

False

Step 6.3.8.1.2.9.2

Test a value on the interval $- 13 < x < 5$ to see if it makes the inequality true.

Tap for more steps...

Step 6.3.8.1.2.9.2.1

Choose a value on the interval $- 13 < x < 5$ and see if this value makes the original inequality true.

$x = 0$

Step 6.3.8.1.2.9.2.2

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

$- ((0) + 13) ((0) - 5) \geq 0$

Step 6.3.8.1.2.9.2.3

The left side $65$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 6.3.8.1.2.9.3

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

Tap for more steps...

Step 6.3.8.1.2.9.3.1

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

$x = 8$

Step 6.3.8.1.2.9.3.2

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

$- ((8) + 13) ((8) - 5) \geq 0$

Step 6.3.8.1.2.9.3.3

The left side $- 63$ is less than the right side $0$ , which means that the given statement is false.

False

Step 6.3.8.1.2.9.4

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

$x < - 13$ False

$- 13 < x < 5$ True

$x > 5$ False

$x < - 13$ False

$- 13 < x < 5$ True

$x > 5$ False

Step 6.3.8.1.2.10

The solution consists of all of the true intervals.

$- 13 \leq x \leq 5$

Step 6.3.8.1.3

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

$[- 13, 5]$

Step 6.3.8.2

Find the intersection of $y < - 1$ and $[- 13, 5]$ .

$- 13 \leq y < - 1$

Step 6.3.9

Write as a piecewise.

${\begin{cases} y + 1 > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3} & - 1 \leq y \leq 5 \\ - (y + 1) > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3} & - 13 \leq y < - 1 \end{cases}$

Step 6.3.10

Simplify $- (y + 1) > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3}$ .

Tap for more steps...

Step 6.3.10.1

Apply the distributive property.

${\begin{cases} y + 1 > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3} & - 1 \leq y \leq 5 \\ - y - 1 \cdot 1 > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3} & - 13 \leq y < - 1 \end{cases}$

Step 6.3.10.2

Multiply $- 1$ by $1$ .

${\begin{cases} y + 1 > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3} & - 1 \leq y \leq 5 \\ - y - 1 > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3} & - 13 \leq y < - 1 \end{cases}$

Step 6.4

Solve $y + 1 > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3}$ when $- 1 \leq y \leq 5$ .

Tap for more steps...

Step 6.4.1

Subtract $1$ from both sides of the inequality.

$y > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3} - 1$

Step 6.4.2

Find the intersection of $y > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3} - 1$ and $- 1 \leq y \leq 5$ .

No solution

Step 6.5

Solve $- y - 1 > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3}$ when $- 13 \leq y < - 1$ .

Tap for more steps...

Step 6.5.1

Solve $- y - 1 > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3}$ for $y$ .

Tap for more steps...

Step 6.5.1.1

Add $1$ to both sides of the inequality.

$- y > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3} + 1$

Step 6.5.1.2

Divide each term in $- y > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3} + 1$ by $- 1$ and simplify.

Tap for more steps...

Step 6.5.1.2.1

Divide each term in $- y > \frac{2 \sqrt{- (x + 13) (x - 5)}}{3} + 1$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} < \frac{\frac{2 \sqrt{- (x + 13) (x - 5)}}{3}}{- 1} + \frac{1}{- 1}$

Step 6.5.1.2.2

Simplify the left side.

Tap for more steps...

Step 6.5.1.2.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} < \frac{\frac{2 \sqrt{- (x + 13) (x - 5)}}{3}}{- 1} + \frac{1}{- 1}$

Step 6.5.1.2.2.2

Divide $y$ by $1$ .

$y < \frac{\frac{2 \sqrt{- (x + 13) (x - 5)}}{3}}{- 1} + \frac{1}{- 1}$

Step 6.5.1.2.3

Simplify the right side.

Tap for more steps...

Step 6.5.1.2.3.1

Combine the numerators over the common denominator.

$y < \frac{\frac{2 \sqrt{- (x + 13) (x - 5)}}{3} + 1}{- 1}$

Step 6.5.1.2.3.2

Write $1$ as a fraction with a common denominator.

$y < \frac{\frac{2 \sqrt{- (x + 13) (x - 5)}}{3} + \frac{3}{3}}{- 1}$

Step 6.5.1.2.3.3

Combine the numerators over the common denominator.

$y < \frac{\frac{2 \sqrt{- (x + 13) (x - 5)} + 3}{3}}{- 1}$

Step 6.5.1.2.3.4

Simplify the expression.

Tap for more steps...

Step 6.5.1.2.3.4.1

Move the negative one from the denominator of $\frac{\frac{2 \sqrt{- (x + 13) (x - 5)} + 3}{3}}{- 1}$ .

$y < - 1 \cdot \frac{2 \sqrt{- (x + 13) (x - 5)} + 3}{3}$

Step 6.5.1.2.3.4.2

Rewrite $- 1 \cdot \frac{2 \sqrt{- (x + 13) (x - 5)} + 3}{3}$ as $- \frac{2 \sqrt{- (x + 13) (x - 5)} + 3}{3}$ .

$y < - \frac{2 \sqrt{- (x + 13) (x - 5)} + 3}{3}$

Step 6.5.2

Find the intersection of $y < - \frac{2 \sqrt{- (x + 13) (x - 5)} + 3}{3}$ and $- 13 \leq y < - 1$ .

$- 13 \leq y < N o (Min i m u m)$

Step 6.6

Find the union of the solutions.

$- 13 \leq y < i N o Min m^{2} u$

Step 7