Precalculus Examples

$\frac{{(x + 4)}^{2}}{9} - \frac{{(y - 1)}^{2}}{16} = 1$

Step 1

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

$- \frac{{(y - 1)}^{2}}{16} = 1 - \frac{{(x + 4)}^{2}}{9}$

Step 2

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

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Step 2.1

Combine into one fraction.

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Step 2.1.1

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

$- \frac{{(y - 1)}^{2}}{16} = \frac{9}{9} - \frac{{(x + 4)}^{2}}{9}$

Step 2.1.2

Combine the numerators over the common denominator.

$- \frac{{(y - 1)}^{2}}{16} = \frac{9 - {(x + 4)}^{2}}{9}$

Step 2.2

Simplify the numerator.

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Step 2.2.1

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

$- \frac{{(y - 1)}^{2}}{16} = \frac{3^{2} - {(x + 4)}^{2}}{9}$

Step 2.2.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3$ and $b = x + 4$ .

$- \frac{{(y - 1)}^{2}}{16} = \frac{(3 + x + 4) (3 - (x + 4))}{9}$

Step 2.2.3

Simplify.

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Step 2.2.3.1

Add $3$ and $4$ .

$- \frac{{(y - 1)}^{2}}{16} = \frac{(x + 7) (3 - (x + 4))}{9}$

Step 2.2.3.2

Apply the distributive property.

$- \frac{{(y - 1)}^{2}}{16} = \frac{(x + 7) (3 - x - 1 \cdot 4)}{9}$

Step 2.2.3.3

Multiply $- 1$ by $4$ .

$- \frac{{(y - 1)}^{2}}{16} = \frac{(x + 7) (3 - x - 4)}{9}$

Step 2.2.3.4

Subtract $4$ from $3$ .

$- \frac{{(y - 1)}^{2}}{16} = \frac{(x + 7) (- x - 1)}{9}$

Step 2.3

Simplify with factoring out.

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Step 2.3.1

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

$- \frac{{(y - 1)}^{2}}{16} = \frac{(x + 7) (- (x) - 1)}{9}$

Step 2.3.2

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

$- \frac{{(y - 1)}^{2}}{16} = \frac{(x + 7) (- (x) - 1 (1))}{9}$

Step 2.3.3

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

$- \frac{{(y - 1)}^{2}}{16} = \frac{(x + 7) (- (x + 1))}{9}$

Step 2.3.4

Simplify the expression.

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Step 2.3.4.1

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

$- \frac{{(y - 1)}^{2}}{16} = \frac{(x + 7) (- 1 (x + 1))}{9}$

Step 2.3.4.2

Move the negative in front of the fraction.

$- \frac{{(y - 1)}^{2}}{16} = - \frac{(x + 7) (x + 1)}{9}$

Step 3

Multiply both sides of the equation by $- 16$ .

$- 16 (- \frac{{(y - 1)}^{2}}{16}) = - 16 (- \frac{(x + 7) (x + 1)}{9})$

Step 4

Simplify both sides of the equation.

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Step 4.1

Simplify the left side.

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Step 4.1.1

Simplify $- 16 (- \frac{{(y - 1)}^{2}}{16})$ .

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Step 4.1.1.1

Cancel the common factor of $16$ .

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Step 4.1.1.1.1

Move the leading negative in $- \frac{{(y - 1)}^{2}}{16}$ into the numerator.

$- 16 \frac{- {(y - 1)}^{2}}{16} = - 16 (- \frac{(x + 7) (x + 1)}{9})$

Step 4.1.1.1.2

Factor $16$ out of $- 16$ .

$16 (- 1) \frac{- {(y - 1)}^{2}}{16} = - 16 (- \frac{(x + 7) (x + 1)}{9})$

Step 4.1.1.1.3

Cancel the common factor.

$16 \cdot - 1 \frac{- {(y - 1)}^{2}}{16} = - 16 (- \frac{(x + 7) (x + 1)}{9})$

Step 4.1.1.1.4

Rewrite the expression.

$- - {(y - 1)}^{2} = - 16 (- \frac{(x + 7) (x + 1)}{9})$

Step 4.1.1.2

Multiply.

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Step 4.1.1.2.1

Multiply $- 1$ by $- 1$ .

$1 {(y - 1)}^{2} = - 16 (- \frac{(x + 7) (x + 1)}{9})$

Step 4.1.1.2.2

Multiply ${(y - 1)}^{2}$ by $1$ .

${(y - 1)}^{2} = - 16 (- \frac{(x + 7) (x + 1)}{9})$

Step 4.2

Simplify the right side.

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Step 4.2.1

Multiply $- 16 (- \frac{(x + 7) (x + 1)}{9})$ .

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Step 4.2.1.1

Multiply $- 1$ by $- 16$ .

${(y - 1)}^{2} = 16 \frac{(x + 7) (x + 1)}{9}$

Step 4.2.1.2

Combine $16$ and $\frac{(x + 7) (x + 1)}{9}$ .

${(y - 1)}^{2} = \frac{16 ((x + 7) (x + 1))}{9}$

${(y - 1)}^{2} = \frac{16 (x + 7) (x + 1)}{9}$

Step 5

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

$y - 1 = \pm \sqrt{\frac{16 (x + 7) (x + 1)}{9}}$

Step 6

Simplify $\pm \sqrt{\frac{16 (x + 7) (x + 1)}{9}}$ .

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Step 6.1

Rewrite $\frac{16 (x + 7) (x + 1)}{9}$ as ${(\frac{2^{2}}{3})}^{2} ((x + 7) (x + 1))$ .

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Step 6.1.1

Factor the perfect power ${(2^{2})}^{2}$ out of $16 (x + 7) (x + 1)$ .

$y - 1 = \pm \sqrt{\frac{{(2^{2})}^{2} ((x + 7) (x + 1))}{9}}$

Step 6.1.2

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

$y - 1 = \pm \sqrt{\frac{{(2^{2})}^{2} ((x + 7) (x + 1))}{3^{2} \cdot 1}}$

Step 6.1.3

Rearrange the fraction $\frac{{(2^{2})}^{2} ((x + 7) (x + 1))}{3^{2} \cdot 1}$ .

$y - 1 = \pm \sqrt{{(\frac{2^{2}}{3})}^{2} ((x + 7) (x + 1))}$

Step 6.2

Pull terms out from under the radical.

$y - 1 = \pm \frac{2^{2}}{3} \sqrt{(x + 7) (x + 1)}$

Step 6.3

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

$y - 1 = \pm \frac{4}{3} \sqrt{(x + 7) (x + 1)}$

Step 6.4

Combine $\frac{4}{3}$ and $\sqrt{(x + 7) (x + 1)}$ .

$y - 1 = \pm \frac{4 \sqrt{(x + 7) (x + 1)}}{3}$

Step 7

The complete solution is the result of both the positive and negative portions of the solution.

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Step 7.1

First, use the positive value of the $\pm$ to find the first solution.

$y - 1 = \frac{4 \sqrt{(x + 7) (x + 1)}}{3}$

Step 7.2

Add $1$ to both sides of the equation.

$y = \frac{4 \sqrt{(x + 7) (x + 1)}}{3} + 1$

Step 7.3

Next, use the negative value of the $\pm$ to find the second solution.

$y - 1 = - \frac{4 \sqrt{(x + 7) (x + 1)}}{3}$

Step 7.4

Add $1$ to both sides of the equation.

$y = - \frac{4 \sqrt{(x + 7) (x + 1)}}{3} + 1$

Step 7.5

The complete solution is the result of both the positive and negative portions of the solution.

$y = \frac{4 \sqrt{(x + 7) (x + 1)}}{3} + 1$

$y = - \frac{4 \sqrt{(x + 7) (x + 1)}}{3} + 1$

$y = \frac{4 \sqrt{(x + 7) (x + 1)}}{3} + 1$

$y = - \frac{4 \sqrt{(x + 7) (x + 1)}}{3} + 1$

Step 8

Set the radicand in $\sqrt{(x + 7) (x + 1)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 7) (x + 1) \geq 0$

Step 9

Solve for $x$ .

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Step 9.1

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

$x + 7 = 0$

$x + 1 = 0$

Step 9.2

Set $x + 7$ equal to $0$ and solve for $x$ .

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Step 9.2.1

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Step 9.2.2

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 9.3

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

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Step 9.3.1

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

$x + 1 = 0$

Step 9.3.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 9.4

The final solution is all the values that make $(x + 7) (x + 1) \geq 0$ true.

$x = - 7, - 1$

Step 9.5

Use each root to create test intervals.

$x < - 7$

$- 7 < x < - 1$

$x > - 1$

Step 9.6

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 9.6.1

Test a value on the interval $x < - 7$ to see if it makes the inequality true.

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Step 9.6.1.1

Choose a value on the interval $x < - 7$ and see if this value makes the original inequality true.

$x = - 10$

Step 9.6.1.2

Replace $x$ with $- 10$ in the original inequality.

$((- 10) + 7) ((- 10) + 1) \geq 0$

Step 9.6.1.3

The left side $27$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 9.6.2

Test a value on the interval $- 7 < x < - 1$ to see if it makes the inequality true.

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Step 9.6.2.1

Choose a value on the interval $- 7 < x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

Step 9.6.2.2

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

$((- 4) + 7) ((- 4) + 1) \geq 0$

Step 9.6.2.3

The left side $- 9$ is less than the right side $0$ , which means that the given statement is false.

False

Step 9.6.3

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

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Step 9.6.3.1

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

$x = 0$

Step 9.6.3.2

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

$((0) + 7) ((0) + 1) \geq 0$

Step 9.6.3.3

The left side $7$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 9.6.4

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

$x < - 7$ True

$- 7 < x < - 1$ False

$x > - 1$ True

$x < - 7$ True

$- 7 < x < - 1$ False

$x > - 1$ True

Step 9.7

The solution consists of all of the true intervals.

$x \leq - 7$ or $x \geq - 1$

Step 10

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

Interval Notation:

$(- \infty, - 7] \cup [- 1, \infty)$

Set-Builder Notation:

${x | x \leq - 7, x \geq - 1}$

Step 11

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${y | y \in ℝ}$

Step 12

Determine the domain and range.

Domain: $(- \infty, - 7] \cup [- 1, \infty), {x | x \leq - 7, x \geq - 1}$

Range: $(- \infty, \infty), {y | y \in ℝ}$

Step 13