Precalculus Examples

${(x - 7)}^{2} + {(y + 3)}^{2} = \frac{49}{4}$

Step 1

Subtract ${(y + 3)}^{2}$ from both sides of the equation.

${(x - 7)}^{2} = \frac{49}{4} - {(y + 3)}^{2}$

Step 2

To write $- {(y + 3)}^{2}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

${(x - 7)}^{2} = \frac{49}{4} - {(y + 3)}^{2} \cdot \frac{4}{4}$

Step 3

Combine $- {(y + 3)}^{2}$ and $\frac{4}{4}$ .

${(x - 7)}^{2} = \frac{49}{4} + \frac{- {(y + 3)}^{2} \cdot 4}{4}$

Step 4

Combine the numerators over the common denominator.

${(x - 7)}^{2} = \frac{49 - {(y + 3)}^{2} \cdot 4}{4}$

Step 5

Multiply $4$ by $- 1$ .

${(x - 7)}^{2} = \frac{49 - 4 {(y + 3)}^{2}}{4}$

Step 6

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

$x - 7 = \pm \sqrt{\frac{49 - 4 {(y + 3)}^{2}}{4}}$

Step 7

Simplify $\pm \sqrt{\frac{49 - 4 {(y + 3)}^{2}}{4}}$ .

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Step 7.1

Write the expression using exponents.

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Step 7.1.1

Rewrite $49$ as $7^{2}$ .

$x - 7 = \pm \sqrt{\frac{7^{2} - 4 {(y + 3)}^{2}}{4}}$

Step 7.1.2

Rewrite $4 {(y + 3)}^{2}$ as ${(2 (y + 3))}^{2}$ .

$x - 7 = \pm \sqrt{\frac{7^{2} - {(2 (y + 3))}^{2}}{4}}$

Step 7.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 7$ and $b = 2 (y + 3)$ .

$x - 7 = \pm \sqrt{\frac{(7 + 2 (y + 3)) (7 - (2 (y + 3)))}{4}}$

Step 7.3

Simplify.

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Step 7.3.1

Apply the distributive property.

$x - 7 = \pm \sqrt{\frac{(7 + 2 y + 2 \cdot 3) (7 - (2 (y + 3)))}{4}}$

Step 7.3.2

Multiply $2$ by $3$ .

$x - 7 = \pm \sqrt{\frac{(7 + 2 y + 6) (7 - (2 (y + 3)))}{4}}$

Step 7.3.3

Add $7$ and $6$ .

$x - 7 = \pm \sqrt{\frac{(2 y + 13) (7 - (2 (y + 3)))}{4}}$

Step 7.3.4

Apply the distributive property.

$x - 7 = \pm \sqrt{\frac{(2 y + 13) (7 - (2 y + 2 \cdot 3))}{4}}$

Step 7.3.5

Multiply $2$ by $3$ .

$x - 7 = \pm \sqrt{\frac{(2 y + 13) (7 - (2 y + 6))}{4}}$

Step 7.3.6

Apply the distributive property.

$x - 7 = \pm \sqrt{\frac{(2 y + 13) (7 - (2 y) - 1 \cdot 6)}{4}}$

Step 7.3.7

Multiply $2$ by $- 1$ .

$x - 7 = \pm \sqrt{\frac{(2 y + 13) (7 - 2 y - 1 \cdot 6)}{4}}$

Step 7.3.8

Multiply $- 1$ by $6$ .

$x - 7 = \pm \sqrt{\frac{(2 y + 13) (7 - 2 y - 6)}{4}}$

Step 7.3.9

Subtract $6$ from $7$ .

$x - 7 = \pm \sqrt{\frac{(2 y + 13) (- 2 y + 1)}{4}}$

Step 7.4

Rewrite $\frac{(2 y + 13) (- 2 y + 1)}{4}$ as ${(\frac{1}{2})}^{2} ((2 y + 13) (- 2 y + 1))$ .

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Step 7.4.1

Factor the perfect power $1^{2}$ out of $(2 y + 13) (- 2 y + 1)$ .

$x - 7 = \pm \sqrt{\frac{1^{2} ((2 y + 13) (- 2 y + 1))}{4}}$

Step 7.4.2

Factor the perfect power $2^{2}$ out of $4$ .

$x - 7 = \pm \sqrt{\frac{1^{2} ((2 y + 13) (- 2 y + 1))}{2^{2} \cdot 1}}$

Step 7.4.3

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

$x - 7 = \pm \sqrt{{(\frac{1}{2})}^{2} ((2 y + 13) (- 2 y + 1))}$

Step 7.5

Pull terms out from under the radical.

$x - 7 = \pm \frac{1}{2} \sqrt{(2 y + 13) (- 2 y + 1)}$

Step 7.6

Combine $\frac{1}{2}$ and $\sqrt{(2 y + 13) (- 2 y + 1)}$ .

$x - 7 = \pm \frac{\sqrt{(2 y + 13) (- 2 y + 1)}}{2}$

Step 8

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

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Step 8.1

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

$x - 7 = \frac{\sqrt{(2 y + 13) (- 2 y + 1)}}{2}$

Step 8.2

Add $7$ to both sides of the equation.

$x = \frac{\sqrt{(2 y + 13) (- 2 y + 1)}}{2} + 7$

Step 8.3

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

$x - 7 = - \frac{\sqrt{(2 y + 13) (- 2 y + 1)}}{2}$

Step 8.4

Add $7$ to both sides of the equation.

$x = - \frac{\sqrt{(2 y + 13) (- 2 y + 1)}}{2} + 7$

Step 8.5

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

$x = \frac{\sqrt{(2 y + 13) (- 2 y + 1)}}{2} + 7$

$x = - \frac{\sqrt{(2 y + 13) (- 2 y + 1)}}{2} + 7$

$x = \frac{\sqrt{(2 y + 13) (- 2 y + 1)}}{2} + 7$

$x = - \frac{\sqrt{(2 y + 13) (- 2 y + 1)}}{2} + 7$