Linear Algebra Examples

$x^{2} + y^{2} = 68$

Step 1

Subtract $x^{2}$ from both sides of the equation.

$y^{2} = 68 - x^{2}$

Step 2

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

$y = \pm \sqrt{68 - x^{2}}$

Step 3

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

Tap for more steps...

Step 3.1

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

$y = \sqrt{68 - x^{2}}$

Step 3.2

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

$y = - \sqrt{68 - x^{2}}$

Step 3.3

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

$y = \sqrt{68 - x^{2}}$

$y = - \sqrt{68 - x^{2}}$

$y = \sqrt{68 - x^{2}}$

$y = - \sqrt{68 - x^{2}}$

Step 4

Set the radicand in $\sqrt{68 - x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$68 - x^{2} \geq 0$

Step 5

Solve for $x$ .

Tap for more steps...

Step 5.1

Subtract $68$ from both sides of the inequality.

$- x^{2} \geq - 68$

Step 5.2

Divide each term in $- x^{2} \geq - 68$ by $- 1$ and simplify.

Tap for more steps...

Step 5.2.1

Divide each term in $- x^{2} \geq - 68$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 68}{- 1}$

Step 5.2.2

Simplify the left side.

Tap for more steps...

Step 5.2.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 68}{- 1}$

Step 5.2.2.2

Divide $x^{2}$ by $1$ .

$x^{2} \leq \frac{- 68}{- 1}$

Step 5.2.3

Simplify the right side.

Tap for more steps...

Step 5.2.3.1

Divide $- 68$ by $- 1$ .

$x^{2} \leq 68$

Step 5.3

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

$\sqrt{x^{2}} \leq \sqrt{68}$

Step 5.4

Simplify the equation.

Tap for more steps...

Step 5.4.1

Simplify the left side.

Tap for more steps...

Step 5.4.1.1

Pull terms out from under the radical.

$| x | \leq \sqrt{68}$

Step 5.4.2

Simplify the right side.

Tap for more steps...

Step 5.4.2.1

Simplify $\sqrt{68}$ .

Tap for more steps...

Step 5.4.2.1.1

Rewrite $68$ as $2^{2} \cdot 17$ .

Tap for more steps...

Step 5.4.2.1.1.1

Factor $4$ out of $68$ .

$| x | \leq \sqrt{4 (17)}$

Step 5.4.2.1.1.2

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

$| x | \leq \sqrt{2^{2} \cdot 17}$

Step 5.4.2.1.2

Pull terms out from under the radical.

$| x | \leq | 2 | \sqrt{17}$

Step 5.4.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$| x | \leq 2 \sqrt{17}$

Step 5.5

Write $| x | \leq 2 \sqrt{17}$ as a piecewise.

Tap for more steps...

Step 5.5.1

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

$x \geq 0$

Step 5.5.2

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

$x \leq 2 \sqrt{17}$

Step 5.5.3

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

$x < 0$

Step 5.5.4

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

$- x \leq 2 \sqrt{17}$

Step 5.5.5

Write as a piecewise.

${\begin{cases} x \leq 2 \sqrt{17} & x \geq 0 \\ - x \leq 2 \sqrt{17} & x < 0 \end{cases}$

Step 5.6

Find the intersection of $x \leq 2 \sqrt{17}$ and $x \geq 0$ .

$0 \leq x \leq 2 \sqrt{17}$

Step 5.7

Solve $- x \leq 2 \sqrt{17}$ when $x < 0$ .

Tap for more steps...

Step 5.7.1

Divide each term in $- x \leq 2 \sqrt{17}$ by $- 1$ and simplify.

Tap for more steps...

Step 5.7.1.1

Divide each term in $- x \leq 2 \sqrt{17}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{2 \sqrt{17}}{- 1}$

Step 5.7.1.2

Simplify the left side.

Tap for more steps...

Step 5.7.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{2 \sqrt{17}}{- 1}$

Step 5.7.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{2 \sqrt{17}}{- 1}$

Step 5.7.1.3

Simplify the right side.

Tap for more steps...

Step 5.7.1.3.1

Move the negative one from the denominator of $\frac{2 \sqrt{17}}{- 1}$ .

$x \geq - 1 \cdot (2 \sqrt{17})$

Step 5.7.1.3.2

Rewrite $- 1 \cdot (2 \sqrt{17})$ as $- (2 \sqrt{17})$ .

$x \geq - (2 \sqrt{17})$

Step 5.7.1.3.3

Multiply $2$ by $- 1$ .

$x \geq - 2 \sqrt{17}$

Step 5.7.2

Find the intersection of $x \geq - 2 \sqrt{17}$ and $x < 0$ .

$- 2 \sqrt{17} \leq x < 0$

Step 5.8

Find the union of the solutions.

$- 2 \sqrt{17} \leq x \leq 2 \sqrt{17}$

Step 6

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

Interval Notation:

$[- 2 \sqrt{17}, 2 \sqrt{17}]$

Set-Builder Notation:

${x | - 2 \sqrt{17} \leq x \leq 2 \sqrt{17}}$

Step 7