Pre-Algebra Examples

$x^{2} + y^{2} = 36$

Step 1

Solve the equation for $y$ .

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Step 1.1

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

$y^{2} = 36 - x^{2}$

Step 1.2

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

$y = \pm \sqrt{36 - x^{2}}$

Step 1.3

Simplify $\pm \sqrt{36 - x^{2}}$ .

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Step 1.3.1

Rewrite $36$ as $6^{2}$ .

$y = \pm \sqrt{6^{2} - x^{2}}$

Step 1.3.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 6$ and $b = x$ .

$y = \pm \sqrt{(6 + x) (6 - x)}$

Step 1.4

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

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Step 1.4.1

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

$y = \sqrt{(6 + x) (6 - x)}$

Step 1.4.2

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

$y = - \sqrt{(6 + x) (6 - x)}$

Step 1.4.3

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

$y = \sqrt{(6 + x) (6 - x)}$

$y = - \sqrt{(6 + x) (6 - x)}$

$y = \sqrt{(6 + x) (6 - x)}$

$y = - \sqrt{(6 + x) (6 - x)}$

$y = \sqrt{(6 + x) (6 - x)}$

$y = - \sqrt{(6 + x) (6 - x)}$

Step 2

A linear equation is an equation of a straight line, which means that the degree of a linear equation must be $0$ or $1$ for each of its variables. In this case, the degree of the variable in the equation violates the linear equation definition, which means that the equation is not a linear equation.

Not Linear