Solve for y (5/13)^2+y^2=1

${(\frac{5}{13})}^{2} + y^{2} = 1$

Simplify each term.

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Apply the product rule to $\frac{5}{13}$ .

$\frac{5^{2}}{13^{2}} + y^{2} = 1$

Raise $5$ to the power of $2$ .

$\frac{25}{13^{2}} + y^{2} = 1$

Raise $13$ to the power of $2$ .

$\frac{25}{169} + y^{2} = 1$

Move all terms not containing $y$ to the right side of the equation.

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Subtract $\frac{25}{169}$ from both sides of the equation.

$y^{2} = 1 - \frac{25}{169}$

Simplify the right side of the equation.

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To write $\frac{1}{1}$ as a fraction with a common denominator, multiply by $\frac{169}{169}$ .

$y^{2} = \frac{1}{1} \cdot \frac{169}{169} - \frac{25}{169}$

Write each expression with a common denominator of $169$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$y^{2} = \frac{1 \cdot 169}{1 \cdot 169} - \frac{25}{169}$

Multiply $169$ by $1$ .

$y^{2} = \frac{1 \cdot 169}{169} - \frac{25}{169}$

Combine the numerators over the common denominator.

$y^{2} = \frac{1 \cdot 169 - 25}{169}$

Simplify the numerator.

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Multiply $1$ by $169$ .

$y^{2} = \frac{169 - 25}{169}$

Subtract $25$ from $169$ .

$y^{2} = \frac{144}{169}$

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

$y = \pm \sqrt{\frac{144}{169}}$

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

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Simplify the right side of the equation.

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Rewrite $\sqrt{\frac{144}{169}}$ as $\frac{\sqrt{144}}{\sqrt{169}}$ .

$y = \pm \frac{\sqrt{144}}{\sqrt{169}}$

Simplify the numerator.

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Rewrite $144$ as $12^{2}$ .

$y = \pm \frac{\sqrt{12^{2}}}{\sqrt{169}}$

Pull terms out from under the radical.

$y = \pm \frac{| 12 |}{\sqrt{169}}$

The absolute value is the distance between a number and zero. The distance between $0$ and $12$ is $12$ .

$y = \pm \frac{12}{\sqrt{169}}$

Simplify the denominator.

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Rewrite $169$ as $13^{2}$ .

$y = \pm \frac{12}{\sqrt{13^{2}}}$

Pull terms out from under the radical.

$y = \pm \frac{12}{| 13 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $13$ is $13$ .

$y = \pm \frac{12}{13}$

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

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

$y = \frac{12}{13}$

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

$y = - \frac{12}{13}$

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

$y = \frac{12}{13}, - \frac{12}{13}$

The result can be shown in multiple forms.

Exact Form:

$y = \frac{12}{13}, - \frac{12}{13}$

Decimal Form:

$y = 0.\overline{923076}, - 0.\overline{923076}$

Solve for y (5/13)^2+y^2=1

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