Trigonometry Examples

$x - 2 y = 0$ , $x^{2} + y^{2} = 125$

Step 1

Add $2 y$ to both sides of the equation.

$x = 2 y$

$x^{2} + y^{2} = 125$

Step 2

Replace all occurrences of $x$ with $2 y$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $x^{2} + y^{2} = 125$ with $2 y$ .

${(2 y)}^{2} + y^{2} = 125$

$x = 2 y$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify ${(2 y)}^{2} + y^{2}$ .

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Step 2.2.1.1

Simplify each term.

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Step 2.2.1.1.1

Apply the product rule to $2 y$ .

$2^{2} y^{2} + y^{2} = 125$

$x = 2 y$

Step 2.2.1.1.2

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

$4 y^{2} + y^{2} = 125$

$x = 2 y$

$4 y^{2} + y^{2} = 125$

$x = 2 y$

Step 2.2.1.2

Add $4 y^{2}$ and $y^{2}$ .

$5 y^{2} = 125$

$x = 2 y$

$5 y^{2} = 125$

$x = 2 y$

$5 y^{2} = 125$

$x = 2 y$

$5 y^{2} = 125$

$x = 2 y$

Step 3

Solve for $y$ in $5 y^{2} = 125$ .

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Step 3.1

Divide each term in $5 y^{2} = 125$ by $5$ and simplify.

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Step 3.1.1

Divide each term in $5 y^{2} = 125$ by $5$ .

$\frac{5 y^{2}}{5} = \frac{125}{5}$

$x = 2 y$

Step 3.1.2

Simplify the left side.

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Step 3.1.2.1

Cancel the common factor of $5$ .

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Step 3.1.2.1.1

Cancel the common factor.

$\frac{5 y^{2}}{5} = \frac{125}{5}$

$x = 2 y$

Step 3.1.2.1.2

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

$y^{2} = \frac{125}{5}$

$x = 2 y$

$y^{2} = \frac{125}{5}$

$x = 2 y$

$y^{2} = \frac{125}{5}$

$x = 2 y$

Step 3.1.3

Simplify the right side.

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Step 3.1.3.1

Divide $125$ by $5$ .

$y^{2} = 25$

$x = 2 y$

$y^{2} = 25$

$x = 2 y$

$y^{2} = 25$

$x = 2 y$

Step 3.2

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

$y = \pm \sqrt{25}$

$x = 2 y$

Step 3.3

Simplify $\pm \sqrt{25}$ .

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Step 3.3.1

Rewrite $25$ as $5^{2}$ .

$y = \pm \sqrt{5^{2}}$

$x = 2 y$

Step 3.3.2

Pull terms out from under the radical, assuming positive real numbers.

$y = \pm 5$

$x = 2 y$

$y = \pm 5$

$x = 2 y$

Step 3.4

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

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Step 3.4.1

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

$y = 5$

$x = 2 y$

Step 3.4.2

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

$y = - 5$

$x = 2 y$

Step 3.4.3

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

$y = 5, - 5$

$x = 2 y$

$y = 5, - 5$

$x = 2 y$

$y = 5, - 5$

$x = 2 y$

Step 4

Replace all occurrences of $y$ with $5$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = 2 y$ with $5$ .

$x = 2 (5)$

$y = 5$

Step 4.2

Simplify the right side.

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Step 4.2.1

Multiply $2$ by $5$ .

$x = 10$

$y = 5$

$x = 10$

$y = 5$

$x = 10$

$y = 5$

Step 5

Replace all occurrences of $y$ with $- 5$ in each equation.

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Step 5.1

Replace all occurrences of $y$ in $x = 2 y$ with $- 5$ .

$x = 2 (- 5)$

$y = - 5$

Step 5.2

Simplify the right side.

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Step 5.2.1

Multiply $2$ by $- 5$ .

$x = - 10$

$y = - 5$

$x = - 10$

$y = - 5$

$x = - 10$

$y = - 5$

Step 6

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(10, 5)$

$(- 10, - 5)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(10, 5), (- 10, - 5)$

Equation Form:

$x = 10, y = 5$

$x = - 10, y = - 5$

Step 8