Precalculus Examples

$x^{2} + y^{2} = 25$ , $y = 2 x$

Step 1

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

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Step 1.1

Replace all occurrences of $y$ in $x^{2} + y^{2} = 25$ with $2 x$ .

$x^{2} + {(2 x)}^{2} = 25$

$y = 2 x$

Step 1.2

Simplify the left side.

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Step 1.2.1

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

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Step 1.2.1.1

Simplify each term.

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Step 1.2.1.1.1

Apply the product rule to $2 x$ .

$x^{2} + 2^{2} x^{2} = 25$

$y = 2 x$

Step 1.2.1.1.2

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

$x^{2} + 4 x^{2} = 25$

$y = 2 x$

$x^{2} + 4 x^{2} = 25$

$y = 2 x$

Step 1.2.1.2

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

$5 x^{2} = 25$

$y = 2 x$

$5 x^{2} = 25$

$y = 2 x$

$5 x^{2} = 25$

$y = 2 x$

$5 x^{2} = 25$

$y = 2 x$

Step 2

Solve for $x$ in $5 x^{2} = 25$ .

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Step 2.1

Divide each term in $5 x^{2} = 25$ by $5$ and simplify.

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Step 2.1.1

Divide each term in $5 x^{2} = 25$ by $5$ .

$\frac{5 x^{2}}{5} = \frac{25}{5}$

$y = 2 x$

Step 2.1.2

Simplify the left side.

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Step 2.1.2.1

Cancel the common factor of $5$ .

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Step 2.1.2.1.1

Cancel the common factor.

$\frac{5 x^{2}}{5} = \frac{25}{5}$

$y = 2 x$

Step 2.1.2.1.2

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

$x^{2} = \frac{25}{5}$

$y = 2 x$

$x^{2} = \frac{25}{5}$

$y = 2 x$

$x^{2} = \frac{25}{5}$

$y = 2 x$

Step 2.1.3

Simplify the right side.

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Step 2.1.3.1

Divide $25$ by $5$ .

$x^{2} = 5$

$y = 2 x$

$x^{2} = 5$

$y = 2 x$

$x^{2} = 5$

$y = 2 x$

Step 2.2

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

$x = \pm \sqrt{5}$

$y = 2 x$

Step 2.3

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

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Step 2.3.1

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

$x = \sqrt{5}$

$y = 2 x$

Step 2.3.2

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

$x = - \sqrt{5}$

$y = 2 x$

Step 2.3.3

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

$x = \sqrt{5}, - \sqrt{5}$

$y = 2 x$

$x = \sqrt{5}, - \sqrt{5}$

$y = 2 x$

$x = \sqrt{5}, - \sqrt{5}$

$y = 2 x$

Step 3

Replace all occurrences of $x$ with $\sqrt{5}$ in each equation.

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Step 3.1

Replace all occurrences of $x$ in $y = 2 x$ with $\sqrt{5}$ .

$y = 2 (\sqrt{5})$

$x = \sqrt{5}$

Step 3.2

Simplify the right side.

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Step 3.2.1

Multiply $2$ by $\sqrt{5}$ .

$y = 2 \sqrt{5}$

$x = \sqrt{5}$

$y = 2 \sqrt{5}$

$x = \sqrt{5}$

$y = 2 \sqrt{5}$

$x = \sqrt{5}$

Step 4

Replace all occurrences of $x$ with $- \sqrt{5}$ in each equation.

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Step 4.1

Replace all occurrences of $x$ in $y = 2 x$ with $- \sqrt{5}$ .

$y = 2 (- \sqrt{5})$

$x = - \sqrt{5}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Multiply $- 1$ by $2$ .

$y = - 2 \sqrt{5}$

$x = - \sqrt{5}$

$y = - 2 \sqrt{5}$

$x = - \sqrt{5}$

$y = - 2 \sqrt{5}$

$x = - \sqrt{5}$

Step 5

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

$(\sqrt{5}, 2 \sqrt{5})$

$(- \sqrt{5}, - 2 \sqrt{5})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(\sqrt{5}, 2 \sqrt{5}), (- \sqrt{5}, - 2 \sqrt{5})$

Equation Form:

$x = \sqrt{5}, y = 2 \sqrt{5}$

$x = - \sqrt{5}, y = - 2 \sqrt{5}$

Step 7