Precalculus Examples

$x^{2} + y^{2} = 128$ , $x + y = 0$

Step 1

Subtract $y$ from both sides of the equation.

$x = - y$

$x^{2} + y^{2} = 128$

Step 2

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

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Step 2.1

Replace all occurrences of $x$ in $x^{2} + y^{2} = 128$ with $- y$ .

${(- y)}^{2} + y^{2} = 128$

$x = - y$

Step 2.2

Simplify the left side.

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Step 2.2.1

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

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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 $- y$ .

${(- 1)}^{2} y^{2} + y^{2} = 128$

$x = - y$

Step 2.2.1.1.2

Raise $- 1$ to the power of $2$ .

$1 y^{2} + y^{2} = 128$

$x = - y$

Step 2.2.1.1.3

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

$y^{2} + y^{2} = 128$

$x = - y$

$y^{2} + y^{2} = 128$

$x = - y$

Step 2.2.1.2

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

$2 y^{2} = 128$

$x = - y$

$2 y^{2} = 128$

$x = - y$

$2 y^{2} = 128$

$x = - y$

$2 y^{2} = 128$

$x = - y$

Step 3

Solve for $y$ in $2 y^{2} = 128$ .

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Step 3.1

Divide each term in $2 y^{2} = 128$ by $2$ and simplify.

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Step 3.1.1

Divide each term in $2 y^{2} = 128$ by $2$ .

$\frac{2 y^{2}}{2} = \frac{128}{2}$

$x = - y$

Step 3.1.2

Simplify the left side.

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Step 3.1.2.1

Cancel the common factor of $2$ .

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Step 3.1.2.1.1

Cancel the common factor.

$\frac{2 y^{2}}{2} = \frac{128}{2}$

$x = - y$

Step 3.1.2.1.2

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

$y^{2} = \frac{128}{2}$

$x = - y$

$y^{2} = \frac{128}{2}$

$x = - y$

$y^{2} = \frac{128}{2}$

$x = - y$

Step 3.1.3

Simplify the right side.

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Step 3.1.3.1

Divide $128$ by $2$ .

$y^{2} = 64$

$x = - y$

$y^{2} = 64$

$x = - y$

$y^{2} = 64$

$x = - y$

Step 3.2

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

$y = \pm \sqrt{64}$

$x = - y$

Step 3.3

Simplify $\pm \sqrt{64}$ .

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Step 3.3.1

Rewrite $64$ as $8^{2}$ .

$y = \pm \sqrt{8^{2}}$

$x = - y$

Step 3.3.2

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

$y = \pm 8$

$x = - y$

$y = \pm 8$

$x = - y$

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.

$y = 8$

$x = - y$

Step 3.4.2

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

$y = - 8$

$x = - y$

Step 3.4.3

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

$y = 8, - 8$

$x = - y$

$y = 8, - 8$

$x = - y$

$y = 8, - 8$

$x = - y$

Step 4

Replace all occurrences of $y$ with $8$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = - y$ with $8$ .

$x = - (8)$

$y = 8$

Step 4.2

Simplify the right side.

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Step 4.2.1

Multiply $- 1$ by $8$ .

$x = - 8$

$y = 8$

$x = - 8$

$y = 8$

$x = - 8$

$y = 8$

Step 5

Replace all occurrences of $y$ with $- 8$ in each equation.

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Step 5.1

Replace all occurrences of $y$ in $x = - y$ with $- 8$ .

$x = - (- 8)$

$y = - 8$

Step 5.2

Simplify the right side.

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Step 5.2.1

Multiply $- 1$ by $- 8$ .

$x = 8$

$y = - 8$

$x = 8$

$y = - 8$

$x = 8$

$y = - 8$

Step 6

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

$(- 8, 8)$

$(8, - 8)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(- 8, 8), (8, - 8)$

Equation Form:

$x = - 8, y = 8$

$x = 8, y = - 8$

Step 8