Precalculus Examples

$3 x^{2} + 3 y^{2} = 3$ , $x^{2} - 3 y^{2} = 33$

Step 1

Simplify the left side.

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Step 1.1

Reorder $3 x^{2}$ and $3 y^{2}$ .

$3 y^{2} + 3 x^{2} = 3$

$x^{2} - 3 y^{2} = 33$

$3 y^{2} + 3 x^{2} = 3$

$x^{2} - 3 y^{2} = 33$

Step 2

Simplify the left side.

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Step 2.1

Reorder $x^{2}$ and $- 3 y^{2}$ .

$- 3 y^{2} + x^{2} = 33$

$3 y^{2} + 3 x^{2} = 3$

$- 3 y^{2} + x^{2} = 33$

$3 y^{2} + 3 x^{2} = 3$

Step 3

Add the two equations together to eliminate $y^{2}$ from the system.

		$3$	$y^{2}$	$+$	$3$	$x^{2}$	$=$		$3$
$+$	$-$	$3$	$y^{2}$	$+$		$x^{2}$	$=$	$3$	$3$
					$4$	$x^{2}$	$=$	$3$	$6$

Step 4

Divide each term in $4 x^{2} = 36$ by $4$ and simplify.

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Step 4.1

Divide each term in $4 x^{2} = 36$ by $4$ .

$\frac{4 x^{2}}{4} = \frac{36}{4}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $4$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{4 x^{2}}{4} = \frac{36}{4}$

Step 4.2.1.2

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

$x^{2} = \frac{36}{4}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $36$ by $4$ .

$x^{2} = 9$

Step 5

Substitute the value found for $x^{2}$ into one of the original equations, then solve for $y^{2}$ .

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Step 5.1

Substitute the value found for $x^{2}$ into one of the original equations to solve for $y^{2}$ .

$3 y^{2} + 3 (9) = 3$

Step 5.2

Multiply $3$ by $9$ .

$3 y^{2} + 27 = 3$

Step 5.3

Move all terms not containing $y^{2}$ to the right side of the equation.

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Step 5.3.1

Subtract $27$ from both sides of the equation.

$3 y^{2} = 3 - 27$

Step 5.3.2

Subtract $27$ from $3$ .

$3 y^{2} = - 24$

Step 5.4

Divide each term in $3 y^{2} = - 24$ by $3$ and simplify.

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Step 5.4.1

Divide each term in $3 y^{2} = - 24$ by $3$ .

$\frac{3 y^{2}}{3} = \frac{- 24}{3}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $3$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{3 y^{2}}{3} = \frac{- 24}{3}$

Step 5.4.2.1.2

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

$y^{2} = \frac{- 24}{3}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Divide $- 24$ by $3$ .

$y^{2} = - 8$

Step 6

This is the final solution to the independent system of equations.

$x^{2} = 9$

$y^{2} = - 8$

Step 7

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

$x = \pm \sqrt{9}$

$y^{2} = - 8$

Step 8

Simplify $\pm \sqrt{9}$ .

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Step 8.1

Rewrite $9$ as $3^{2}$ .

$x = \pm \sqrt{3^{2}}$

$y^{2} = - 8$

Step 8.2

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

$x = \pm 3$

$y^{2} = - 8$

$x = \pm 3$

$y^{2} = - 8$

Step 9

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

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Step 9.1

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

$x = 3$

$y^{2} = - 8$

Step 9.2

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

$x = - 3$

$y^{2} = - 8$

Step 9.3

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

$x = 3, - 3$

$y^{2} = - 8$

$x = 3, - 3$

$y^{2} = - 8$

Step 10

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

$y = \pm \sqrt{- 8}$

$x = 3, - 3$

Step 11

Simplify $\pm \sqrt{- 8}$ .

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Step 11.1

Rewrite $- 8$ as $- 1 (8)$ .

$y = \pm \sqrt{- 1 \cdot 8}$

$x = 3, - 3$

Step 11.2

Rewrite $\sqrt{- 1 (8)}$ as $\sqrt{- 1} \cdot \sqrt{8}$ .

$y = \pm \sqrt{- 1} \cdot \sqrt{8}$

$x = 3, - 3$

Step 11.3

Rewrite $\sqrt{- 1}$ as $i$ .

$y = \pm i \cdot \sqrt{8}$

$x = 3, - 3$

Step 11.4

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 11.4.1

Factor $4$ out of $8$ .

$y = \pm i \cdot \sqrt{4 (2)}$

$x = 3, - 3$

Step 11.4.2

Rewrite $4$ as $2^{2}$ .

$y = \pm i \cdot \sqrt{2^{2} \cdot 2}$

$x = 3, - 3$

$y = \pm i \cdot \sqrt{2^{2} \cdot 2}$

$x = 3, - 3$

Step 11.5

Pull terms out from under the radical.

$y = \pm i \cdot (2 \sqrt{2})$

$x = 3, - 3$

Step 11.6

Move $2$ to the left of $i$ .

$y = \pm 2 i \sqrt{2}$

$x = 3, - 3$

$y = \pm 2 i \sqrt{2}$

$x = 3, - 3$

Step 12

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

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Step 12.1

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

$y = 2 i \sqrt{2}$

$x = 3, - 3$

Step 12.2

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

$y = - 2 i \sqrt{2}$

$x = 3, - 3$

Step 12.3

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

$y = 2 i \sqrt{2}, - 2 i \sqrt{2}$

$x = 3, - 3$

$y = 2 i \sqrt{2}, - 2 i \sqrt{2}$

$x = 3, - 3$

Step 13

The final result is the combination of all values of $x$ with all values of $y$ .

$(3, 2 i \sqrt{2}), (3, - 2 i \sqrt{2}), (- 3, 2 i \sqrt{2}), (- 3, - 2 i \sqrt{2})$

Step 14