Precalculus Examples

$2 x^{2} + y^{2} = 17$ , $3 x^{2} - 2 y^{2} = - 6$

Step 1

Simplify the left side.

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Step 1.1

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

$y^{2} + 2 x^{2} = 17$

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

$y^{2} + 2 x^{2} = 17$

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

Step 2

Simplify the left side.

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Step 2.1

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

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

$y^{2} + 2 x^{2} = 17$

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

$y^{2} + 2 x^{2} = 17$

Step 3

Multiply each equation by the value that makes the coefficients of $y^{2}$ opposite.

$2 \cdot (y^{2} + 2 x^{2}) = (2) (17)$

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

Step 4

Simplify.

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Step 4.1

Simplify the left side.

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Step 4.1.1

Simplify $2 \cdot (y^{2} + 2 x^{2})$ .

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Step 4.1.1.1

Apply the distributive property.

$2 y^{2} + 2 (2 x^{2}) = (2) (17)$

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

Step 4.1.1.2

Multiply $2$ by $2$ .

$2 y^{2} + 4 x^{2} = (2) (17)$

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

$2 y^{2} + 4 x^{2} = (2) (17)$

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

$2 y^{2} + 4 x^{2} = (2) (17)$

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

Step 4.2

Simplify the right side.

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Step 4.2.1

Multiply $2$ by $17$ .

$2 y^{2} + 4 x^{2} = 34$

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

$2 y^{2} + 4 x^{2} = 34$

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

$2 y^{2} + 4 x^{2} = 34$

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

Step 5

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

		$2$	$y^{2}$	$+$	$4$	$x^{2}$	$=$	$3$	$4$
$+$	$-$	$2$	$y^{2}$	$+$	$3$	$x^{2}$	$=$	$-$	$6$
					$7$	$x^{2}$	$=$	$2$	$8$

Step 6

Divide each term in $7 x^{2} = 28$ by $7$ and simplify.

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Step 6.1

Divide each term in $7 x^{2} = 28$ by $7$ .

$\frac{7 x^{2}}{7} = \frac{28}{7}$

Step 6.2

Simplify the left side.

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Step 6.2.1

Cancel the common factor of $7$ .

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Step 6.2.1.1

Cancel the common factor.

$\frac{7 x^{2}}{7} = \frac{28}{7}$

Step 6.2.1.2

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

$x^{2} = \frac{28}{7}$

Step 6.3

Simplify the right side.

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Step 6.3.1

Divide $28$ by $7$ .

$x^{2} = 4$

Step 7

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

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Step 7.1

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

$2 y^{2} + 4 (4) = 34$

Step 7.2

Multiply $4$ by $4$ .

$2 y^{2} + 16 = 34$

Step 7.3

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

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Step 7.3.1

Subtract $16$ from both sides of the equation.

$2 y^{2} = 34 - 16$

Step 7.3.2

Subtract $16$ from $34$ .

$2 y^{2} = 18$

Step 7.4

Divide each term in $2 y^{2} = 18$ by $2$ and simplify.

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Step 7.4.1

Divide each term in $2 y^{2} = 18$ by $2$ .

$\frac{2 y^{2}}{2} = \frac{18}{2}$

Step 7.4.2

Simplify the left side.

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Step 7.4.2.1

Cancel the common factor of $2$ .

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Step 7.4.2.1.1

Cancel the common factor.

$\frac{2 y^{2}}{2} = \frac{18}{2}$

Step 7.4.2.1.2

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

$y^{2} = \frac{18}{2}$

Step 7.4.3

Simplify the right side.

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Step 7.4.3.1

Divide $18$ by $2$ .

$y^{2} = 9$

Step 8

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

$x^{2} = 4$

$y^{2} = 9$

Step 9

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

$x = \pm \sqrt{4}$

$y^{2} = 9$

Step 10

Simplify $\pm \sqrt{4}$ .

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Step 10.1

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

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

$y^{2} = 9$

Step 10.2

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

$x = \pm 2$

$y^{2} = 9$

$x = \pm 2$

$y^{2} = 9$

Step 11

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

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Step 11.1

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

$x = 2$

$y^{2} = 9$

Step 11.2

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

$x = - 2$

$y^{2} = 9$

Step 11.3

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

$x = 2, - 2$

$y^{2} = 9$

$x = 2, - 2$

$y^{2} = 9$

Step 12

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

$y = \pm \sqrt{9}$

$x = 2, - 2$

Step 13

Simplify $\pm \sqrt{9}$ .

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Step 13.1

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

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

$x = 2, - 2$

Step 13.2

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

$y = \pm 3$

$x = 2, - 2$

$y = \pm 3$

$x = 2, - 2$

Step 14

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

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Step 14.1

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

$y = 3$

$x = 2, - 2$

Step 14.2

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

$y = - 3$

$x = 2, - 2$

Step 14.3

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

$y = 3, - 3$

$x = 2, - 2$

$y = 3, - 3$

$x = 2, - 2$

Step 15

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

$(2, 3), (2, - 3), (- 2, 3), (- 2, - 3)$

Step 16

The result can be shown in multiple forms.

Point Form:

$(2, 3), (2, - 3), (- 2, 3), (- 2, - 3)$

Equation Form:

$x = 2, y = 3$

$x = 2, y = - 3$

$x = - 2, y = 3$

$x = - 2, y = - 3$

Step 17