Solve by Addition/Elimination 2x^2-5y^2=-12 5x^2+2y^2=28

$2 x^{2} - 5 y^{2} = - 12$ $5 x^{2} + 2 y^{2} = 28$

Step 1

Simplify the left side.

Reorder $2 x^{2}$ and $- 5 y^{2}$ .

$- 5 y^{2} + 2 x^{2} = - 12$

$5 x^{2} + 2 y^{2} = 28$

$- 5 y^{2} + 2 x^{2} = - 12$

$5 x^{2} + 2 y^{2} = 28$

Step 2

Simplify the left side.

Reorder $5 x^{2}$ and $2 y^{2}$ .

$2 y^{2} + 5 x^{2} = 28$

$- 5 y^{2} + 2 x^{2} = - 12$

$2 y^{2} + 5 x^{2} = 28$

$- 5 y^{2} + 2 x^{2} = - 12$

Step 3

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

$2 \cdot (- 5 y^{2} + 2 x^{2}) = (2) (- 12)$

$5 \cdot (2 y^{2} + 5 x^{2}) = (5) (28)$

Step 4

Simplify.

Simplify the left side.

Simplify $2 \cdot (- 5 y^{2} + 2 x^{2})$ .

Apply the distributive property.

$2 (- 5 y^{2}) + 2 (2 x^{2}) = (2) (- 12)$

$5 \cdot (2 y^{2} + 5 x^{2}) = (5) (28)$

Multiply.

Multiply $- 5$ by $2$ .

$- 10 y^{2} + 2 (2 x^{2}) = (2) (- 12)$

$5 \cdot (2 y^{2} + 5 x^{2}) = (5) (28)$

Multiply $2$ by $2$ .

$- 10 y^{2} + 4 x^{2} = (2) (- 12)$

$5 \cdot (2 y^{2} + 5 x^{2}) = (5) (28)$

$- 10 y^{2} + 4 x^{2} = (2) (- 12)$

$5 \cdot (2 y^{2} + 5 x^{2}) = (5) (28)$

$- 10 y^{2} + 4 x^{2} = (2) (- 12)$

$5 \cdot (2 y^{2} + 5 x^{2}) = (5) (28)$

$- 10 y^{2} + 4 x^{2} = (2) (- 12)$

$5 \cdot (2 y^{2} + 5 x^{2}) = (5) (28)$

Simplify the right side.

Multiply $2$ by $- 12$ .

$- 10 y^{2} + 4 x^{2} = - 24$

$5 \cdot (2 y^{2} + 5 x^{2}) = (5) (28)$

$- 10 y^{2} + 4 x^{2} = - 24$

$5 \cdot (2 y^{2} + 5 x^{2}) = (5) (28)$

Simplify the left side.

Simplify $5 \cdot (2 y^{2} + 5 x^{2})$ .

Apply the distributive property.

$- 10 y^{2} + 4 x^{2} = - 24$

$5 (2 y^{2}) + 5 (5 x^{2}) = (5) (28)$

Multiply.

Multiply $2$ by $5$ .

$- 10 y^{2} + 4 x^{2} = - 24$

$10 y^{2} + 5 (5 x^{2}) = (5) (28)$

Multiply $5$ by $5$ .

$- 10 y^{2} + 4 x^{2} = - 24$

$10 y^{2} + 25 x^{2} = (5) (28)$

$- 10 y^{2} + 4 x^{2} = - 24$

$10 y^{2} + 25 x^{2} = (5) (28)$

$- 10 y^{2} + 4 x^{2} = - 24$

$10 y^{2} + 25 x^{2} = (5) (28)$

$- 10 y^{2} + 4 x^{2} = - 24$

$10 y^{2} + 25 x^{2} = (5) (28)$

Simplify the right side.

Multiply $5$ by $28$ .

$- 10 y^{2} + 4 x^{2} = - 24$

$10 y^{2} + 25 x^{2} = 140$

$- 10 y^{2} + 4 x^{2} = - 24$

$10 y^{2} + 25 x^{2} = 140$

$- 10 y^{2} + 4 x^{2} = - 24$

$10 y^{2} + 25 x^{2} = 140$

Step 5

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

	$-$	$1$	$0$	$y^{2}$	$+$		$4$	$x^{2}$	$=$	$-$	$2$	$4$
$+$		$1$	$0$	$y^{2}$	$+$	$2$	$5$	$x^{2}$	$=$	$1$	$4$	$0$
						$2$	$9$	$x^{2}$	$=$	$1$	$1$	$6$

Step 6

Divide each term in $29 x^{2} = 116$ by $29$ and simplify.

Divide each term in $29 x^{2} = 116$ by $29$ .

$\frac{29 x^{2}}{29} = \frac{116}{29}$

Simplify the left side.

Cancel the common factor of $29$ .

Cancel the common factor.

$\frac{29 x^{2}}{29} = \frac{116}{29}$

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

$x^{2} = \frac{116}{29}$

Simplify the right side.

Divide $116$ by $29$ .

$x^{2} = 4$

Step 7

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

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

$- 10 y^{2} + 4 (4) = - 24$

Multiply $4$ by $4$ .

$- 10 y^{2} + 16 = - 24$

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

Subtract $16$ from both sides of the equation.

$- 10 y^{2} = - 24 - 16$

Subtract $16$ from $- 24$ .

$- 10 y^{2} = - 40$

Divide each term in $- 10 y^{2} = - 40$ by $- 10$ and simplify.

Divide each term in $- 10 y^{2} = - 40$ by $- 10$ .

$\frac{- 10 y^{2}}{- 10} = \frac{- 40}{- 10}$

Simplify the left side.

Cancel the common factor of $- 10$ .

Cancel the common factor.

$\frac{- 10 y^{2}}{- 10} = \frac{- 40}{- 10}$

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

$y^{2} = \frac{- 40}{- 10}$

Simplify the right side.

Divide $- 40$ by $- 10$ .

$y^{2} = 4$

Step 8

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

$x^{2} = 4$

$y^{2} = 4$

Step 9

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

$x = \pm \sqrt{4}$

$y^{2} = 4$

Step 10

Simplify $\pm \sqrt{4}$ .

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

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

$y^{2} = 4$

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

$x = \pm 2$

$y^{2} = 4$

$x = \pm 2$

$y^{2} = 4$

Step 11

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

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

$x = 2$

$y^{2} = 4$

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

$x = - 2$

$y^{2} = 4$

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

$x = 2, - 2$

$y^{2} = 4$

$x = 2, - 2$

$y^{2} = 4$

Step 12

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

$y = \pm \sqrt{4}$

$x = 2, - 2$

Step 13

Simplify $\pm \sqrt{4}$ .

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

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

$x = 2, - 2$

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

$y = \pm 2$

$x = 2, - 2$

$y = \pm 2$

$x = 2, - 2$

Step 14

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

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

$y = 2$

$x = 2, - 2$

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

$y = - 2$

$x = 2, - 2$

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

$y = 2, - 2$

$x = 2, - 2$

$y = 2, - 2$

$x = 2, - 2$

Step 15

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

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

Step 16

The result can be shown in multiple forms.

Point Form:

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

Equation Form:

$x = 2, y = 2$

$x = 2, y = - 2$

$x = - 2, y = 2$

$x = - 2, y = - 2$

Step 17

Solve by Addition/Elimination 2x^2-5y^2=-12 5x^2+2y^2=28

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