Solve by Addition/Elimination 2x^2-4y^2=-8 , 5x^2+3y^2=32

$2 x^{2} - 4 y^{2} = - 8$ , $5 x^{2} + 3 y^{2} = 32$

Step 1

Simplify the left side.

Reorder $2 x^{2}$ and $- 4 y^{2}$ .

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

$5 x^{2} + 3 y^{2} = 32$

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

$5 x^{2} + 3 y^{2} = 32$

Step 2

Simplify the left side.

Reorder $5 x^{2}$ and $3 y^{2}$ .

$3 y^{2} + 5 x^{2} = 32$

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

$3 y^{2} + 5 x^{2} = 32$

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

Step 3

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

$- 5 \cdot (- 4 y^{2} + 2 x^{2}) = (- 5) (- 8)$

$2 \cdot (3 y^{2} + 5 x^{2}) = (2) (32)$

Step 4

Simplify.

Simplify the left side.

Simplify $- 5 \cdot (- 4 y^{2} + 2 x^{2})$ .

Apply the distributive property.

$- 5 (- 4 y^{2}) - 5 (2 x^{2}) = (- 5) (- 8)$

$2 \cdot (3 y^{2} + 5 x^{2}) = (2) (32)$

Multiply.

Multiply $- 4$ by $- 5$ .

$20 y^{2} - 5 (2 x^{2}) = (- 5) (- 8)$

$2 \cdot (3 y^{2} + 5 x^{2}) = (2) (32)$

Multiply $2$ by $- 5$ .

$20 y^{2} - 10 x^{2} = (- 5) (- 8)$

$2 \cdot (3 y^{2} + 5 x^{2}) = (2) (32)$

$20 y^{2} - 10 x^{2} = (- 5) (- 8)$

$2 \cdot (3 y^{2} + 5 x^{2}) = (2) (32)$

$20 y^{2} - 10 x^{2} = (- 5) (- 8)$

$2 \cdot (3 y^{2} + 5 x^{2}) = (2) (32)$

$20 y^{2} - 10 x^{2} = (- 5) (- 8)$

$2 \cdot (3 y^{2} + 5 x^{2}) = (2) (32)$

Simplify the right side.

Multiply $- 5$ by $- 8$ .

$20 y^{2} - 10 x^{2} = 40$

$2 \cdot (3 y^{2} + 5 x^{2}) = (2) (32)$

$20 y^{2} - 10 x^{2} = 40$

$2 \cdot (3 y^{2} + 5 x^{2}) = (2) (32)$

Simplify the left side.

Simplify $2 \cdot (3 y^{2} + 5 x^{2})$ .

Apply the distributive property.

$20 y^{2} - 10 x^{2} = 40$

$2 (3 y^{2}) + 2 (5 x^{2}) = (2) (32)$

Multiply.

Multiply $3$ by $2$ .

$20 y^{2} - 10 x^{2} = 40$

$6 y^{2} + 2 (5 x^{2}) = (2) (32)$

Multiply $5$ by $2$ .

$20 y^{2} - 10 x^{2} = 40$

$6 y^{2} + 10 x^{2} = (2) (32)$

$20 y^{2} - 10 x^{2} = 40$

$6 y^{2} + 10 x^{2} = (2) (32)$

$20 y^{2} - 10 x^{2} = 40$

$6 y^{2} + 10 x^{2} = (2) (32)$

$20 y^{2} - 10 x^{2} = 40$

$6 y^{2} + 10 x^{2} = (2) (32)$

Simplify the right side.

Multiply $2$ by $32$ .

$20 y^{2} - 10 x^{2} = 40$

$6 y^{2} + 10 x^{2} = 64$

$20 y^{2} - 10 x^{2} = 40$

$6 y^{2} + 10 x^{2} = 64$

$20 y^{2} - 10 x^{2} = 40$

$6 y^{2} + 10 x^{2} = 64$

Step 5

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

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

Step 6

Divide each term in $26 y^{2} = 104$ by $26$ and simplify.

Divide each term in $26 y^{2} = 104$ by $26$ .

$\frac{26 y^{2}}{26} = \frac{104}{26}$

Simplify the left side.

Cancel the common factor of $26$ .

Cancel the common factor.

$\frac{26 y^{2}}{26} = \frac{104}{26}$

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

$y^{2} = \frac{104}{26}$

Simplify the right side.

Divide $104$ by $26$ .

$y^{2} = 4$

Step 7

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

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

$20 (4) - 10 x^{2} = 40$

Multiply $20$ by $4$ .

$80 - 10 x^{2} = 40$

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

Subtract $80$ from both sides of the equation.

$- 10 x^{2} = 40 - 80$

Subtract $80$ from $40$ .

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

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

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

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

Simplify the left side.

Cancel the common factor of $- 10$ .

Cancel the common factor.

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

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

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

Simplify the right side.

Divide $- 40$ by $- 10$ .

$x^{2} = 4$

Step 8

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

$y^{2} = 4$

$x^{2} = 4$

Step 9

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

$y = \pm \sqrt{4}$

$x^{2} = 4$

Step 10

Simplify $\pm \sqrt{4}$ .

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

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

$x^{2} = 4$

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

$y = \pm 2$

$x^{2} = 4$

$y = \pm 2$

$x^{2} = 4$

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.

$y = 2$

$x^{2} = 4$

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

$y = - 2$

$x^{2} = 4$

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

$y = 2, - 2$

$x^{2} = 4$

$y = 2, - 2$

$x^{2} = 4$

Step 12

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

$x = \pm \sqrt{4}$

$y = 2, - 2$

Step 13

Simplify $\pm \sqrt{4}$ .

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

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

$y = 2, - 2$

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

$x = \pm 2$

$y = 2, - 2$

$x = \pm 2$

$y = 2, - 2$

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.

$x = 2$

$y = 2, - 2$

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

$x = - 2$

$y = 2, - 2$

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

$x = 2, - 2$

$y = 2, - 2$

$x = 2, - 2$

$y = 2, - 2$

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^2-4y^2=-8 , 5x^2+3y^2=32

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