Solve by Addition/Elimination 2x^2-3y^2=-19 , 3x^2+2y^2=30

$2 x^{2} - 3 y^{2} = - 19$ , $3 x^{2} + 2 y^{2} = 30$

Step 1

Simplify the left side.

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

$- 3 y^{2} + 2 x^{2} = - 19$

$3 x^{2} + 2 y^{2} = 30$

$- 3 y^{2} + 2 x^{2} = - 19$

$3 x^{2} + 2 y^{2} = 30$

Step 2

Simplify the left side.

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

$2 y^{2} + 3 x^{2} = 30$

$- 3 y^{2} + 2 x^{2} = - 19$

$2 y^{2} + 3 x^{2} = 30$

$- 3 y^{2} + 2 x^{2} = - 19$

Step 3

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

$2 \cdot (- 3 y^{2} + 2 x^{2}) = (2) (- 19)$

$3 \cdot (2 y^{2} + 3 x^{2}) = (3) (30)$

Step 4

Simplify.

Simplify the left side.

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

Apply the distributive property.

$2 (- 3 y^{2}) + 2 (2 x^{2}) = (2) (- 19)$

$3 \cdot (2 y^{2} + 3 x^{2}) = (3) (30)$

Multiply.

Multiply $- 3$ by $2$ .

$- 6 y^{2} + 2 (2 x^{2}) = (2) (- 19)$

$3 \cdot (2 y^{2} + 3 x^{2}) = (3) (30)$

Multiply $2$ by $2$ .

$- 6 y^{2} + 4 x^{2} = (2) (- 19)$

$3 \cdot (2 y^{2} + 3 x^{2}) = (3) (30)$

$- 6 y^{2} + 4 x^{2} = (2) (- 19)$

$3 \cdot (2 y^{2} + 3 x^{2}) = (3) (30)$

$- 6 y^{2} + 4 x^{2} = (2) (- 19)$

$3 \cdot (2 y^{2} + 3 x^{2}) = (3) (30)$

$- 6 y^{2} + 4 x^{2} = (2) (- 19)$

$3 \cdot (2 y^{2} + 3 x^{2}) = (3) (30)$

Simplify the right side.

Multiply $2$ by $- 19$ .

$- 6 y^{2} + 4 x^{2} = - 38$

$3 \cdot (2 y^{2} + 3 x^{2}) = (3) (30)$

$- 6 y^{2} + 4 x^{2} = - 38$

$3 \cdot (2 y^{2} + 3 x^{2}) = (3) (30)$

Simplify the left side.

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

Apply the distributive property.

$- 6 y^{2} + 4 x^{2} = - 38$

$3 (2 y^{2}) + 3 (3 x^{2}) = (3) (30)$

Multiply.

Multiply $2$ by $3$ .

$- 6 y^{2} + 4 x^{2} = - 38$

$6 y^{2} + 3 (3 x^{2}) = (3) (30)$

Multiply $3$ by $3$ .

$- 6 y^{2} + 4 x^{2} = - 38$

$6 y^{2} + 9 x^{2} = (3) (30)$

$- 6 y^{2} + 4 x^{2} = - 38$

$6 y^{2} + 9 x^{2} = (3) (30)$

$- 6 y^{2} + 4 x^{2} = - 38$

$6 y^{2} + 9 x^{2} = (3) (30)$

$- 6 y^{2} + 4 x^{2} = - 38$

$6 y^{2} + 9 x^{2} = (3) (30)$

Simplify the right side.

Multiply $3$ by $30$ .

$- 6 y^{2} + 4 x^{2} = - 38$

$6 y^{2} + 9 x^{2} = 90$

$- 6 y^{2} + 4 x^{2} = - 38$

$6 y^{2} + 9 x^{2} = 90$

$- 6 y^{2} + 4 x^{2} = - 38$

$6 y^{2} + 9 x^{2} = 90$

Step 5

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

	$-$	$6$	$y^{2}$	$+$	$4$	$x^{2}$	$=$	$-$	$3$	$8$
$+$		$6$	$y^{2}$	$+$	$9$	$x^{2}$	$=$		$9$	$0$
				$1$	$3$	$x^{2}$	$=$		$5$	$2$

Step 6

Divide each term in $13 x^{2} = 52$ by $13$ and simplify.

Divide each term in $13 x^{2} = 52$ by $13$ .

$\frac{13 x^{2}}{13} = \frac{52}{13}$

Simplify the left side.

Cancel the common factor of $13$ .

Cancel the common factor.

$\frac{13 x^{2}}{13} = \frac{52}{13}$

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

$x^{2} = \frac{52}{13}$

Simplify the right side.

Divide $52$ by $13$ .

$x^{2} = 4$

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 $y^{2}$ .

$- 6 y^{2} + 4 (4) = - 38$

Multiply $4$ by $4$ .

$- 6 y^{2} + 16 = - 38$

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

Subtract $16$ from both sides of the equation.

$- 6 y^{2} = - 38 - 16$

Subtract $16$ from $- 38$ .

$- 6 y^{2} = - 54$

Divide each term in $- 6 y^{2} = - 54$ by $- 6$ and simplify.

Divide each term in $- 6 y^{2} = - 54$ by $- 6$ .

$\frac{- 6 y^{2}}{- 6} = \frac{- 54}{- 6}$

Simplify the left side.

Cancel the common factor of $- 6$ .

Cancel the common factor.

$\frac{- 6 y^{2}}{- 6} = \frac{- 54}{- 6}$

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

$y^{2} = \frac{- 54}{- 6}$

Simplify the right side.

Divide $- 54$ by $- 6$ .

$y^{2} = 9$

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 $\pm \sqrt{4}$ .

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

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

$y^{2} = 9$

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.

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

$x = 2$

$y^{2} = 9$

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

$x = - 2$

$y^{2} = 9$

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 $\pm \sqrt{9}$ .

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

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

$x = 2, - 2$

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.

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

$y = 3$

$x = 2, - 2$

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

$y = - 3$

$x = 2, - 2$

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

Solve by Addition/Elimination 2x^2-3y^2=-19 , 3x^2+2y^2=30

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