Solve by Addition/Elimination 4x^2-3y^2=-8 , 5x^2+2y^2=13

$4 x^{2} - 3 y^{2} = - 8$ , $5 x^{2} + 2 y^{2} = 13$

Step 1

Simplify the left side.

Reorder $4 x^{2}$ and $- 3 y^{2}$ .

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

$5 x^{2} + 2 y^{2} = 13$

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

$5 x^{2} + 2 y^{2} = 13$

Step 2

Simplify the left side.

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

$2 y^{2} + 5 x^{2} = 13$

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

$2 y^{2} + 5 x^{2} = 13$

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

Step 3

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

$2 \cdot (- 3 y^{2} + 4 x^{2}) = (2) (- 8)$

$3 \cdot (2 y^{2} + 5 x^{2}) = (3) (13)$

Step 4

Simplify.

Simplify the left side.

Simplify $2 \cdot (- 3 y^{2} + 4 x^{2})$ .

Apply the distributive property.

$2 (- 3 y^{2}) + 2 (4 x^{2}) = (2) (- 8)$

$3 \cdot (2 y^{2} + 5 x^{2}) = (3) (13)$

Multiply.

Multiply $- 3$ by $2$ .

$- 6 y^{2} + 2 (4 x^{2}) = (2) (- 8)$

$3 \cdot (2 y^{2} + 5 x^{2}) = (3) (13)$

Multiply $4$ by $2$ .

$- 6 y^{2} + 8 x^{2} = (2) (- 8)$

$3 \cdot (2 y^{2} + 5 x^{2}) = (3) (13)$

$- 6 y^{2} + 8 x^{2} = (2) (- 8)$

$3 \cdot (2 y^{2} + 5 x^{2}) = (3) (13)$

$- 6 y^{2} + 8 x^{2} = (2) (- 8)$

$3 \cdot (2 y^{2} + 5 x^{2}) = (3) (13)$

$- 6 y^{2} + 8 x^{2} = (2) (- 8)$

$3 \cdot (2 y^{2} + 5 x^{2}) = (3) (13)$

Simplify the right side.

Multiply $2$ by $- 8$ .

$- 6 y^{2} + 8 x^{2} = - 16$

$3 \cdot (2 y^{2} + 5 x^{2}) = (3) (13)$

$- 6 y^{2} + 8 x^{2} = - 16$

$3 \cdot (2 y^{2} + 5 x^{2}) = (3) (13)$

Simplify the left side.

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

Apply the distributive property.

$- 6 y^{2} + 8 x^{2} = - 16$

$3 (2 y^{2}) + 3 (5 x^{2}) = (3) (13)$

Multiply.

Multiply $2$ by $3$ .

$- 6 y^{2} + 8 x^{2} = - 16$

$6 y^{2} + 3 (5 x^{2}) = (3) (13)$

Multiply $5$ by $3$ .

$- 6 y^{2} + 8 x^{2} = - 16$

$6 y^{2} + 15 x^{2} = (3) (13)$

$- 6 y^{2} + 8 x^{2} = - 16$

$6 y^{2} + 15 x^{2} = (3) (13)$

$- 6 y^{2} + 8 x^{2} = - 16$

$6 y^{2} + 15 x^{2} = (3) (13)$

$- 6 y^{2} + 8 x^{2} = - 16$

$6 y^{2} + 15 x^{2} = (3) (13)$

Simplify the right side.

Multiply $3$ by $13$ .

$- 6 y^{2} + 8 x^{2} = - 16$

$6 y^{2} + 15 x^{2} = 39$

$- 6 y^{2} + 8 x^{2} = - 16$

$6 y^{2} + 15 x^{2} = 39$

$- 6 y^{2} + 8 x^{2} = - 16$

$6 y^{2} + 15 x^{2} = 39$

Step 5

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

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

Step 6

Divide each term in $23 x^{2} = 23$ by $23$ and simplify.

Divide each term in $23 x^{2} = 23$ by $23$ .

$\frac{23 x^{2}}{23} = \frac{23}{23}$

Simplify the left side.

Cancel the common factor of $23$ .

Cancel the common factor.

$\frac{23 x^{2}}{23} = \frac{23}{23}$

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

$x^{2} = \frac{23}{23}$

Simplify the right side.

Divide $23$ by $23$ .

$x^{2} = 1$

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 $y^{2}$ .

$- 6 y^{2} + 8 (1) = - 16$

Multiply $8$ by $1$ .

$- 6 y^{2} + 8 = - 16$

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

Subtract $8$ from both sides of the equation.

$- 6 y^{2} = - 16 - 8$

Subtract $8$ from $- 16$ .

$- 6 y^{2} = - 24$

Divide each term in $- 6 y^{2} = - 24$ by $- 6$ and simplify.

Divide each term in $- 6 y^{2} = - 24$ by $- 6$ .

$\frac{- 6 y^{2}}{- 6} = \frac{- 24}{- 6}$

Simplify the left side.

Cancel the common factor of $- 6$ .

Cancel the common factor.

$\frac{- 6 y^{2}}{- 6} = \frac{- 24}{- 6}$

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

$y^{2} = \frac{- 24}{- 6}$

Simplify the right side.

Divide $- 24$ by $- 6$ .

$y^{2} = 4$

Step 8

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

$x^{2} = 1$

$y^{2} = 4$

Step 9

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

$x = \pm \sqrt{1}$

$y^{2} = 4$

Step 10

Any root of $1$ is $1$ .

$x = \pm 1$

$y^{2} = 4$

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.

$x = 1$

$y^{2} = 4$

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

$x = - 1$

$y^{2} = 4$

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

$x = 1, - 1$

$y^{2} = 4$

$x = 1, - 1$

$y^{2} = 4$

Step 12

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

$y = \pm \sqrt{4}$

$x = 1, - 1$

Step 13

Simplify $\pm \sqrt{4}$ .

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

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

$x = 1, - 1$

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

$y = \pm 2$

$x = 1, - 1$

$y = \pm 2$

$x = 1, - 1$

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.

$y = 2$

$x = 1, - 1$

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

$y = - 2$

$x = 1, - 1$

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

$y = 2, - 2$

$x = 1, - 1$

$y = 2, - 2$

$x = 1, - 1$

Step 15

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

$(1, 2), (1, - 2), (- 1, 2), (- 1, - 2)$

Step 16

The result can be shown in multiple forms.

Point Form:

$(1, 2), (1, - 2), (- 1, 2), (- 1, - 2)$

Equation Form:

$x = 1, y = 2$

$x = 1, y = - 2$

$x = - 1, y = 2$

$x = - 1, y = - 2$

Step 17

Solve by Addition/Elimination 4x^2-3y^2=-8 , 5x^2+2y^2=13

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