Solve by Addition/Elimination 3x^2-y^2=11 , x^2+4y^2=8

$3 x^{2} - y^{2} = 11$ , $x^{2} + 4 y^{2} = 8$

Simplify the left side.

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

$- y^{2} + 3 x^{2} = 11$

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

$- y^{2} + 3 x^{2} = 11$

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

Simplify the left side.

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

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

$- y^{2} + 3 x^{2} = 11$

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

$- y^{2} + 3 x^{2} = 11$

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

$- y^{2} + 3 x^{2} = 11$

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

Simplify.

Simplify the left side.

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

Apply the distributive property.

$- y^{2} + 3 x^{2} = 11$

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

Multiply $4$ by $- 3$ .

$- y^{2} + 3 x^{2} = 11$

$- 12 y^{2} - 3 x^{2} = (- 3) (8)$

$- y^{2} + 3 x^{2} = 11$

$- 12 y^{2} - 3 x^{2} = (- 3) (8)$

$- y^{2} + 3 x^{2} = 11$

$- 12 y^{2} - 3 x^{2} = (- 3) (8)$

Simplify the right side.

Multiply $- 3$ by $8$ .

$- y^{2} + 3 x^{2} = 11$

$- 12 y^{2} - 3 x^{2} = - 24$

$- y^{2} + 3 x^{2} = 11$

$- 12 y^{2} - 3 x^{2} = - 24$

$- y^{2} + 3 x^{2} = 11$

$- 12 y^{2} - 3 x^{2} = - 24$

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

			$-$	$y^{2}$	$+$	$3$	$x^{2}$	$=$		$1$	$1$
$+$	$-$	$1$	$2$	$y^{2}$	$-$	$3$	$x^{2}$	$=$	$-$	$2$	$4$
	$-$	$1$	$3$	$y^{2}$				$=$	$-$	$1$	$3$

Divide each term in $- 13 y^{2} = - 13$ by $- 13$ and simplify.

Divide each term in $- 13 y^{2} = - 13$ by $- 13$ .

$\frac{- 13 y^{2}}{- 13} = \frac{- 13}{- 13}$

Simplify the left side.

Cancel the common factor of $- 13$ .

Cancel the common factor.

$\frac{- 13 y^{2}}{- 13} = \frac{- 13}{- 13}$

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

$y^{2} = \frac{- 13}{- 13}$

Simplify the right side.

Divide $- 13$ by $- 13$ .

$y^{2} = 1$

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 $x^{2}$ .

$- (1) + 3 x^{2} = 11$

Multiply $- 1$ by $1$ .

$- 1 + 3 x^{2} = 11$

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

Add $1$ to both sides of the equation.

$3 x^{2} = 11 + 1$

Add $11$ and $1$ .

$3 x^{2} = 12$

Divide each term in $3 x^{2} = 12$ by $3$ and simplify.

Divide each term in $3 x^{2} = 12$ by $3$ .

$\frac{3 x^{2}}{3} = \frac{12}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x^{2}}{3} = \frac{12}{3}$

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

$x^{2} = \frac{12}{3}$

Simplify the right side.

Divide $12$ by $3$ .

$x^{2} = 4$

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

$y^{2} = 1$

$x^{2} = 4$

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

$y = \pm \sqrt{1}$

$x^{2} = 4$

Any root of $1$ is $1$ .

$y = \pm 1$

$x^{2} = 4$

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.

$y = 1$

$x^{2} = 4$

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

$y = - 1$

$x^{2} = 4$

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

$y = 1, - 1$

$x^{2} = 4$

$y = 1, - 1$

$x^{2} = 4$

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

$x = \pm \sqrt{4}$

$y = 1, - 1$

Simplify $\pm \sqrt{4}$ .

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

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

$y = 1, - 1$

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

$x = \pm 2$

$y = 1, - 1$

$x = \pm 2$

$y = 1, - 1$

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.

$x = 2$

$y = 1, - 1$

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

$x = - 2$

$y = 1, - 1$

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

$x = 2, - 2$

$y = 1, - 1$

$x = 2, - 2$

$y = 1, - 1$

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

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

The result can be shown in multiple forms.

Point Form:

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

Equation Form:

$x = 2, y = 1$

$x = 2, y = - 1$

$x = - 2, y = 1$

$x = - 2, y = - 1$

Solve by Addition/Elimination 3x^2-y^2=11 , x^2+4y^2=8

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