Solve by Addition/Elimination 4x^2-2y^2=-4 , 3x^2+5y^2=23 - MathMaster

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$4 x^{2} - 2 y^{2} = - 4$ , $3 x^{2} + 5 y^{2} = 23$

Reorder the polynomial.

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

$3 x^{2} + 5 y^{2} = 23$

Reorder the polynomial.

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

$5 y^{2} + 3 x^{2} = 23$

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

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

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

$(2) \cdot (5 y^{2} + 3 x^{2}) = (2) (23)$

Simplify.

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

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Apply the distributive property.

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

$(2) \cdot (5 y^{2} + 3 x^{2}) = (2) (23)$

Multiply.

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Multiply $- 2$ by $5$ .

$- 10 y^{2} + 5 (4 x^{2}) = (5) (- 4)$

$(2) \cdot (5 y^{2} + 3 x^{2}) = (2) (23)$

Multiply $4$ by $5$ .

$- 10 y^{2} + 20 x^{2} = (5) (- 4)$

$(2) \cdot (5 y^{2} + 3 x^{2}) = (2) (23)$

$- 10 y^{2} + 20 x^{2} = (5) (- 4)$

$(2) \cdot (5 y^{2} + 3 x^{2}) = (2) (23)$

$- 10 y^{2} + 20 x^{2} = (5) (- 4)$

$(2) \cdot (5 y^{2} + 3 x^{2}) = (2) (23)$

Multiply $5$ by $- 4$ .

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

$(2) \cdot (5 y^{2} + 3 x^{2}) = (2) (23)$

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

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Apply the distributive property.

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

$2 (5 y^{2}) + 2 (3 x^{2}) = (2) (23)$

Multiply.

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Multiply $5$ by $2$ .

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

$10 y^{2} + 2 (3 x^{2}) = (2) (23)$

Multiply $3$ by $2$ .

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

$10 y^{2} + 6 x^{2} = (2) (23)$

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

$10 y^{2} + 6 x^{2} = (2) (23)$

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

$10 y^{2} + 6 x^{2} = (2) (23)$

Multiply $2$ by $23$ .

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

$10 y^{2} + 6 x^{2} = 46$

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

$10 y^{2} + 6 x^{2} = 46$

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

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

Divide each term by $26$ and simplify.

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

$\frac{26 x^{2}}{26} = \frac{26}{26}$

Cancel the common factor of $26$ .

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Cancel the common factor.

$\frac{26 x^{2}}{26} = \frac{26}{26}$

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

$x^{2} = \frac{26}{26}$

$x^{2} = \frac{26}{26}$

Divide $26$ by $26$ .

$x^{2} = 1$

$x^{2} = 1$

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

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

$- 10 y^{2} + 20 (1) = - 20$

Multiply $20$ by $1$ .

$- 10 y^{2} + 20 = - 20$

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

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Subtract $20$ from both sides of the equation.

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

Subtract $20$ from $- 20$ .

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

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

Divide each term by $- 10$ and simplify.

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

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

Cancel the common factor of $- 10$ .

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Cancel the common factor.

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

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

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

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

Divide $- 40$ by $- 10$ .

$y^{2} = 4$

$y^{2} = 4$

$y^{2} = 4$

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

$x^{2} = 1$

$y^{2} = 4$

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

$x = \pm \sqrt{1}$

$y^{2} = 4$

Any root of $1$ is $1$ .

$x = \pm 1$

$y^{2} = 4$

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

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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.

$x = 1, - 1$

$y^{2} = 4$

$x = 1, - 1$

$y^{2} = 4$

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

$x = 1, - 1$

$y = \pm \sqrt{4}$

$x = 1, - 1$

Simplify the right side of the equation.

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

$x = 1, - 1$

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

$x = 1, - 1$

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

$x = 1, - 1$

$y = \pm 2$

$x = 1, - 1$

$x = 1, - 1$

$y = \pm 2$

$x = 1, - 1$

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

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

$x = 1, - 1$

$y = 2$

$x = 1, - 1$

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

$x = 1, - 1$

$y = - 2$

$x = 1, - 1$

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

$x = 1, - 1$

$y = 2, - 2$

$x = 1, - 1$

$x = 1, - 1$

$y = 2, - 2$

$x = 1, - 1$

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

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

The result can be shown in multiple forms.

Point Form:

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

Equation Form:

$x = 1, y = 2$

$x = 1, y = - 2$

$x = - 1, y = 2$

$x = - 1, y = - 2$

Solve by Addition/Elimination 4x^2-2y^2=-4 , 3x^2+5y^2=23

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