Precalculus Examples

$x + y = 0$ , $x - y = 0$

Step 1

Solve the system of equations.

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Multiply each equation by the value that makes the coefficients of $x$ opposite.

$x + y = 0$

$(- 1) \cdot (x - y) = (- 1) (0)$

Simplify.

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Simplify the left side.

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Simplify $(- 1) \cdot (x - y)$ .

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Apply the distributive property.

$x + y = 0$

$- 1 x - 1 (- y) = (- 1) (0)$

Rewrite $- 1 x$ as $- x$ .

$x + y = 0$

$- x - 1 (- y) = (- 1) (0)$

Multiply $- 1 (- y)$ .

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Multiply $- 1$ by $- 1$ .

$x + y = 0$

$- x + 1 y = (- 1) (0)$

Multiply $y$ by $1$ .

$x + y = 0$

$- x + y = (- 1) (0)$

$x + y = 0$

$- x + y = (- 1) (0)$

$x + y = 0$

$- x + y = (- 1) (0)$

$x + y = 0$

$- x + y = (- 1) (0)$

Simplify the right side.

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Multiply $- 1$ by $0$ .

$x + y = 0$

$- x + y = 0$

$x + y = 0$

$- x + y = 0$

$x + y = 0$

$- x + y = 0$

Add the two equations together to eliminate $x$ from the system.

		$x$	$+$	$y$	$=$	$0$
$+$	$-$	$x$	$+$	$y$	$=$	$0$
			$2$	$y$	$=$	$0$

Divide each term in $2 y = 0$ by $2$ and simplify.

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Divide each term in $2 y = 0$ by $2$ .

$\frac{2 y}{2} = \frac{0}{2}$

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 y}{2} = \frac{0}{2}$

Divide $y$ by $1$ .

$y = \frac{0}{2}$

Simplify the right side.

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Divide $0$ by $2$ .

$y = 0$

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

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

$x + 0 = 0$

Add $x$ and $0$ .

$x = 0$

The solution to the independent system of equations can be represented as a point.

$(0, 0)$

Step 2

Since the system has a point of intersection, the system is independent.

Independent

Step 3

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