Determine if Dependent, Independent, or Inconsistent y=x y=-x

$y = x$ $y = - x$

Step 1

Solve the system of equations.

Move all terms containing variables to the left.

Subtract $x$ from both sides of the equation.

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

Add $x$ to both sides of the equation.

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

Reorder the polynomial.

$- x + y = 0$

$y + x = 0$

Reorder the polynomial.

$x + y = 0$

$- x + y = 0$

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.

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Divide $0$ by $2$ .

$y = 0$

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

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

$- x + 0 = 0$

Add $- x$ and $0$ .

$- x = 0$

Divide each term in $- x = 0$ by $- 1$ and simplify.

Divide each term in $- x = 0$ by $- 1$ .

$\frac{- x}{- 1} = \frac{0}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{0}{- 1}$

Divide $x$ by $1$ .

$x = \frac{0}{- 1}$

Simplify the right side.

Divide $0$ by $- 1$ .

$x = 0$

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

Determine if Dependent, Independent, or Inconsistent y=x y=-x

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